UBC Wiki - User contributions [en]

User:RaphaelTan

2011-01-25T03:16:49Z

RaphaelTan:

Name: Raphael Tan
Faculty:Arts
Class: Math110 003

'''Pythagorean theorem'''
The Pythagorean theorem is often associated with a right triangle wherein its sides are of relative lengths. We can say that the sum of the area of the two sides opposite the right angle is equal to the area of the right angle. this can be represented by the equation a²+b²=c². In this equation both a and b represent the sides opposite to the right angle while c represents the right angle. The Pythagorean theorem was discovered by a mathematician named Pythagoras and has been used and applied in everyday life such as architecture and research. There are numerous proofs to the Pythagorean theorem some of which are proof using similar triangles, proof by subtraction and proof by rotation. Ever since its discovery the Pythagorean theorem has had multiple consequences and uses toward our everyday understanding.

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

'''Application of calculus in business'''
In maths we often use functions as the basis of raw calculus. There are a number of different functions which serve different purposes such as graphing, mapping, illustrating relations between different things and a lot more. An example of a function would be y = m x + b this can be used to demonstrate simple concepts and calculations in business. For example, when calculating revenues and costs. Now let's put the equation into business terms. Lets say that i am running a factory that produces paper my initial cost for all the machinery and rent(fixed costs) are equal to $500 my variable cost now for producing a single piece of paper is $2. Using this i can calculate my total costs(TC) which will allow me to see how i will price my product. TC = 500 + 2(Q) where Q represents the quantity i will produce. Aside from this, there are many other ways in which we can apply calculus into business such as in calculating profits and marginal costings.

http://www.scribd.com/doc/9686225/Applied-Calculus-for-Business-Students

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T00:05:03Z

RaphaelTan:

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha from Group 10 for replying. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we 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 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5
====Example 7====
* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)
5x + 7 < 3x + 3
2x + 7 < 3
2x < –4
x < –2

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Quadratic Inequalities=
References:

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=Videos teaching Inequality=

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains quadratic Inequalities.

=Useful Links=

http://www.purplemath.com/modules/ineqsolv.htm

=Group 10=

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

One of the first steps to solving inequalities is the symbol and meanings of the inequalities.

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-12-03T00:03:57Z

RaphaelTan:

One of the first steps to solving inequalities is the symbol and meanings of the inequalities.

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-30T22:39:47Z

RaphaelTan:

'''What is an inequality?'''
It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".
1. It may be that "x" is greater than "y"
which can be written as x>y.
2. x is less than y
x < y
3. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation (such as the notation used for defining a domain), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

'''Addition/Subtraction'''
The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.
'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to the addition

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -2 or x > 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-30T22:38:25Z

RaphaelTan:

'''What is an inequality?'''
It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".
1. It may be that "x" is greater than "y"
which can be written as x>y.
2. x is less than y
x < y
3. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation (such as the notation used for defining a domain), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

'''Addition/Subtraction'''
The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.
'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to the addition

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2+9 < 0

Our first step is to associate this with and equation. Therefore:

y=-x2+9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2+9=0

Now we can solve:

-x2+9=0

x2+9=0

(x+3)(x-3)=0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x<-3

2.) -3<x<3

3.) x>3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y=-x2+9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2+9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x<-2 or x>2

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-30T22:37:39Z

RaphaelTan:

'''What is an inequality?'''
It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".
1. It may be that "x" is greater than "y"
which can be written as x>y.
2. x is less than y
x < y
3. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation (such as the notation used for defining a domain), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

'''Addition/Subtraction'''
The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.
'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to the addition

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2+9 < 0

Our first step is to associate this with and equation. Therefore:

y=-x^2+9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x^2+9=0

Now we can solve:

-x^2+9=0

x^2+9=0

(x+3)(x-3)=0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x<-3

2.) -3<x<3

3.) x>3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y=-x^2+9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x^2+9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x<-2 or x>2

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-30T22:34:52Z

RaphaelTan:

'''What is an inequality?'''
It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".
1. It may be that "x" is greater than "y"
which can be written as x>y.
2. x is less than y
x < y
3. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation (such as the notation used for defining a domain), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

'''Addition/Subtraction'''
The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.
'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to the addition

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x^2+9 < 0

Our first step is to associate this with and equation. Therefore:

y=-x^2+9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x^2+9=0

Now we can solve:

-x^2+9=0

x^2+9=0

(x+3)(x-3)=0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x<-3

2.) -3<x<3

3.) x>3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y=-x^2+9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x^2+9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x<-2 or x>2

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-29T17:53:28Z

RaphaelTan:

What is an inequality?

How to represent the solutions of inequalities?

How to solve linear inequalities?

How to solve quadratic inequalities?

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-11-29T17:53:13Z

RaphaelTan: Created page with 'What is an inequality? How to represent the solutions of inequalities? How to solve linear inequalities? How to solve quadratic inequalities?'

What is an inequality?
How to represent the solutions of inequalities?
How to solve linear inequalities?
How to solve quadratic inequalities?

Course:MATH110/Archive/2010-2011/003/Groups/Group 10

2010-11-29T17:51:52Z

RaphaelTan: /* Homework Subpages: */

{{Infobox MATH110 Groups
| group number = 10
| member 1 = Anna Koniuhova
| member 2 = [[User:MichelleGutmanis|Michelle Gutmanis]]
| member 3 = Hyun Lee
| member 4 = [[User:AgnesLuong|Agnes Luong]]
| member 5 = [[User:TrevorShumka|Trevor Shumka]]
| member 6 = [[User:RaphaelTan|Raphael Tan]]
}}

<pre>Hi guys! Do you want to work with us (Group 9) for the group project? :) Our emails are on our group page.

-- Ellen | ellentsang.nl@hotmail.com</pre>
[[User:EllenTsang|EllenTsang]] 07:21, 18 November 2010 (UTC)

=='''Homework 3 - Third Part - Problem Solving Skills - Due date: October 13, 2010'''==

'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

In this problem, two different methods of telling time are used, one is by hours and minutes and the other is just minutes. Since an hour is 60 min, and the problem states that it took an hour and 20 min, this would mean it took 80 min. Therefore, it took 80 min to travel from the terminal to the airport and 80 min to travel from the terminal to the airport each way traveling at a speed of 30mi/hr.

'''2. A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.'''

This problem states that a lady did not have her drivers license, failed to stop at a stop sign, and wen three blocks down a one-way street the wrong way without getting stopped - which all are assumed to be things that you should pay attention to while driving. This problem however never mentioned that the woman was driving, so she could have been walking or running. Therefore, the driving rules would not apply to her and she was not breaking the law, so that is why the policeman did not need to stop her.

'''3. One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.'''

Since we already know that the labels are incorrect, we could select a fruit the box that says APPLES & ORANGES because that way if an apple is selected, we would know that that box is apples, and the one that says oranges would have to be APPLES & ORANGES because it is labeled incorrectly so it cannot be oranges.

Ex.

Box1 : APPLES & ORANGES
(must be APPLES because it cannot be APPLES & ORANGES, and apples can't go in the ORANGES box)

Box2 : ORANGES
(must be named APPLES & ORANGES because: it cannot be apples since an apple was selected from original APPLES & ORANGES box (box1), and cannot be ORANGES because it is labelled incorrectly)

Box3 : APPLES
(must be ORANGES because: it cannot be APPLES because it is labelled incorrectly, and the original ORANGES box (box2) is now APPLES & ORANGES)

'''4. I am the brother of the blind fiddler, but brothers I have none. How can this be?'''

Being a brother could mean that he is the brother of a boy or a girl. Therefore, since he does not have any brothers, the blind fiddler must be his sister.

'''5. Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?'''

Since the quarter has to go all the way around the other quarter, if you try it, you will find that it revolves twice. One rotation will get it halfway around the quarter, and the other rotation will bring it to its original position.

'''6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?

Considering if there were only 6 apples in the basket with two of each kind, then after six draws one can be sure of getting at least two of one kind.

'''7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?

i) If the socks were drawn one by one and the pair had to be drawn consecutively, there is a possibility that that may never happen if the socks drawn come out alternatively (considering that once the sock is drawn it is not put back in.

ii) Again, using the same method in part i (drawing the socks one at a time) if all the socks drawn up until the 40th sock were the same color then by the 41st pick you would have drawn a pair of mismatched socks.

'''8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.

Reuben was born on January 1st at 12am. Then two days later it would be a new year, he would have turned 21 from 20. Then later in the year he celebrates his birthday again at 12am, turning 22. Then it becomes the next year which means he is 23.

'''9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?

If the boat is at least 15 feet tall then all the rungs on the ladder would show, as well if the ladder is hung horizontally then all the rungs would show as well.

'''10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.
Patterns (i) and (ii) will not follow, as we are using the same group of "all people" where one half are all women and one half are all chocolate eaters does not make this exclusive. Meaning that the chocolate eaters can be included in the group of women and vice versa. If the question was phrased as "one half of all people are women while the OTHER half are chocolate eaters" then patterns (i) and (ii) may follow.

''10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.''

(i) It does not follow that ¼ of all of the people are women chocolate eaters because the facts above state that ½ of the total people are women, it does not mean that they all eat chocolate. (ii) Similarly, it does not follow that ½ of men are chocolate eaters. Being a man or woman is independent of preference for chocolate.

''11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?''

This is not possible because the facts are inconsistent such that the situation doesn’t make sense with the question.

''12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.''

If for instance the Bronx train arrives at 10:00, 10:10, 10:20, and the Brooklyn train arrives at 9:59, 10:09, 10:19 then he would take preference over the Brooklyn train and never visit because it arrives earlier.

''13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?''

If you consider the time between chimes then you know it takes 45/4 seconds to strike 10:00.

''14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?''

(i) There are six ways that two of the four babies can be correctly tagged. (ii) There are no ways that ¾ of the four babies can be correctly tagged and 1 incorrectly.

''15. Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?''

Alex is correct, you should not accept his bet because naturally the amount of red cards is 50/50 to the amount of black cards in the deck.

'''16. Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

D = Daughter
S = Son

Daughter has same number of brothers as sisters. Therefore,
D = S + 1

Son has twice as many brothers as sisters. Therefore,
S - 1 = D/2

Combine the equations together to solve for S:

2(S+1 = D/2) = 2S - 2 = D

Since we already know that D = S+1, S + 1 = 2S - 2

To solve for S, isolate S: 2S-S = 1+2
S = 3

We can find D by substituting S into one of the above equations:

Using the equation D = S + 1, D = 3 + 1 D = 4

It can therefore be seen that there are four daughters and three sons.

'''17. The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

This problem can easily be solved by considering what happens if Dan stands on the scale while sarah is already on it. The scale will read 50 kg with Sarah on the scale and then 110 kg when Dan gets on as well. Since this reading is 5 kg higher than the total reading in the example (105 kg), the scale reads 5kg too high.

'''18. Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In this situation, Alice leaves two thirds of a whole, Bret leaves two thirds of two thirds (4/9), and Carla leaves two thirds of four ninths (16/81).

Adding the fractions of the total amount left in the jar, we find that it equals 8/27. Now we need to determine what value of x satisfies the equation that (8/27)x = 40

This can be done by dividing 40 by 8/27. The value for this is 135. Therefore, the jar started with 135 pennies originally.

'''19. One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

M = Total Milk
C = Total Coffee
A = Angela's Drink
F = Family

A = (1/4)M + (1/6)C
F = (3/4)M + (5/6)C

The least number of people in the family is the number of people required to drink the remaining 5/6 of coffee (5 People)

'''20. Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

F = Fast Clock
S = Slow Clock

Both F and S = 5 minutes.

One Hour = 60 Minutes

60 = x (F + S)

aka. the clocks become 10 minutes apart every hour.

60/x = 10

60/10 = x

x = 6

The clocks are therefore an hour apart after six hours.

'''Basic Skills Project:'''

As a group we plan to contribute to the inequalities section. By creating subsections of:
solving linear inequalities
quadratic
rational
radical
trigonometric
exponential
logarithmic inequalities

and making examples of each of them we can make it clearer for other students as well as adding other useful online links.

An example can be shown with:
'''Subsection 1: Linear inequalities'''

online reference: http://www.purplemath.com/modules/ineqgrph.htm

video explanation: http://www.youtube.com/watch?v=0X-bMeIN53I

=='''Homework Subpages:'''==

**not sure if this is how you make the homework page ..

Homework 4:[ http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_10/Homework_4]

Basic Skills Project:[ http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_10/Basic_skills_project]

Course:MATH110/Archive/2010-2011/003/Groups/Group 10

2010-11-29T17:51:37Z

RaphaelTan:

{{Infobox MATH110 Groups
| group number = 10
| member 1 = Anna Koniuhova
| member 2 = [[User:MichelleGutmanis|Michelle Gutmanis]]
| member 3 = Hyun Lee
| member 4 = [[User:AgnesLuong|Agnes Luong]]
| member 5 = [[User:TrevorShumka|Trevor Shumka]]
| member 6 = [[User:RaphaelTan|Raphael Tan]]
}}

<pre>Hi guys! Do you want to work with us (Group 9) for the group project? :) Our emails are on our group page.

-- Ellen | ellentsang.nl@hotmail.com</pre>
[[User:EllenTsang|EllenTsang]] 07:21, 18 November 2010 (UTC)

=='''Homework 3 - Third Part - Problem Solving Skills - Due date: October 13, 2010'''==

'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

In this problem, two different methods of telling time are used, one is by hours and minutes and the other is just minutes. Since an hour is 60 min, and the problem states that it took an hour and 20 min, this would mean it took 80 min. Therefore, it took 80 min to travel from the terminal to the airport and 80 min to travel from the terminal to the airport each way traveling at a speed of 30mi/hr.

'''2. A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.'''

This problem states that a lady did not have her drivers license, failed to stop at a stop sign, and wen three blocks down a one-way street the wrong way without getting stopped - which all are assumed to be things that you should pay attention to while driving. This problem however never mentioned that the woman was driving, so she could have been walking or running. Therefore, the driving rules would not apply to her and she was not breaking the law, so that is why the policeman did not need to stop her.

'''3. One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.'''

Since we already know that the labels are incorrect, we could select a fruit the box that says APPLES & ORANGES because that way if an apple is selected, we would know that that box is apples, and the one that says oranges would have to be APPLES & ORANGES because it is labeled incorrectly so it cannot be oranges.

Ex.

Box1 : APPLES & ORANGES
(must be APPLES because it cannot be APPLES & ORANGES, and apples can't go in the ORANGES box)

Box2 : ORANGES
(must be named APPLES & ORANGES because: it cannot be apples since an apple was selected from original APPLES & ORANGES box (box1), and cannot be ORANGES because it is labelled incorrectly)

Box3 : APPLES
(must be ORANGES because: it cannot be APPLES because it is labelled incorrectly, and the original ORANGES box (box2) is now APPLES & ORANGES)

'''4. I am the brother of the blind fiddler, but brothers I have none. How can this be?'''

Being a brother could mean that he is the brother of a boy or a girl. Therefore, since he does not have any brothers, the blind fiddler must be his sister.

'''5. Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?'''

Since the quarter has to go all the way around the other quarter, if you try it, you will find that it revolves twice. One rotation will get it halfway around the quarter, and the other rotation will bring it to its original position.

'''6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?

Considering if there were only 6 apples in the basket with two of each kind, then after six draws one can be sure of getting at least two of one kind.

'''7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?

i) If the socks were drawn one by one and the pair had to be drawn consecutively, there is a possibility that that may never happen if the socks drawn come out alternatively (considering that once the sock is drawn it is not put back in.

ii) Again, using the same method in part i (drawing the socks one at a time) if all the socks drawn up until the 40th sock were the same color then by the 41st pick you would have drawn a pair of mismatched socks.

'''8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.

Reuben was born on January 1st at 12am. Then two days later it would be a new year, he would have turned 21 from 20. Then later in the year he celebrates his birthday again at 12am, turning 22. Then it becomes the next year which means he is 23.

'''9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?

If the boat is at least 15 feet tall then all the rungs on the ladder would show, as well if the ladder is hung horizontally then all the rungs would show as well.

'''10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.
Patterns (i) and (ii) will not follow, as we are using the same group of "all people" where one half are all women and one half are all chocolate eaters does not make this exclusive. Meaning that the chocolate eaters can be included in the group of women and vice versa. If the question was phrased as "one half of all people are women while the OTHER half are chocolate eaters" then patterns (i) and (ii) may follow.

''10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.''

(i) It does not follow that ¼ of all of the people are women chocolate eaters because the facts above state that ½ of the total people are women, it does not mean that they all eat chocolate. (ii) Similarly, it does not follow that ½ of men are chocolate eaters. Being a man or woman is independent of preference for chocolate.

''11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?''

This is not possible because the facts are inconsistent such that the situation doesn’t make sense with the question.

''12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.''

If for instance the Bronx train arrives at 10:00, 10:10, 10:20, and the Brooklyn train arrives at 9:59, 10:09, 10:19 then he would take preference over the Brooklyn train and never visit because it arrives earlier.

''13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?''

If you consider the time between chimes then you know it takes 45/4 seconds to strike 10:00.

''14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?''

(i) There are six ways that two of the four babies can be correctly tagged. (ii) There are no ways that ¾ of the four babies can be correctly tagged and 1 incorrectly.

''15. Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?''

Alex is correct, you should not accept his bet because naturally the amount of red cards is 50/50 to the amount of black cards in the deck.

'''16. Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

D = Daughter
S = Son

Daughter has same number of brothers as sisters. Therefore,
D = S + 1

Son has twice as many brothers as sisters. Therefore,
S - 1 = D/2

Combine the equations together to solve for S:

2(S+1 = D/2) = 2S - 2 = D

Since we already know that D = S+1, S + 1 = 2S - 2

To solve for S, isolate S: 2S-S = 1+2
S = 3

We can find D by substituting S into one of the above equations:

Using the equation D = S + 1, D = 3 + 1 D = 4

It can therefore be seen that there are four daughters and three sons.

'''17. The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

This problem can easily be solved by considering what happens if Dan stands on the scale while sarah is already on it. The scale will read 50 kg with Sarah on the scale and then 110 kg when Dan gets on as well. Since this reading is 5 kg higher than the total reading in the example (105 kg), the scale reads 5kg too high.

'''18. Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In this situation, Alice leaves two thirds of a whole, Bret leaves two thirds of two thirds (4/9), and Carla leaves two thirds of four ninths (16/81).

Adding the fractions of the total amount left in the jar, we find that it equals 8/27. Now we need to determine what value of x satisfies the equation that (8/27)x = 40

This can be done by dividing 40 by 8/27. The value for this is 135. Therefore, the jar started with 135 pennies originally.

'''19. One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

M = Total Milk
C = Total Coffee
A = Angela's Drink
F = Family

A = (1/4)M + (1/6)C
F = (3/4)M + (5/6)C

The least number of people in the family is the number of people required to drink the remaining 5/6 of coffee (5 People)

'''20. Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

F = Fast Clock
S = Slow Clock

Both F and S = 5 minutes.

One Hour = 60 Minutes

60 = x (F + S)

aka. the clocks become 10 minutes apart every hour.

60/x = 10

60/10 = x

x = 6

The clocks are therefore an hour apart after six hours.

'''Basic Skills Project:'''

As a group we plan to contribute to the inequalities section. By creating subsections of:
solving linear inequalities
quadratic
rational
radical
trigonometric
exponential
logarithmic inequalities

and making examples of each of them we can make it clearer for other students as well as adding other useful online links.

An example can be shown with:
'''Subsection 1: Linear inequalities'''

online reference: http://www.purplemath.com/modules/ineqgrph.htm

video explanation: http://www.youtube.com/watch?v=0X-bMeIN53I

=='''Homework Subpages:'''==

**not sure if this is how you make the homework page ..

Homework 4:[ http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_10/Homework_4]
Basic Skills Project:[ http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_10/Basic_skills_project]

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Homework 4

2010-10-19T02:02:34Z

RaphaelTan:

'''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?

During the 9 days of vacation, Homer will be able to sleep on Tuesday. In order to have come to this answer the problem must have been analysed. Based on the problem given we can immediately rule out the first Saturday,Sunday,Monday,Wednesday and the following Saturday and Sunday. This is because Homer is waken in all of these days. This leaves us with the Tuesday,Thursday and Friday however the problem also mentions that none of the noisemakers was quiet for three consecutive days and no pair made noise on more than one day. Looking at this we can slowly rule out each noisemaker based on which days they have already waken Homer. The construction workers would have waken Homer up on Thursday since none were quiet for three straight days while the Dog may have waken Homer on either Thursday,Friday or Saturday. This leaves us with the Salesman who must have waken Homer up on Friday since, like the other noisemakers it cannot be silent for 3 consecutive days. This leaves Tuesday free of any obstructions.

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Homework 4

2010-10-19T02:01:57Z

RaphaelTan:

'''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?

During the 9 days of vacation, Homer will be able to sleep on Tuesday. In order to have come to this answer the problem must have been analysed. Based on the problem given we can immediately rule out the first Saturday,Sunday,Monday,Wednesday and the following Saturday and Sunday. This is because Homer is waken in all of these days. This leaves us with the Tuesday,Thursday and Friday however the problem also mentions that none of the noisemakers was quiet for three consecutive days and no pair made noise on more than one day. Looking at this we can slowly rule out each noisemaker based on which days they have already waken Homer. The construction workers would have waken Homer up on Thursday since none were quiet for three straight days while the Dog may have waken Homer on either Thursday,Friday or Saturday. This leaves us with the Salesman who must have waken Homer up on Friday since, like the other noisemakers it cannot be silent for 3 consecutive days. This leaves Tuesday free of any obstructions.

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Homework 4

2010-10-19T02:01:32Z

RaphaelTan: Created page with ''''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 la…'

'''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?

During the 9 days of vacation, Homer will be able to sleep on Tuesday. In order to have come to this answer the problem must have been analysed. Based on the problem given we can immediately rule out the first Saturday,Sunday,Monday,Wednesday and the following Saturday and Sunday. This is because Homer is waken in all of these days. This leaves us with the Tuesday,Thursday and Friday however the problem also mentions that none of the noisemakers was quiet for three consecutive days and no pair made noise on more than one day. Looking at this we can slowly rule out each noisemaker based on which days they have already waken Homer. The construction workers would have waken Homer up on Thursday since none were quiet for three straight days while the Dog may have waken Homer on either Thursday,Friday or Saturday. This leaves us with the Salesman who must have waken Homer up on Friday since, like the other noisemakers it cannot be silent for 3 consecutive days. This leaves Tuesday free of any obstructions.

Course talk:MATH110/003/Groups/Group 10

2010-10-13T00:12:21Z

RaphaelTan: Created page with '21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race? since …'

21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?

since Sven placed exactly in the middle among all runners and Dan came in slower than Sven in 10th place this would mean that Sven would have to be in 9th place or higher. Let's say that Sven came in 9th place this would mean that there are a total of 17 runners since 9 is midway of 1 and 17. This shows that there were a total of 17 runners in the race since Sven coming in at 8th place would mean a total of 15 runners. However, since Lars came in at 16th place this would mean that there are more than 15 runners.

22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?

There would be a total of 23 days in the vacation since 11+12=23

23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today's date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula's children? Explain.

No, it would not be possible for Paul to determine the ages of each specific child since there is a lack of information given. In addition, qualitative information such as the oldest child having red hair would have no bearing on the situation.

24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?

One candle was exactly twice as long as the other exactly 2 hours after being lit. this is because one candle took twice as long to burn out than the other which would mean that 2/3 of the 6 hour candle=4 and 1/3 of the 3 hour candle=2

25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.

In order to solve this problem we used the equation length/time to determine how long it would take for a candle to burn out. The candle with length L took 4 hours to burn out (10:00-6:00) while the candle L+1 took 6 hours to burn out (10:30-4:30). If we apply this to the formula we get L/4 and L+1/6. In order to find L we must equate these to each other L/4=L+1/6. This would equal 6L=4L+4. 2L=4 which means the length of L=2.

User:RaphaelTan

2010-09-16T23:31:02Z

RaphaelTan:

Name: Raphael Tan
Faculty:Arts
Class: Math110 003

Pythagorean theorem
The Pythagorean theorem is often associated with a right triangle wherein its sides are of relative lengths. We can say that the sum of the area of the two sides opposite the right angle is equal to the area of the right angle. this can be represented by the equation a²+b²=c². In this equation both a and b represent the sides opposite to the right angle while c represents the right angle. The Pythagorean theorem was discovered by a mathematician named Pythagoras and has been used and applied in everyday life such as architecture and research. There are numerous proofs to the Pythagorean theorem some of which are proof using similar triangles, proof by subtraction and proof by rotation. Ever since its discovery the Pythagorean theorem has had multiple consequences and uses toward our everyday understanding.

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

User:RaphaelTan

2010-09-16T21:42:27Z

RaphaelTan: Created page with 'Name: Raphael Tan Faculty:Arts Class: Math110 003 Pythagorean theorem The Pythagorean theorem is often associated with a right triangle.'

Name: Raphael Tan
Faculty:Arts
Class: Math110 003

Pythagorean theorem
The Pythagorean theorem is often associated with a right triangle.