English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

suppose "y" is directly proportional to the second power of "x". if "x" is halved, what happens to "y"? considering that the constant proportionality is positive. thanks for the help.

2006-10-29 12:56:19 · 5 answers · asked by shih rips 6 in Science & Mathematics Mathematics

5 answers

Hi,

So if y is proportional to x^2

y= ax^2 where a is some constant of proportionality

so if we change x to x/2, the new relationship (y') is

y'= a(x/2)^2 = (ax^2)/4

so y' is 1/4 as large as y.

Hope that helps,
Matt

2006-10-29 13:06:39 · answer #1 · answered by Matt 3 · 0 0

Your verbal statement turns into the following:

y = kx^2

If you replace x with (1/2x) you get

y' = k * 1/4 x^2

So, the new y' is 1/4 of the old y. Thus, halving x multiplies y by 1/4.

2006-10-29 21:02:18 · answer #2 · answered by tbolling2 4 · 0 0

If y prop to x^2, then apply the 1/2 to x --> (x/2)^2 = (1/4)x^2.

To preserve the proportionality:

Apply the 1/4 to both sides, meaning you apply as follows:
y --> (1/4)y, so y/4 is prop to (x/2)^2.

2006-10-29 21:01:39 · answer #3 · answered by Action 4 · 0 0

I answered one of your questions tonight. I've noticed that you are having trouble understanding Direct Variation.

I found a site that will help to clear things up for you.

Go to:

http://regentsprep.org/Regents/Math/math-topic.cfm?TopicCode=variation

Guido

2006-10-29 21:06:34 · answer #4 · answered by Anonymous · 0 0

well, y= x^2

so y will be made less by 1/4

2006-10-29 21:01:53 · answer #5 · answered by sjalt 1 · 0 0

fedest.com, questions and answers