UBC Wiki - User contributions [en]

User:AlbertKonig

2011-01-28T08:43:21Z

AlbertKonig:

Hey everyone, my name is Albert and well I'm a first year science student; my major is Food, Nutrition and Health in I am in faculty of Land/Food systems (short LFS). I'm new in Canada and I have been to different places here (deep cove, Victoria Island, and some other places), but I reckon that I just have seen a small portion of what this so called "beautiful place on earth" has to offer. Hence I appreciate any kind of suggestion where I can go or what places are worth to visit ^_^

Though I am in LFS I am very interested in cell biology and life sciences in general. As far of now, I can’t anymore count how many time I encountered different formulas of interpreting, predicting and analyzing different factors of colonization of bacteria, cell reproductions and how long a healing process of a small cut can last, medical developments (producing billions of antibacterial in only hours) etc etc. Hence, all these functions: "algorithms, arithmetic functions, graphs, Pythagorean theorem etc", that we are studying are not only helpful for engineering guys (to build safe housetops above our heads) but also for making us having a longer life span.

--> Calculus in our life

As we all may be aware of (and as we talked in class) math is a course and a subject that is involved in almost everything around us. When we go to the supermarket to do grocery shopping, when we try to tank our car, when we play games, and many other things. But one application in math is a very important one. This part is Calculus. Calculus, which generally saying, is the fundamental idea of algebra and geometry. And these two basic ideas in mathematics are used in different fields such as biology, physics, astronomy, architecture, etc.
One of the very interesting application of calculus in my opinion is the weather forecast that is used for different natural phenomena such as lightening or storm. This field in particular is interesting because even when we have a model that may be able to predict specific scenarios it may not be always right, because not every storm hits the same tree on the same place at the same time.
Calculus, in particular, in this field can be used to predict the rate of change of location of the storm (dx) from one point to another point in a period of time (dt); in physical terms this means the speed of the storm. In addition if we want to calculate the acceleration of the storm we have to take the second derivative of dx (location of the storm) at that period of time (dt); explained by the equation: (d(dx/dt)/dt).
Another application of calculus is to calculate the energy of the storm during a specific distance that it leaves behind. This can be calculated by integration. This means that we have to calculate the area under the curve of the force versus distance and hence find out how much destruction the storm has caused, as an example.
There are many other applications that can be used to calculate different effects of calculus. Possible examples are to calculate the different rotation within the storm and the different force of destruction that may apply at different altitudes and regions.

File:The equation of the graph.pdf

2011-01-28T06:49:19Z

AlbertKonig:

File:Wolframalpha-20110128001836852.gif

2011-01-28T06:26:24Z

AlbertKonig:

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 11

2011-01-19T04:01:37Z

AlbertKonig:

team valais homework 11

1) The model that we are going to use in this part is:

C(x)= 100 + 7(x-20), x>20

The above models is a linear equation. At first we are told that we have to spend 100$ (fixed cost) for the production of the first 20 units. Later on we are told that for any extra unit we have to pay 7$ (Variable cost). The variable 'x' in the above equation indicates the amount of units bought minus the initial 20 units that have to be subtracted because they are worth 100$. Or to describe it short: (x-20) is for the number of units to be multiplied by 7 to get the addition cost of producing 1 more unit.

2) Using the above equation our model predicts the following value for 150 units:
C(150) = 100 + 7(150-20)= 1010

What we can conclude from the above equation is that for 150 units we will have a Total cost of 1010$.

3) Now in order to find the Average Cost of per item produced we have to do some modification to the model introduced in no.1. Average Cost (A(x)) is the derivative of C(x).

C(x)= 100 + 7(x-20)

A(x) = 40x^(-2)

The average cost per item increases as production levels increase.

Other models (not necessarily linear) for which you get other behaviors such as:

1. The average cost remains constant as production increases.
A(x) = 5
C(x) = 5x

2. The average cost diminishes as production increases.
A(x) = 1/x^2
C(x) = 1/x

3. The average cost increases as production increases.
A(x) = x
C(x) = x^2

4.You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

A(x) = (x-10)^2 + 10
C(x) = x(x-10)^2 + 100
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 11

2011-01-19T04:00:00Z

AlbertKonig:

team valais homework 11

1) The model that we are going to use in this part is:

C(x)= 100 + 7(x-20), x>20

The above models is a linear equation. At first we are told that we have to spend 100$ (fixed cost) for the production of the first 20 units. Later on we are told that for any extra unit we have to pay 7$ (Variable cost). The variable 'x' in the above equation indicates the amount of units bought minus the initial 20 units that have to be subtracted because they are worth 100$. Or to describe it short: (x-20) is for the number of units to be multiplied by 7 to get the addition cost of producing 1 more unit.
2) Using the above equation our model predicts the following value for 150 units:
C(150) = 100 + 7(150-20)= 1010

What we can conclude from the above equation is that for 150 units we will have a Total cost of 1010$.

3) Now in order to find the Average Cost of per item produced we have to do some modification to the model introduced in no.1. Average Cost (A(x)) is the derivative of C(x).

C(x)= 100 + 7(x-20)

A(x) = 40x^(-2)

The average cost per item increases as production levels increase.

Other models (not necessarily linear) for which you get other behaviors such as:

1. The average cost remains constant as production increases.
A(x) = 5
C(x) = 5x

2. The average cost diminishes as production increases.
A(x) = 1/x^2
C(x) = 1/x

3. The average cost increases as production increases.
A(x) = x
C(x) = x^2

4.You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

A(x) = (x-10)^2 + 10
C(x) = x(x-10)^2 + 100
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

Sandbox:DavidKohler/Schedule

2010-11-02T07:38:20Z

AlbertKonig:

Simply write your name or student number next to an empty time slot that suits you. You're not allowed to move others of course. -- [[User:DavidKohler|DavidKohler]]

Tuesday November 2
* 3:00 - 3:20-Ghita Youssefi
* 3:20 - 3:40- Avi Harry
* 3:40 - 4:00-

Wednesday November 3
* 9:20 - 9:40 Gracie Mann
* 9:40 - 10:00 - Albert Konig
* 10:00 - 10:20 Agnes Luong
* 10:20 - 10:40
* 10:40 - 11:00

Thursday November 4
* 2:40 - 3:00- Charly Huxford
* 3:00 - 3:20
* 3:20 - 3:40
* 3:40 - 4:00
* 4:00 - 4:20
* 4:20 - 4:40
* 4:40 - 5:00

Friday November 5
* 9:20 - 9:40- Sabrina Pannu
* 9:40 - 10:00- Shauna Maty
* 10:00 - 10:20 Paige Holloway
* 10:20 - 10:40
* 10:40 - 11:00

Sandbox:DavidKohler/Schedule

2010-11-01T18:08:29Z

AlbertKonig:

Simply write your name or student number next to an empty time slot that suits you. You're not allowed to move others of course. -- [[User:DavidKohler|DavidKohler]]

Tuesday November 2
* 3:00 - 3:20-Ghita Youssefi-56808108
* 3:20 - 3:40
* 3:40 - 4:00-Albert Koenig - 56042104
* 4:00 - 4:20
* 4:20 - 4:40
* 4:40 - 5:00

Wednesday November 3
* 9:20 - 9:40
* 9:40 - 10:00
* 10:00 - 10:20
* 10:20 - 10:40
* 10:40 - 11:00

Thursday November 4
* 2:40 - 3:00
* 3:00 - 3:20
* 3:20 - 3:40
* 3:40 - 4:00
* 4:00 - 4:20
* 4:20 - 4:40
* 4:40 - 5:00

Friday November 5
* 9:20 - 9:40
* 9:40 - 10:00
* 10:00 - 10:20
* 10:20 - 10:40
* 10:40 - 11:00

Course:MATH110/Archive/2010-2011/003/Groups/Group 01

2010-10-13T06:56:30Z

AlbertKonig: /* Working on Solving Problems */

Group members:
* Catherine Chen
* Curtis Doucette
* Tanya Jacob
* Albert König
* Shauna Maty

----

== Deriving to the area of the pentagon using squares ==

[[File:P1120678.JPG]]

[[Media:P1120679.jpg]]

Unable to show steps on wiki. Hand written work will be submitted.

== Working on Solving Problems ==

1) There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal

2) As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only "might have" her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car

3) The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.

4) If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.

5) From different point of views there different numbers of rotationsa.

a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees.

b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.

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.
Let's say Reuben's birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.

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?

Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn't an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.

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.

It is impossible to determine the ages of Paula's children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don't know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36.
The second piece of information
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?

Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.

After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.

Course:MATH110/Archive/2010-2011/003/Groups/Group 01

2010-10-13T06:48:51Z

AlbertKonig: /* Working on Solving Problems */

Group members:
* Catherine Chen
* Curtis Doucette
* Tanya Jacob
* Albert König
* Shauna Maty

----

== Deriving to the area of the pentagon using squares ==

[[File:P1120678.JPG]]

[[Media:P1120679.jpg]]

Unable to show steps on wiki. Hand written work will be submitted.

== Working on Solving Problems ==

1) There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal

2) As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only saw her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.

3) The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.

4) If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.

5) From different point of views there different numbers of circumambulation.

a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0O and one time at 180O.

b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.

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.
Let's say Reuben's birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.

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?

Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn't an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.

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.

It is impossible to determine the ages of Paula's children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don't know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36.
The second piece of information
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?

Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.

After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.

Course:MATH110/Archive/2010-2011/003/Groups/Group 01

2010-10-13T06:45:40Z

AlbertKonig: /* Working on Solving Problems */

Group members:
* Catherine Chen
* Curtis Doucette
* Tanya Jacob
* Albert König
* Shauna Maty

----

== Deriving to the area of the pentagon using squares ==

[[File:P1120678.JPG]]

[[Media:P1120679.jpg]]

Unable to show steps on wiki. Hand written work will be submitted.

== Working on Solving Problems ==

1) There are several reason for this time difference.

a)One might be due to the different route the bus was taking when returning from the airport to the terminal. He might have used some different routes in order to collect other passengers on his way back.

b)Another reason could be due to some passengers. There might be several disabled people who had to have a special care and hence the bus had to wait longer than usual in order for those people enter the bus

2) As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only saw her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.

3) The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.

4) If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.

5) From different point of views there different numbers of circumambulation.

a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0O and one time at 180O.

b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.

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.
Let's say Reuben's birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.

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?

Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn't an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.

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.

It is impossible to determine the ages of Paula's children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don't know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36.
The second piece of information
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?

Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.

After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.

User:AlbertKonig

2010-09-26T21:46:31Z

AlbertKonig: Created page with 'Hey everyone, my name is Albert and well I'm a first year science student; my major is Food, Nutrition and Health in I am in faculty of Land/Food systems (short LFS). I'm new in …'

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:30:02Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^(-1) + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:29:40Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^ (-1) + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:29:02Z

AlbertKonig: /* Re: WebWorks Questions */

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:28:51Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^^-1 + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:28:29Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^-^1 + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:28:17Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^-1) + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:28:03Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^2 - 7 + 6 = 0</math>
v) <math>x^2 - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^2 * x^-1 + 13 = 14</math>
iii) <math>x^1 = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:27:10Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>((x^4/3)) - (x^2/3) + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^(2) - 7 + 6 = 0</math>
v) <math>x^(2) - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^(2) * x^(-1) + 13 = 14</math>
iii) <math>x^(1) = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:26:33Z

AlbertKonig: /* Re: WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>x^4/3 - x^2/3 + 6 = 0</math>
ii) Take a factor from <math>x^2/3</math>
iii) <math>x^2/3 * (x^2 - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^(2) - 7 + 6 = 0</math>
v) <math>x^(2) - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^(2) * x^(-1) + 13 = 14</math>
iii) <math>x^(1) = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

2010-09-26T21:25:30Z

AlbertKonig: /* WebWorks Questions */

==WebWorks Questions==
There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : <math>P(x)= x^{4/3} - 7x^{2/3}+6</math>

The second is : <math> x^2 / x+13=14</math> ; solve for x.

Thanks!

==Re: WebWorks Questions==

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

1) i) Solve for x: <math>x^(4/3) - x^(2/3) + 6 = 0</math>
ii) Take a factor from <math>x^(2/3)</math>
iii) <math>x^(2/3) * (x^(2) - 7) + 6 = 0</math>
iv) Take x^(2/3) to the other side, which gives us: <math>x^(2) - 7 + 6 = 0</math>
v) <math>x^(2) - 1 = 0</math>
vi) x = +/- 1

2) i) Solve for x: <math>x^2 / x+13=14</math>
ii) Do the following: <math>x^(2) * x^(-1) + 13 = 14</math>
iii) <math>x^(1) = -1</math>
iv) x = -1

Lets c if this is right!

Albert

==Question 17==

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for <math>\sqrt{3V\over h}</math>.

I'm still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]

Try to use exponent notation instead of the square root. So write <math>x^{1/2}</math> instead of <math>\sqrt{x}</math>.

-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide <math>\frac {x^{1}}{2}</math>. Took me a few tries to figure this one out.

[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)

==Question 26==

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

<math>{6\over x-1}-{6\over x}\geqq 1</math>

I still believe the correct answer is <math>(-\infty, -2]\cup[3,\infty)</math> but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]

==Question 15==

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

===Re: Question 15===

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are ''A (7,6) and B (13,0)''.

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to:
'''
((y2+y1)/(2)) , ((x2+x1)/(2))'''

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2))
= (y,x)
Therefore midpoint '''(x,y) = (10,3)'''

''Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:''

3=1(10) + b
Now I just isolate b and solve:
b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

==Question 7==

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

<math>\sqrt{(2-0)^2+(2-b)^2}</math>

and with point (6,-5):

<math>\sqrt{(6-0)^2+(-5-b)^2}</math>

and then I made them equal to each other to solve for b.

This simplifies to

<math>2+(2-b)=6+(-5-b)</math>

b is then found to be

<math>-4.25</math>

But when I plugged in <math>-4.25</math> into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)

Adams, I think you've got the sets of points wrong. The question is supposed to be "Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (5, -4). I got the right answer with the method you used, so it's just a matter of using the correct data!

[[User:BellaTory|BellaTory]]

The PDF Version that I printed out says (2,2) and (6,-5) :(
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

===Re: Question 7===
Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point ''equidistant'' from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck!
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks!
LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany