UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-02T02:52:49Z

EmilyOates:

====Links====
http://en.wikipedia.org/wiki/Logarithmic_scale

----

The mathematical definition of pH (power of Hydrogen) is the negative logarithmic value of the Hydrogen ion (H+) concentration.

The acidity or alkalinity of a substance (ph) can be written as the equation:

<math>pH = -log [H+]</math>

When <math>H+</math> is the hydrogen ion solution in moles per litre.

The pH graph looks like this, courtesy of WolframAlpha

[[File: Log_Graph.gif]]

==Example==

Suppose we test tomato juice and we find that the <math>H+</math> is equal to 0.0001.
How do we find the pH value and determine whether the tomato juice is acidic or alkaline?

'''1) Substitute into our logarithmic substance formula'''

In this case:
<math>pH = -log[0.0001] (base 10)
10^-4=0.0001</math>
Hence:
pH = 4

'''This means that the tomato juice is acidic with a pH of 4.'''

Since the negative log of H+ is used in the pH scale, we will usually get positive values (as long as H+ is below 1). Additionally, the larger our pH, the smaller our H+.

The pH scale turns a concise scale into a large scale between the values 0 and 1.

We know that there is an infinite number of values between 0 and 1 which shows that using a log scale allows us to covert small quantities into large quantities.

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

[[File:Ph Environment Canada elmhurts.edu.gif]]

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Pure (neutral) water has a pH around 7 at 25 °C (77 °F); this value varies with temperature. When an acid is dissolved in water the pH will be less than 7 (if at 25 °C (77 °F)) and when a base, or alkali is dissolved in water the pH will be greater than 7 (if at 25 °C (77 °F)). A solution of a strong acid, such as hydrochloric acid, at concentration 1 mol dm−3 has a pH of 0. A solution of a strong alkali, such as sodium hydroxide, at concentration 1 mol dm−3 has a pH of 14. Thus, measured pH values will mostly lie in the range 0 to 14.

Since pH is a logarithmic scale a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.

http://en.wikipedia.org/wiki/PH

[[File:Ph - wwwchem.csustan.edu.gif]]

http://wwwchem.csustan.edu/chem3070/3070m07.htm

File:Log Graph.gif

2011-02-02T02:33:51Z

EmilyOates:

User:EmilyOates

2011-01-26T03:50:48Z

EmilyOates:

Emily Oates- Land and Food Systems

Pythagorean Theorem

The Pythagorean Theorem, otherwise known as the Pythagorean Equation, can be written as the equation : a²+b²=c². This equation was formulated after a Greek Mathematician named Pythagoras, although there is much debate to the actual origins of the theorem. By knowing two sides of a right angled triangle, this equation allows us to figure out the third side. An important fact to note is that the C is represented by the hypotenuse, which is the longest side of a triangle, and A and B refer to the lengths of the other two sides. It is also important to realize that whenever referring to the theorem, one is relating the area of squares on the sides of the triangle. In addition, the Pythagorean Theorem can be used throughout various aspects of mathematics. Such topics can vary from the law of cosines, three dimensional geometry, hyperbolic geometry and differential geometry, just to name a few.

Calculus Applications:

Wanting to pursue a career in the sciences, calculus plays an integral role in my getting there. It is seen throughout various occupations within the field, and is needed for various applications. For instance, for medical purposes, calculus is used to measure certain injuries of the head caused by impact. This can be seen when testing new cars, they use calculus to determine the crash test dummies injuries, and where they can improve the models. Calculus is important to asses the damage an individual would be subjected to. In addition, in biology and chemistry, calculus is used constantly in determining half lives. Also, it is used in medical research and analysis daily. When dealing with the human body it allows us to measure angles of blood vessels, and shows the rate at which a particular drug is eliminated from a patient. It helps determine the correct dosage of medication. As well, It allows us to study graphs, determine reaction rates and decay. When used in conjunction with population dynamics, it allows us to correlate between birth and death rates, and shows a model of population change. All in all, calculus paints a picture for is in which we can view, relate and analyze extremely important data related to the health sector. Without it we would surely not be where we are today.

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-24T16:12:55Z

EmilyOates: Blanked the page

Course:MATH110/Archive/2010-2011/003/Teams/Zurich

2011-01-24T16:11:52Z

EmilyOates:

{{
Infobox MATH110 Teams
| team name = Zurich
| member 1 = [[User:EllenTsang|Ellen Tsang]]
| member 2 = [[User:EmilyOates|Emily Oates]]
| member 3 = [[User:PhilipLauFaiWong|Philip Wong]]
| member 4 = [[User:RaphaelTan|Raphael Tan]]
}}

Our keywords:
Monkey, Kiwi, Pig, Bubble tea.

==Subpages==

<dpl>
titlematch={{PAGENAME}}/%
namespace={{NAMESPACE}}
shownamespace=false
</dpl>

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork Week 11

2011-01-19T04:07:34Z

EmilyOates: /* Question 1 */

==Question 1==

How do I put 'degrees celsius' into Webwork for the second part of the question? [[User:EllenTsang|EllenTsang]]

Yea Same Question.

I also have the same question!

I think you write degC, hope this helps!

==Question 2==

Does anyone know how to do problem 11?

==Question 3==

How do you do number 9?

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:37:11Z

EmilyOates:

'''Bold text'''{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines, [[Media:Example.ogg]]
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==

'''Example of Solving Linear Equations'''

''Definition of a linear equation'': A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

''Definition of a quadratic equation'': a quadratic equation is a polynomial equation of the second degree.

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

''Definition of a linear inequality:'' an inequality which involves a linear function.

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

''Definition of a quadratic inequality'': an inequality in which one side is a quadratic polynomial and the other side is zero.

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which are very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:33:17Z

EmilyOates:

'''Bold text'''{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines, [[Media:Example.ogg]]
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Definition of a linear equation: A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Definition of a quadratic equation: a quadratic equation is a polynomial equation of the second degree.

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Definition of a linear inequality: an inequality which involves a linear function.

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Definition of a quadratic inequality: an inequality in which one side is a quadratic polynomial and the other side is zero.

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which are very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:32:12Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Definition of a linear equation: A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Definition of a quadratic equation: a quadratic equation is a polynomial equation of the second degree.

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Definition of a linear inequality: an inequality which involves a linear function.

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Definition of a quadratic inequality: an inequality in which one side is a quadratic polynomial and the other side is zero.

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which are very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:31:09Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

[[Definition of a linear equation:]] A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

[[Definition of a quadratic equation]]: a quadratic equation is a polynomial equation of the second degree.

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

[[Definition of a linear inequality:]] an inequality which involves a linear function.

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

[[Definition of a quadratic inequality]]: an inequality in which one side is a quadratic polynomial and the other side is zero.

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which are very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:25:50Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which are very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:20:31Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which is very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube|Y6x06SBbEcA}}
{{#ev:youtube|9IUEk9fn2Vs}}

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:17:19Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which is very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
{{#ev:youtube | Y6x06SBbEcA | 400}
http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:11:24Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

'''''A website which we found extremely helpful in understanding equations can be found here:'''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which is very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:10:48Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

''''''A website which we found extremely helpful in understanding equations can be found here:''''''

http://www.sosmath.com/algebra/solve/solve0/solve0.html

This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which is very easy to understand.

''
'''In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:
'''''
http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-12-03T02:08:39Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

A website which we found extremely helpful in understanding equations can be found here:
http://www.sosmath.com/algebra/solve/solve0/solve0.html
This website contains questions involving linear equations, equations containing radicals, equations containing absolute values, quadratic equations, equations involving fractions, exponential equations, logarithmic equations and trigonometric equations. In addition to all these questions, it provides you with the solution in a step by step manner, which is very easy to understand.

In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:

http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

The first video is about solving rational equations, while the second is about linear equations.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-11-24T23:03:40Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

A website which we found extremely helpful in understanding equations can be found here:

http://www.sosmath.com/algebra/solve/solve0/solve0.html

In addition, there are two youtube videos which also helped contribute to our understanding. They can be found at these links:

http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-11-24T23:03:27Z

EmilyOates:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

=Homework 5=

==Those That Pose No Problem==
Basic functions,
Basic Properties of Functions,
Equations,
Polynomial Long Division,
Graphs of Functions,
Intersections of Functions,
Reading Graphs of functions,
Distances and Lines,
Operations on Graphs of Functions,
Construction of Graphs,
Trigonometry and the Pythagoras Theorem,
Areas and Volumes,
Mathematical Writing.

==Those That Just Need Practice==
Inequalities and
Composition of Functions

=Basic Skills Learning Guide Project=

==Learning Outcomes With Regards To Equations==
'''Example of Solving Linear Equations'''

Example 1:

3/4x +5/6 = 5x - 125/3

Multiply both sides by the lowest common multiple of 4, 6, and 3, or 12.

12(3/4x +5/6) = 12(5x -125/3)

Simplify:

12(3/4x) + 12(5/6) = 12(5x) - 12(125/3)

3(3x) + 2(5) = 60x - 500

9x + 10 = 60x - 500

Subtract 9x from both sides of the equation:

10 = 51x - 500

Add 500 to both sides of the equation:

510 = 51x

Divide both sides by 51:

x = 510/51 = 10

'''Example of Solving Quadratic Equation'''

Example 1:

Solve 6x^2 + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

-b + or - sqrt(b^2 - 4ac)/2a

x = -(11) + or - sqrt(11)^2 - (4)(6)(-35)/(2)(6)

x = -11 + or - sqrt(121 + 840)/12

x = -11 + or - sqrt961/12

x = -11 + or - 31/12

x = -11 - 31 / 12, -11 + -31/12

x = -42/12, 20/12

x = -7/2, 5/3

The solution is x = –7/2, 5/3

'''Example of Solving Linear Inequalities'''

Example 1:

Solve (2x – 3)/4 < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

(2x – 3)/4 < 2

(4) × (2x – 3)/4 < (4)(2)

2x – 3 < 8

2x < 11

x < 11/2 = 5.5

'''Example of Solving Quadratic Inequalities'''

Example 1:

Solve x^2 + 2x – 8 < 0.
First, find the zeroes:

x^2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = –4 or x = 2

These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.

Then the solution is: –4 < x < 2

A website which we found extremely helpful in understanding equations can be found here:

http://www.sosmath.com/algebra/solve/solve0/solve0.html

In addition, there are two youbtube videos which also helped contribute to our understanding. They can be found at these links:

http://www.youtube.com/watch?v=Y6x06SBbEcA

http://www.youtube.com/watch?v=9IUEk9fn2Vs&feature=channel

User:EmilyOates

2010-09-17T06:00:50Z

EmilyOates:

User:EmilyOates

2010-09-17T05:36:26Z

EmilyOates: Created page with 'Emily Oates- Land and Food Systems'

Emily Oates- Land and Food Systems