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Subject: The equation y=a(x-h)^2-k


Author:
Kayanachan
[Edit]
Date Posted: 16:40:55 04/24/07 Tue
I'm having some trouble with my math homework. I can't seem to figure what h and k are in the equation y=a(x-h)^2-k. Any one know?

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[> Subject: Re: The equation y=a(x-h)^2-k


Author:
Suzie
[Edit]
Date Posted: 15:57:54 10/28/08 Tue
>I'm having some trouble with my math homework. I can't
>seem to figure what h and k are in the equation
>y=a(x-h)^2-k. Any one know?

(h,k) is the vertex of the parabola, h is the point on the x- axis and k is the point on the y-axis

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Subject: Geometry- proof


Author:
LV
[Edit]
Date Posted: 14:21:25 06/03/04 Thu
Hello
I am needing help with a proof... actually just needing some guidance as to where to start! I have to prove the following in as many ways as possible: Given a triangle ABC with median AD, prove that 2((AB^2)+(AC^2))= BC^2 + 4(AD^2). I have proven this using analytic methods... and that will barely get me a passing mark. Any ideas?

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[> Subject: Re: Geometry- proof


Author:
mik
[Edit]
Date Posted: 16:26:28 11/18/05 Fri
hi.. i am having trouble figuring out what iam trying to prove in this question: In triangle ABC, A(o,a) B(0,0) and C (b,c) , using hte analytic method, prove that the right bisectors of the sides meet at a common point.

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[> [> Subject: Re: Geometry- proof


Author:
QUITTNER
[Edit]
Date Posted: 08:23:07 11/21/05 Mon
>>> In triangle ABC, A(o,a) B(0,0) and C (b,c) , using hte analytic method, prove that the right bisectors of the sides meet at a common point. <<<
..... A to B is a vertical line, C is further to the right.
..... Find the equation of the line AC, then the equation of its right-angled bisector.
.....Do the same with AB (a horizontal line). At which point does it meet the bisector line you found?
..... Then do the same with line BC. At which point does it meet the other 2? Prove that it is the SAME point.

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[> Subject: Re: Geometry- proof


Author:
mik
[Edit]
Date Posted: 16:26:37 11/18/05 Fri
hi.. i am having trouble figuring out what iam trying to prove in this question: In triangle ABC, A(o,a) B(0,0) and C (b,c) , using hte analytic method, prove that the right bisectors of the sides meet at a common point.

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[> Subject: Re: Geometry- proof


Author:
Ladricka Brown
[Edit]
Date Posted: 12:41:22 02/12/06 Sun
>Hello
>I am needing help with a proof... actually just
>needing some guidance as to where to start! I have to
>prove the following in as many ways as possible: Given
>a triangle ABC with median AD, prove that
>2((AB^2)+(AC^2))= BC^2 + 4(AD^2). I have proven this
>using analytic methods... and that will barely get me
>a passing mark. Any ideas?

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[> Subject: Re: Geometry- proof


Author:
Ladricka Brown
[Edit]
Date Posted: 12:44:52 02/12/06 Sun
>Hello
>I am needing help with finding numbers in 30-60-90 right triangle and a 45-45-90 triangle as well as angles of elevation and depression, think u could help me out? It would mean a great deal to me if u could help me out>

Thanks, Ladricka Brown

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[> [> Subject: Re: Geometry- proof


Author:
Esperanza Arriaza
[Edit]
Date Posted: 19:59:16 02/16/06 Thu
>>Hello
>>I am needing help with finding numbers in 30-60-90
>right triangle and a 45-45-90 triangle as well as
>angles of elevation and depression, think u could help
>me out? It would mean a great deal to me if u could
>help me out>
>
> Thanks, Ladricka Brown

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[> [> Subject: Re: Geometry- proof


Author:
Mohammed Adams (angles of elevation and depression)
[Edit]
Date Posted: 08:14:38 07/23/07 Mon
i am having problem in solving a particular topic in mathematics and this topic include Angles of elevation and depression.




Thanks,from Adams

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Subject: Need help with Quadratic Equations


Author:
Robin
[Edit]
Date Posted: 16:55:58 11/20/06 Mon
For the function y=x^2-4x-5 perform the following task:

Put the function in the form y=a(x-h)^2+k

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Subject: parametric equation


Author:
adeela (required equation for rounded arch)
[Edit]
Date Posted: 08:37:39 09/11/06 Mon

i want to draw a figure in OPENGL computer graphics. may task is to make rounded arch pointed arch and an ogee arch for that i required a parametric equation.

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Subject: Amount


Author:
Randy
[Edit]
Date Posted: 23:59:40 07/14/06 Fri
Calculate the amount of rainfall (in inches) needed to put 100,000 gallons of water over an area of 3.76 acres.

Thanks for your assistance.

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Subject: Area and Percent


Author:
Randy
[Edit]
Date Posted: 19:12:54 07/06/06 Thu
Calculate the area of the largest circle that can be cut from a rectangular sheet of metal and the percent of the rectangle wasted.

Dimensions of sheet: Length- 11 in.

Width- 8.5 in.

Thanks for your assistance.

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[> Subject: Re: Area and Percent


Author:
QUITTNER
[Edit]
Date Posted: 11:20:30 07/10/06 Mon
Calculate the area of the largest circle that can be cut from a rectangular sheet of metal and the percent of the rectangle wasted.
Dimensions of sheet: Length- 11 in., Width- 8.5 in.
..... The diameter of the largest possible circle is the same as length of the width of the rectangle.
Calculate the area of the circle. (ANSWER 1)
Calculate the area of the rectangle.
Deduct the area of the circle from the area of the rectangle, that difference is the wasted area.
Get that wasted area in terms of a percentage of the total area of the rectangle. (ANSWER 2)
..... Good luck!

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Subject: Geometry


Author:
Heather C.
[Edit]
Date Posted: 13:54:31 04/25/06 Tue
If two polygons are similar with lengths of corresponding sides in the ratio of a:b, then the ratio of their perimeters is ________and the ratio of their areas is
__________.

Please help me fill in the blanks!

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Subject: What is the equation for understanding message boards?


Author:
Jim Jefferson
[Edit]
Date Posted: 07:23:04 12/08/05 Thu
This is a test


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[> Subject: Re: What is the equation for understanding message boards?


Author:
QUITTNER
[Edit]
Date Posted: 11:19:16 12/08/05 Thu
Calculate the time required to drain the whole body of enough blood to stop the bloody picture from repeating itself. Plot your results.
..... This test is due yesterday. [LAUGHS]

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Subject: coordinate geometry: linear equations in two variables


Author:
Misty Bowman
[Edit]
Date Posted: 18:06:25 06/18/05 Sat
what is the equation of the line that contains the points with (x,y)coordinates(-3,7) and (5,-1)? I need to know how to do this problem step by step.

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[> Subject: Re: coordinate geometry: linear equations in two variables


Author:
QUITTNER
[Edit]
Date Posted: 07:17:31 06/20/05 Mon
>>> what is the equation of the line that contains the points with (x,y)coordinates(-3,7) and (5,-1)? <<<
..... The equation of a straight line is y=ax+b. Just insert the given values into the two simultaneous equations, and solve them for a and for b.
7=(-3)a + b
(-1)= 5a + b

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[> Subject: Re: coordinate geometry: linear equations in two variables


Author:
fizixx
[Edit]
Date Posted: 18:01:22 12/01/05 Thu
>what is the equation of the line that contains the
>points with (x,y)coordinates(-3,7) and (5,-1)? I need
>to know how to do this problem step by step.

The equation of a line is:

y = mx + b

m is the slope
x is the independent variable
b is the 'y' intercept
y is the dependent variable

You essentially have to supply 'm' and 'b'.

You can use the following formula:

y - y1 = mx - x1

This is the pont-slope equation, and you have all this information at hand.

m = (y2 - y1)/(x2 - x1)

Let

y1 = -3
y2 = 5
x1 = 7
x2 = -1

It really doesn't matter which ones you pick, as long as the x values correspond to the actual x coordinate given....same for y.. in other words. If you have (8,5)....x = 8 and y = 5. You can't switch this. Understand?

So, plug this in and you will find that m = -8/8 = -1

Then the point-slope equation becomes:

y - 7 = -x - (-3)
==> y - 7 = -x + 3
==> y = -x + 3 + 7
==> y = -x + 10

That's your equation. Hope this makes sense.

fizixx

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Subject: Vertex of Parabola


Author:
Wira Maya Tio
[Edit]
Date Posted: 19:54:08 07/21/05 Thu
Please help?! I have 2 horizontally aligned points, and need to find the vertex of the parabola. Additional information known is that the half curve length of the parabola must be L.

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[> Subject: Re: Vertex of Parabola


Author:
fizixx
[Edit]
Date Posted: 17:52:41 12/01/05 Thu
>Please help?! I have 2 horizontally aligned points,
>and need to find the vertex of the parabola.
>Additional information known is that the half curve
>length of the parabola must be L.

This is not enough information for anyone to solve this for you. You need to supply more detail. Saying that you have two points aligned along a horizontal is insufficient. Do you have values for these points? In other words, do you know their cardinal points? For example; (2,4).....or....(-1,5). Like that.

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Subject: I like your website. Good design, pretty navigations.


Author:
Juli
[Edit]
Date Posted: 13:59:04 07/22/05 Fri
I like your website. Good design, pretty navigations.
mortgage

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Subject: Parabola


Author:
FB
[Edit]
Date Posted: 15:39:19 06/06/05 Mon
Ok i am having a parabola problem i think i drew it right but im not sure. Anyways here is the question:

A masonry bridge over a river has an arch in the form of a parabola with a vertical axis. The widths of the arch at water level are 5m above water level and 12m and 8m respectively. Find the height of the arch above the point at water level which is 2m from the middle.

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[> Subject: Re: Parabola


Author:
QUITTNER
[Edit]
Date Posted: 08:17:07 06/10/05 Fri
>>> A masonry bridge over a river has an arch in the form of a parabola with a vertical axis. The widths of the arch at water level are 5m above water level and 12m and 8m respectively. Find the height of the arch above the point at water level which is 2m from the middle. <<<
..... A shortened form of the equation with the origin (x=0 and y=0) at the top of the arch is
y = p * x^2 with all y being negative.
..... Put y1 at x=2, put y2 at x=4, and y3 at x=6. The x of y2 and y3 are half the total widths.
..... We also know that (y2 - y3) = 5 (IS THAT RIGHT?) where y3 is the water level.
..... Insert into the equations y=p*x^2 where * means times, and ^2 means squared:
y3=6*6*p=36p, y2=16p, and y2-y3=5. Solve for p and you then can get y1, y2 and y3. The required height above water level is y1-y3 where x=2.

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Subject: i need the area of the overlapping parts


Author:
randy
[Edit]
Date Posted: 19:19:51 03/24/05 Thu
two overlapping squares, one at a 59-degree angle to the other. The larger of the two squares is 180 meters on a side, and the smaller of the two is 162 meters on a side. The corner point of the larger square is at the center point of the smaller square.

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Subject: Finding the Equation of a Parabola using the vertex and 1 point


Author:
Ed
[Edit]
Date Posted: 22:13:15 11/05/03 Wed
I seem to not be able to come up with the right answers for the following problem:

vertex (5,12), Point (7,15)
I know that standard form is f(x)=a(x-h)^2+k
how do i plug these in to get the proper answer.
The book gives the answer as -3/4(x-5)^2+12

How do I get to that answer?

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[> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
[Edit]
Date Posted: 09:11:49 11/10/03 Mon
>vertex (5,12), Point (7,15)
>I know that standard form is f(x)=a(x-h)^2+k
>how do i plug these in to get the proper answer.
>The book gives the answer as -3/4(x-5)^2+12
..... We have two points (x,y), with the vertex being one of them. We also know that
y=a(x-h)^2+k This is (y-k)=a*(x-h)^2
Note that we deal here with the differences in x and in y, distances from the (given) vertex. Therefore k=the y of the vertex, and h is the x of the vertex.
Therefore (15-12)=a*(7-5)^2
3=a*(2)^2 and 3=a*4 giving a=3/4
The parabola's equation is therefore for ANY x and y:
y=(3/4)*(x-5)^2 + 12 COMPARE with the book's answer.

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[> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
masnah binti abd.rahim
[Edit]
Date Posted: 02:59:18 09/09/04 Thu
>>vertex (5,12), Point (7,15)
>>I know that standard form is f(x)=a(x-h)^2+k
>>how do i plug these in to get the proper answer.
>>The book gives the answer as -3/4(x-5)^2+12
>..... We have two points (x,y), with the vertex being
>one of them. We also know that
>y=a(x-h)^2+k This is (y-k)=a*(x-h)^2
>Note that we deal here with the differences in x and
>in y, distances from the (given) vertex. Therefore
>k=the y of the vertex, and h is the x of the vertex.
>Therefore (15-12)=a*(7-5)^2
>3=a*(2)^2 and 3=a*4 giving a=3/4
>The parabola's equation is therefore for ANY x and y:
>y=(3/4)*(x-5)^2 + 12 COMPARE with the book's answer.

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
[Edit]
Date Posted: 10:17:49 09/09/04 Thu
masnah binti abd.rahim, WHAT IS YOUR QUESTION?

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[> [> [> Subject: Re: Finding the Equation of a Parabola


Author:
Justin
[Edit]
Date Posted: 15:55:44 02/21/05 Mon
if the problem says express each equation in the form y=a(x-h)²+k or x=a(y-k)²+h, then how do you do y=2x²-12x+6?

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[> [> [> [> Subject: Re: Finding the Equation of a Parabola


Author:
QUITTNER
[Edit]
Date Posted: 10:00:46 02/22/05 Tue
>>> if the problem says express each equation in the form y=a(x-h)²+k or x=a(y-k)²+h, then how do you do y=2x²-12x+6?
..... First you selcet the proper equation. Here the left side is y, therefore use y=a(x-h)^2+k. Expand this to get
y=a*(x^2-2hx+h^2) + k or
y=a(x^2) - 2ahx +a(h^2) + k (1) We also know
y=2x^2) - 12x + 6 (2)
Compare the coeficients of equation (1) with those of (2):
a=2 4h=12, h=3 18+k=6, k=-12

Always check your answers:
y=2*(x-3)^2 - 12 = 2*(x^2-6x+9) - 12 =
y=2*(x^2)-12x+18-12 OK

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[> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
Kristen
[Edit]
Date Posted: 09:33:14 10/18/04 Mon
y=a(h-h)* +k (*=^2)
fill in your points
15=a(7-5)*+12
15-12=a(2)*
3=4a
a=3/4

Standard, fill in a and vertex
a(x-h)*+k
3/4(x-5)*+12

thus that is your equation in standard form

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
Kristen
[Edit]
Date Posted: 09:34:19 10/18/04 Mon
>y=a(h-h)* +k (*=^2)
>fill in your points
>15=a(7-5)*+12
>15-12=a(2)*
>3=4a
>a=3/4
>
>Standard, fill in a and vertex
>a(x-h)*+k
>3/4(x-5)*+12
>
>thus that is your equation in standard form

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
[Edit]
Date Posted: 07:49:56 10/19/04 Tue
Given that the vertex has coordinates (a,b), and the one give point has coordinats (x1,y1)
Starting with the equation of a parabola:
(y-a)^2 = k * (x-b)
We know a and b (the vertex), and we can then use
(y1-a)^2 = k * (x1-b) to find k

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[> [> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
[Edit]
Date Posted: 08:05:59 10/19/04 Tue
CORRECTION: (Reversing a and b in my previous post):

Given that the vertex has coordinates (a,b), and the
one given point has coordinates (x1,y1)

Starting with the equation of a parabola:
(y-b)^2 = k * (x-a)
We know a and b (the vertex), and we can then use
(y1-b)^2 = k * (x1-a) to find k

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Subject: Books on Maple and Statistics


Author:
Academy
[Edit]
Date Posted: 05:57:01 12/19/04 Sun
A carefully prepared and well-written book

http://writers.fultus.com/aladjev/book01.html

without doubt can be recommended as a very valuable manual and textbook for a Maple user who is looking for a mainly textbook which combines programming manual with a presentation of widely applicable new Maple means.

The following web site:

http://writers.fultus.com/aladjev/book02.html

presents a new book on General Theory of Statistics.

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[> Subject: Re: Books on Maple and Statistics


Author:
QUITTNER
[Edit]
Date Posted: 06:35:26 12/20/04 Mon
Thanks for the info. Do have a MERRY CHRISTMAS and/or other happiness at this greetings season, and a healthy and happy whole NEW YEAR 2005! Health is more important than money.

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Subject: finding equation of parabola going through 3 points


Author:
soule
[Edit]
Date Posted: 23:11:21 11/30/04 Tue
im having a hard time solving this problem. Can somebody help? it says. Find an equation for the parabola with horizontal axis of symmetry if it passes though points (5,0) (14,1) and (5,-2)

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[> Subject: Re: finding equation of parabola going through 3 points


Author:
QUITTNER
[Edit]
Date Posted: 12:12:25 12/03/04 Fri
>>> Find an equation for the parabola with horizontal axis of symmetry if it passes though points (5,0) (14,1) and (5,-2) <<<
..... As I remember it, the equation for a parabola with HORIZONTAL axis of symmetry is
(y-a) squared = k times (x-b)

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[> [> Subject: Re: finding equation of parabola going through 3 points


Author:
QUITTNER
[Edit]
Date Posted: 07:25:39 12/06/04 Mon
>>>> Find an equation for the parabola with horizontal
>axis of symmetry if it passes though points (5,0)
>(14,1) and (5,-2) <<<<
>>> ..... As I remember it, the equation for a parabola with HORIZONTAL axis of symmetry is (y-a) squared = k times (x-b) <<<
..... Normally from this point on you substitute the 3 x and the 3 y of the 3 points given into 3 simultaneous equations, and then solve for the 3 unknown variables a,b,and k.
HOWEVER, in this particular example there is an easier way:
..... Note that two of the given points have the SAME x, with their y being 0 and -2. This means that the axis of symmetry is half-way between these y, namely -1. But the vertex is on this axis, so that the vertex, too, has y=-1
..... In the equation on its left side it says (y-a)^2, and the a represents the y of the vertex, which we know now as being -1 = a. And on the right side of the equation b is the x of the vertex. That means that the new equation, for this example only, is:
(y+1) squared = k times (x-b) which is (y+1)^2=k*(x-b)
Now there are only 2 unknown variables, k and b, and you need now only 2 equations to solve for them:
(0+1)^2 = k*(5-b) for point (5,0)
(1+1)^2 = k*(14-b) for point (14,1)
Always check your results.

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Subject: calculate a parabola from 5 diff points


Author:
guy
[Edit]
Date Posted: 02:07:44 05/26/04 Wed
hi i need some help im trying to find an algorithm for
finding the equation of a parabola from 5 diff points
thanks for your time!!!
guy

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[> Subject: Re: calculate a parabola from 5 diff points


Author:
QUITTNER
[Edit]
Date Posted: 10:19:03 05/27/04 Thu
>>> hi i need some help im trying to find an algorithm for finding the equation of a parabola from 5 diff points thanks for your time!!! guy <<<
..... A parabola is fully described by only 3 points. If you are given more than 3 points then you could use "the least squares method" to find the closest approximate parabola, meaning finding a parabola whose value at a given x, call it y, is a distance, call it d, away from a given point at that x, with the sum of all the (d squared) being at a minimum. Is that what you want?

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[> [> Subject: Re: calculate a parabola from 5 diff points


Author:
QUITTNER
[Edit]
Date Posted: 11:36:38 06/17/04 Thu
By the way, some people use A+Bx+Cx^2 as a general equation of what they call a parabola.
..... A REAL parabola is NOT the same as that, but is defined as a curve, all of whose points are at the SAME distance from a given line and from a given point, called the "Focal Point". The equation of a special kind of parabola is (y-A)^2=K*(x-B), with, in an even more special case, A and B are both zero.
..... If there are 3 unknowns then you need 3 simultaneous equations to solve (can be quite dificult!) for these unknowns.

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Subject: Inverse of Polynomial Function


Author:
tnewton
[Edit]
Date Posted: 17:40:43 02/17/04 Tue
What I would really like to do is find the coefficients of INV(f(x)) from the coefficients of f(x) of an nth degree polynomial (1-variable).

f(x)= SUM(ai * x^i) for i = 0..n)
INV(f(x)) = ???


Any and all help is greatly appreciated. Thanks!

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[> Subject: Re: Inverse of Polynomial Function


Author:
QUITTNER
[Edit]
Date Posted: 10:22:18 04/23/04 Fri
>>>What I would really like to do is find the coefficients of INV(f(x)) from the coefficients of f(x) of an nth degree polynomial (1-variable).<<<

>>> f(x)= SUM(ai * x^i) for i = 0..n)
>INV(f(x)) = ??? <<<
..... As I remember it, if y is a function of the sum of various degrees of x, then the inverse is (simply - oh yeah?) x as a function of y. That may be quite difficult to find is many cases involving higher than the 3rd degree of x.

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[> [> Subject: Re: Inverse of Polynomial Function


Author:
tnewton
[Edit]
Date Posted: 19:39:51 05/04/04 Tue
Yes, that's the definition of inverse, but it requires you to know the order of the function first in order to solve. I believe there may be some algorithm to find the reverse coefficients regardless of degree, and may require very deep recursion (related to the polynomial order); at least that's the direction I seem to keep finding myself heading. I've seen a process that finds the solution to the inverse function; it's a little tough to follow and I haven't been able to use it as a basis for determining the coefficients. Maybe if I figure this out I'll get my name in the history books...
>>>>What I would really like to do is find the
>coefficients of INV(f(x)) from the coefficients of
>f(x) of an nth degree polynomial (1-variable).<<<
>
>>>> f(x)= SUM(ai * x^i) for i = 0..n)
>>INV(f(x)) = ??? <<<
>..... As I remember it, if y is a function of the sum
>of various degrees of x, then the inverse is (simply -
>oh yeah?) x as a function of y. That may be quite
>difficult to find is many cases involving higher than
>the 3rd degree of x.

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[> [> [> Subject: Re: Inverse of Polynomial Function


Author:
QUITTNER
[Edit]
Date Posted: 10:42:17 05/06/04 Thu
tnewton, unfortunately I can't help you with that. But maybe you'll come up with some useful solution(s). Good luck, and keep well!

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Subject: Finding equation of a parabola with the use of 3 points


Author:
Joanna
[Edit]
Date Posted: 23:51:55 05/11/03 Sun
I'm doing some tutorial questions on matrices and one asked you to find the equation of a parabola that passes thru' 3 points....(3,-1), (1,-7) and (-2,14). I was wondering if anyone knows what processes/formula i should use to find the answer? Thanx.

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[> Subject: Re: Finding equation of a parabola with the use of 3 points


Author:
QUITTNER
[Edit]
Date Posted: 06:21:51 05/23/03 Fri
Joanna, I assume that you know how to invert a matrix. One of he equations of a parabola is A+Bx+Cx^2=y, which can also be written as Ax^0+Bx^1+Cx^2=y, where x^0 is always 1. A Polynomial of nth degree needs n+1 points; here we have a 2nd degree polynomial and 3 points, that's OK.
..... Let matrix M1 equal the left side of the equation, and fill in all the x's as follows:
Take one of the points (in any order), (3,-1) first. Its x is 3, therefore the top line of M1 is 1, for x'0, 3 for x, and 9 for x'2. The 2nd line is for point (1,-7): 1,1,1. The 3rd line for point (-2,14) is 1,-2,4. Therefore, to repeat, M1 is

| 1 3 9 | for point 1,
| 1 1 1 } 2,
| 1 =2 4 | 3.

The right side of the equation of the parabola is only y, we then make Matrix M2 equal to
| -1 | for point 1,
| -7 | 2,
| 14 | 3.

Next invert M1 to get matrix M3:
| 1/5 1 1/5 |
| 1/10 1/6 -15/4 |
| 1/10 -1/6 1/15 |

To get the results matrix, multiply M2 by M3, giving
| -4 |
| -5 |
| 2 | Therefore the equation of the parabola is
A+Bx+Cx^2=y where A=-4, B=-5 and C=2. Make sure you check this by evaluating the equation for each of the given x values of each of the 3 points and see that you get the given y's.

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[> [> Subject: Finding equation of a parabola with vertex at Oy and passing thru (2,2)


Author:
lanie
[Edit]
Date Posted: 18:00:29 11/06/03 Thu
>Joanna, I assume that you know how to invert a matrix.
>One of he equations of a parabola is A+Bx+Cx^2=y,
>which can also be written as Ax^0+Bx^1+Cx^2=y, where
>x^0 is always 1. A Polynomial of nth degree needs n+1
>points; here we have a 2nd degree polynomial and 3
>points, that's OK.
>..... Let matrix M1 equal the left side of the
>equation, and fill in all the x's as follows:
>Take one of the points (in any order), (3,-1) first.
>Its x is 3, therefore the top line of M1 is 1, for
>x'0, 3 for x, and 9 for x'2. The 2nd line is for point
>(1,-7): 1,1,1. The 3rd line for point (-2,14) is
>1,-2,4. Therefore, to repeat, M1 is
>
>| 1 3 9 | for point 1,
>| 1 1 1 } 2,
>| 1 =2 4 | 3.
>
>The right side of the equation of the parabola is only
>y, we then make Matrix M2 equal to
>| -1 | for point 1,
>| -7 | 2,
>| 14 | 3.
>
>Next invert M1 to get matrix M3:
>| 1/5 1 1/5 |
>| 1/10 1/6 -15/4 |
>| 1/10 -1/6 1/15 |
>
>To get the results matrix, multiply M2 by M3, giving
>| -4 |
>| -5 |
>| 2 | Therefore the equation of the parabola is
>A+Bx+Cx^2=y where A=-4, B=-5 and C=2. Make sure
>you check this by evaluating the equation for each of
>the given x values of each of the 3 points and see
>that you get the given y's.

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[> [> [> Subject: Re: Finding equation of a parabola with vertex at Oy and passing thru (2,2)


Author:
QUITTNER
[Edit]
Date Posted: 11:19:31 11/11/03 Tue
You wrote: "Finding equation of a parabola with vertex at Oy and passing thru (2,2)"
..... I suppose this means that the vertex is at the ORIGIN (0,0), and that one point is at (2,2)
..... We have the general equation y-k=a(x-h)^2 where, in our case k and h are both zero. Therefore
y=a*x^2 and in our case 2=a*2^2=4a. Therefore a=1/2 and
y=(1/2)*x^2
..... Check it out: if x=2 then y=1/2 of 4, y=2 OK

[ Post a Reply to This Message ]
[> Subject: Finding equation of a parabola


Author:
Maria
[Edit]
Date Posted: 23:03:22 04/24/04 Sat
hi, could you please tell me how to find the equation of a line (not a straightline), with points (8,11) and the other one is (x,35) where x is an unknown coordinate. the slope is 0. This is a parabola. Thank you very much.

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[> [> Subject: Re: Finding equation of a parabola


Author:
QUITTNER
[Edit]
Date Posted: 12:57:38 04/26/04 Mon
>hi, could you please tell me how to find the equation
>of a line (not a straightline), with points (8,11) and
>the other one is (x,35) where x is an unknown
>coordinate. the slope is 0. This is a parabola. Thank
>you very much.
..... A parabola has very many points, and there is a slope at every one of them. You say the slope is 0, but you don't say at which point that slope is zero.

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Subject: Hints for mathematical Topic wanted, Solution heavenly!


Author:
Peter
[Edit]
Date Posted: 06:14:20 09/05/03 Fri
I got a geometric Problem which explodes my skills.
The Problem can be described by 2 nonlinear equations with 2 unknowns (trigonometric polynoms).
I got a numerical solution for the equations.

But the real problem is, the parameters of the trigonometric polynoms are statistically distributed and not exact. I don't know, how to consider non exact parameters in solving a system of nonlinear equations??
Does this lead into the theory of Optimisation??

It would be of great benefit for me if I get hints for the right topic in mathematics where I have to look for solutions!

Please contact me per email, if you're interested in my problem. Then I will explain further details!

thanks!

fpierre@everymail.net

[ Post a Reply to This Message ]
[> Subject: please help me!


Author:
Rawan Kilo
[Edit]
Date Posted: 18:47:48 02/09/04 Mon
The volume of the rectangular solid is given by the expression 34.944x(cubed)+44.684x(squared)-8.29x-16.8. The surface area of one side of the rectangular solid is given by the expression 6.72x(squared)+12.47+5.6. The sum of the dimensions of the rectangular solid to the nearest hundredth.
You must utilize handheld technology to answer the problem and provide sketches of the graphs representing the cubic and quadratic expressions seen on handheld screen to justify your explanations.

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[> [> Subject: Re: please help me!


Author:
QUITTNER
[Edit]
Date Posted: 12:31:45 02/13/04 Fri
>The volume of the rectangular solid is given by the
>expression 34.944x(cubed)+44.684x(squared)-8.29x-16.8.
>The surface area of one side of the rectangular solid
>is given by the expression 6.72x(squared)+12.47+5.6.
>The sum of the dimensions of the rectangular solid to
>the nearest hundredth.
..... What is the difficulty? You are given a volume of a rectangular solid in terms of an equation, call it y3 (3-dimensional).
..... You are also given the surface area of one side in terms of an equation, call it y2 (2-dimensional).
..... The volume of a rectangular solid is one side's length times the surface area of one side, so that y3 divided by y2 should give you the equation of the side's length, call it y1 (1-dimensional).
..... The graphs of these equations are partly negative, so you must first mention, then reject the negative parts. y1 appears to be sufficiently linear (depends on how many decimal places are used in the calculations), so that any two points can be used to get the equation of that line, such as the (approximate) root of y1 and its (-ve) intercept on the y-axis. Can you take it from here?

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Subject: conics


Author:
cheekz
[Edit]
Date Posted: 15:02:13 01/18/04 Sun
The cables of the suspension bridge hang in a curve that approximates a parabola. The road passes through the vertex. The supporting towers are 550 m apart and 50 m high. Write and equation for the parabola

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[> Subject: Re: conics


Author:
QUITTNER
[Edit]
Date Posted: 12:37:21 01/19/04 Mon
>>> The cables of the suspension bridge hang in a curve that approximates a parabola. The road passes through the vertex. The supporting towers are 550 m apart and 50 m high. Write and equation for the parabola. <<<
..... Assuming that the vertex is at the origin (0,0), the equation of the parabola is
y^2 = 2 a x - Find a
..... The vertex is where the cables hit the road level, and it's a symmetic graph. The top of the tower on the right is at (275,50). Substitute into the equation to find a.

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Subject: Mathematics related to taxes


Author:
Maria
[Edit]
Date Posted: 18:32:51 10/16/03 Thu
I have been working on an LCX assignement for the past few days, and yet, I'm still stumped. the problem is quite simple, it consists of talking the total price of an object and find the initial value; before taxes have been added to the price. If the Total is 80.20$ and we know that the initial price is taxed once at 7%, compounded and taxed again, at a rate of 7.5%, what is the initial price of the item?

[ Post a Reply to This Message ]
[> Subject: Re: Mathematics related to taxes


Author:
Maria
[Edit]
Date Posted: 18:35:22 10/16/03 Thu
>I have been working on an LCX assignement for the past
>few days, and yet, I'm still stumped. the problem is
>quite simple, it consists of taking the total price
>of an object and find the initial value; before taxes
>have been added to the price. If the Total is 80.20$
>and we know that the initial price is taxed once at
>7%, compounded and taxed again, at a rate of 7.5%,
>what is the initial price of the item?

By the way, if you could include a detailed solution, i would very much appreciateit.

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[> [> Subject: Re: Mathematics related to taxes


Author:
QUITTNER
[Edit]
Date Posted: 08:37:50 10/31/03 Fri
Subject: Mathematics related to taxes
Author: Maria Date Posted: 18:32:51 10/16/03 Thu
>>> I have been working on an LCX assignement for the past few days, and yet, I'm still stumped. the problem is quite simple, it consists of talking the total price of an object and find the initial value; before taxes have been added to the price. If the Total is 80.20$ and we know that the initial price is taxed once at 7%, compounded and taxed again, at a rate of 7.5%, what is the initial price of the item? <<<
..... If the item would have been taxed once at 7%, then to find the initial price before taxes if the total price is 80.20, then divide 80.20 by one plus 7 hundreds. That is 80.20/1.07=74.95. Check this by using a calculator 74.95 + 7%=80.20 (rounded), OK
..... If something is compounded then you have to know also the time(s) involved, which is NOT given here.

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Subject: + Save 75% on printer ink! Click Here!


Author:
LowCostInk
[Edit]
Date Posted: 13:42:56 10/05/03 Sun





Don't keep paying for inkjet cartridges when you can refill them for a fraction
of the price!



CLICK HERE!

[ Post a Reply to This Message ]
Subject: + Save 75% on printer ink! Click Here!


Author:
LowCostInk
[Edit]
Date Posted: 13:26:14 10/05/03 Sun





Don't keep paying for inkjet cartridges when you can refill them for a fraction
of the price!



CLICK HERE!

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Subject: Distance/ Time


Author:
Randy
[Edit]
Date Posted: 14:13:05 12/31/02 Tue
A steel railroad track is struck with a hammer at one end and sound waves are observed at the other, both along the track and through the air. What is the time interval (sec.)between the two sounds as heard by the observer? How long is the track? (The speed of sound in steel is assumed to be 16,000 fps.)

Please give step by step solution. Thanks for your assistance.

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[> Subject: Re: Distance/ Time


Author:
QUITTNER
[Edit]
Date Posted: 10:49:33 02/18/03 Tue
You wrote: >>> How long is the track? (The speed of sound
>in steel is assumed to be 16,000 fps.) <<<
..... Some information is missing. You also must know: What is the speed of sound in the air? The difference between the two sound speeds is a clue; it's also in feet per second.
1) If you know the length of the track then you divide that length by the speed difference (feet/(feet/second)) to get the difference between the sounds in seconds.
2)If you know the time difference between the two sounds then you multiply the speed difference by the time difference between the two sounds to get the track length in feet (seconds times (feet per second)).

[ Post a Reply to This Message ]
[> Subject: Re: Distance/ Time


Author:
Enrico D.
[Edit]
Date Posted: 23:32:33 03/15/03 Sat
>A steel railroad track is struck with a hammer at one
>end and sound waves are observed at the other, both
>along the track and through the air. What is the time
>interval (sec.)between the two sounds as heard by the
>observer? How long is the track? (The speed of sound
>in steel is assumed to be 16,000 fps.)
>
>Please give step by step solution. Thanks for your
>assistance.
If the railroad makes a curve, better a circle, the sound will arrive forst by air than by steel....

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[> Subject: rotation


Author:
Tina
[Edit]
Date Posted: 12:22:14 10/04/03 Sat
A car goes on a 20,00-mile trip. the tires, including the spare, are rotated every 4,000 miles, so that the fire tires were used for an identical numer of miles. How many miles was each tire monted on the care, in use and is there another way of simplifying the problem?
>
>Please give step by step solution. Thanks for your
>assistance.

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Subject: Calculus


Author:
xncuiw
[Edit]
Date Posted: 21:39:35 10/02/03 Thu
Find the x-coordinates of the points on the hyperbola xy=1 where the tangets from teh point (1,-1) intersect the curve.

-plz give a step solution, I know what the answer is(from back of book), but I don't know how to do it. answer = -1+-root2

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Subject: What numbers add up to $655.11


Author:
Mike Oxlong
[Edit]
Date Posted: 07:48:16 04/24/03 Thu
I need to figure out which combination of these numbers add up to $655.11. It could be three, four, or any number of the numbers that add up to it.

44.80
26.00
44.80
26.00
61.05
20.25
186.34
57.74
185.12
31.37
39.49
64.00
71.74
212.43
138.00
207.00

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[> Subject: Re: What numbers add up to $655.11


Author:
Jim Fox
[Edit]
Date Posted: 09:36:26 09/06/03 Sat
Answer: see numbers followed by "YES".
I've been able to handle up to 10,000 numbers, so far.
Jim Fox sends -- 6 September 2003.

>I need to figure out which combination of these
>numbers add up to $655.11. It could be three, four, or
>any number of the numbers that add up to it.
>
>44.80 YES
>26.00 YES
>44.80
>26.00
>61.05
>20.25 YES
>186.34 YES
>57.74 YES
>185.12 YES
>31.37 YES
>39.49 YES
>64.00 YES
>71.74
>212.43
>138.00
>207.00

[ Post a Reply to This Message ]
Subject: Help in understanding set theory problems


Author:
Pete Snyder
[Edit]
Date Posted: 04:38:25 05/05/03 Mon
Hello- I would like to ask you if you would please help me solve the two following problems. I have looked through many textbooks at my community college library. The frustrating part is that the books present information in an abstract sense, where no real numbers are used to solve the problems. Highly abbreviated equations are used and then 40-50 problems are given at the end of each section. No detailed examples are given in order to use as templates to solve the exercise problems. Would somebody please be able to provide detailed solutions to the following two problems. I would then be able to use these as examples (or templates) in order to understand what is being presented in the mathematics textbooks for these kinds of exercises. Also-please let me know how to actually say or verbalize the following equations. Thank you very much!

1. Let y and z be non-zero integers.
Show mathematical proof that if (3y, 10z) = 9, then x|(y,z). I need a proof and not just a numerical answer.

2. Let G be the following group of permutations.
G = {(3), (1234), (13)(24), (1432), (24), (12)(34), (13), (14)(23)}.
Let H = {(1), (1234), (13)(24), (1432)}.
Prove H is cyclic and find all the cosets of H in G.

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[> Subject: Re: Help in understanding set theory problems


Author:
QUITTNER
[Edit]
Date Posted: 15:28:09 05/05/03 Mon
Sorry, I don't know how to solve these problems. Are you aware of the many books in the SCHAUM OUTLINE SERIES on different mathematical topics? I like their many worked examples showing you how to solve problems, in addition to the theory. Good luck and have fun!

[ Post a Reply to This Message ]
[> Subject: Re: Help in understanding set theory problems


Author:
Lynn
[Edit]
Date Posted: 14:37:34 07/16/03 Wed
I assume that since you posted a couple of months ago, that if I'm answering assignment questions, it's too late. So, just so you can actually see how this is done, here we go:
(Incidentally, Shawm's is not too good at showing worked out solutions to problems that need proofs rather than numbers.)

>1. Let y and z be non-zero integers.
>Show mathematical proof that if (3y, 10z) = 9, then
>x|(y,z). I need a proof and not just a numerical
>answer.

Verbally: "If the greatest common divisor of 3y and 10z is nine, then find a factor of the greatest common divisor of y and z? (I'm paraphrasing a bit since as you typed it, it makes no sense.)

x = 3.

Proof (in words, sort of):

The fact that 9 is the greatest common divisor of 3y and 10z means two things: 1) that 9 divides both 3y and 10z.
and 2) that nothing bigger than 9 (no multiple of 9) also divides them.

So, let's see what we get out of consequence 1:
If 9|3y then it's clear that 3|y.
If 9|10z then 9 must divide z since 9 and 10 share no common factors.

Since 3|y and 3|9|z, we have that 3|(y,z).

In point of fact, using consequence 2 and the same sort of logic, it's possible to prove that 3 = (y,z)

Now, that's not how I would hand it in (although if marking it, I would sigh and give full marks).

What I would hand in:
Claim: 3|(y,z).
Proof:
9 | 3y (by hypothesis)
Thus, 3 | y (by dividing both sides by 9)

9 | 10z
(9,10) = 1
Thus, 9 | z (using the unique factorization theorem)
(This step may require more work, depending on how many consequences of the UFT you've seen in class.)
3 | 9
Thus 3 | z (transitivity)

Since 3|y and 3|z, then 3|(y,z).
QED
>
>2. Let G be the following group of permutations.
>G = {(3), (1234), (13)(24), (1432), (24), (12)(34),
>(13), (14)(23)}.
>Let H = {(1), (1234), (13)(24), (1432)}.
>Prove H is cyclic and find all the cosets of H in G.

(First, I assume that the first permutation in G is (1). Otherwise, it's not a group and nothing makes sense.)

The first part is fairly simple. To prove that a group is cyclic, all you need is to find a generator and then prove that that generator works. Since H has only 4 elements, trial and error should be quite quick.

So my solution would go as follows:
Claim: g:=(1234) is a generator of H.
Proof: Let this group act on the "word" abcd.

g(abcd) = dabc
g^2(abcd) = g(dabc) = cdab = [(13)(24)](abcd)
(ie, g^2 = (13)(24) )
g^3(abcd) = g(cdab) = bcda = [(1432)](abcd)
(ie, g^3 = (1432) )
g^4(abcd) = g(bcda) = abcd = [(1)](abcd)
(ie, g^4 = (1))
QED

The second part is also comparatively straight forward. A subgroup always divides the big group into equal parts when you take cosets. Furthermore, these equal parts are the same size as the subgroup. So, since H has 4 elements, it divides the 8-element group G into 2 parts:

{(1),(1234),(13)(24),(1432)} = H and
{(24), (12)(34),(13), (14)(23)} (the rest).

If I were grading this, I would accept what I've written here as a decent proof although I might like references to theorems for each of the sentences which I just stated.

Hope that helps.

Lynn

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Subject: derivative


Author:
Justin McHugh
[Edit]
Date Posted: 15:34:33 06/19/03 Thu
f(x)= xsin(x) is the derivative of this 2cos x - xsin(x)?

[ Post a Reply to This Message ]
[> Subject: Re: derivative


Author:
QUITTNER
[Edit]
Date Posted: 12:29:47 06/30/03 Mon
>f(x)= xsin(x) is the derivative of this 2cos x -
>xsin(x)?
No, it isn't.

[ Post a Reply to This Message ]
[> Subject: Re: derivative


Author:
Lynn
[Edit]
Date Posted: 13:52:10 07/16/03 Wed
>f(x)= xsin(x) is the derivative of this 2cos x -
>xsin(x)?

No. By the product rule, the derivative of f(x) is xcos(x) + sin(x).

That is, x times (sin x)' + (x)' times sin x.

Incidentally, there is no way to use trig. identities to change one into the other either.

Lynn

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Subject: EASY MONEY READ$$$$$$$$$$$$$$$$$$$$$$$$$$$$READ


Author:
nick
[Edit]
Date Posted: 15:42:36 04/26/03 Sat
quick easy cash
--------------------------------------------------------------------------------
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newsgroups (on-line forums, message boards, chat sites, discussions.)
.......5)Visit these message boards and post this article as a new message by highlighting the
text of this letter and selecting paste from the edit menu. Fill in the Subject, this will be the
header that everyone sees as they scroll through the list of postings in a particular group,
click the post message button.You're done with your first one!Congratulations..THAT'S IT!
All you have to do is jump to different newsgroups and post away, after you get the hang of it,
it will take about 30 seconds for each newsgroup!
**REMEMBER, THE MORE NEWSGROUPS YOU POST IN, THE MORE MONEY YOU WILL MAKE!! BUT YOU HAVE TO POST
A MINIMUM OF 200**
That's it! You will begin receiving money from around the world within days! You may eventually
want to rent a P.O.Box due to the large amount of mail you will receive. If you wish to stay
anonymous, you can invent a name to use, as long as the postman will deliver it.
**JUST MAKE SURE ALL THE ADDRESSES ARE CORRECT.**
**Now the WHY part:
Out of 200 postings, say I receive only 5 replies (a very low example). So then I made $5.00 with
my name at #6 on the letter. Now, each of the 5 persons who just sent me $1.00 make the MINIMUM
200 postings, each with my name at #5 and only 5 persons respond to each of the original 5, that
is another $25.00 for me, now those 25 each make 200 MINIMUM posts with my name at #4 and only 5
replies each, I will bring in an additional $125 ! Now, those 125 persons turn around and post
the MINIMUM 200 with my name at #3 and only receive 5 replies each, I will make an additional
$626 ! OK, now here is the fun part, each of those 625 persons post a MINIMUM 200 letters with
my name at #2 and they each only receive 5 replies, that just made me $3,125 !!!
Those 3,125 persons will all deliver this message to 200 newsgroups with my name at #1 and if
still ONLY 5 persons per 200 newsgroups react I will receive $15,625 ! With an original
investment of only $6.00! AMAZING! When your name is no longer on the list, you just take the
latest posting in the newsgroups, and send out another $6.00 to names on the list, putting your
name at number 6 again. And start posting again.
** By the way, if you try to deceive people by posting the messages with your name in the list
and not sending
the CASH to the rest of the people already on the list, you will NOT get as much.i know someone
who did this and he only made about $150.00, and that's after seven or eight weeks! Then he
sent the 6 $1.00 bills, people added him to their lists, and in 4-5 weeks he had over $5.000.
This is the fairest and most honest way I have ever seen to share the wealth of the world without
costing
anything but our time!!! You also may want to buy mailing and e-mail lists for future CASH.
Make sure you print this article out RIGHT NOW! Also, try to keep a list of everyone that sends
you CASH and
always keep an eye on the newsgroups to make sure everyone is playing fairly. Remember, HONESTY
IS THE
BEST POLICY. You don't need to cheat the basic idea to make the CASH!!
**The thing to remember is: do you realize that thousands of people all over the world are
joining the internet and reading these articles everyday?, JUST LIKE YOU are now!! So, can you
afford $6.00 and see if it really works?? I think so...
Remember, play FAIRLY and HONESTLY and this will really work. GOOD LUCK!!!!

[ Post a Reply to This Message ]
Subject: Area of the sidewalk


Author:
QUITTNER
[Edit]
Date Posted: 12:00:41 11/21/02 Thu
The first and third side of the rectangle's sidewalk have both an area of 30x3 sq.m, and the other 2 sides have both an area of 20x3 sq.m. Then you have to add the total area of the sidewalks at all of the 4 corners, each of them being a square 3x3 sq.m. Can you take it from here? Good luck!

[ Post a Reply to This Message ]
Subject: Math Problem (easy, but I don't know)


Author:
Joe Francis
[Edit]
Date Posted: 19:46:17 11/18/02 Mon
A rectangle lot 30m by 20m is surrounded on all four sides by a concrete walk 3m wide. If you need to concrete only the sidewalk, how much concrete will you need? (surface area)

[ Post a Reply to This Message ]

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