Please Explain!

[krisolaw]

If y varies inversely with the square root of x and y = 15 when x = 9, find y when x = 25.

I do not understand this problem. I know the formula is y=k/x^2 but I am not sure how to approach. This is what I get:

15=k/9^2
15=k/81
1215=k

y=1215/25^2
y=1215/625
y=1.944 is what I get, but the answer is suppose to be 9. HOW? Thanks in advance!!!

 

[stapel]

The exercise says that y varies inversely with the square ROOT of x, but you say that you "know" the equation involves the SQUARE of x. Which is it?

Eliz.

 

[krisolaw]

According to the book the formula is y = k/x^2 but when I plug in the numbers, I do not get the correct answer. Perhaps I need some clarity of the origial problem.

 

[Guest]

Maybe i have it wrong, but.....
if y varies with the inverse squre root of x, why have you used the inverse square of x?

y= k/ ((x)^(1/2))

 

[Denis]

According to the book the formula is y = k/x^2 but when I plug in the numbers, I do not get the correct answer. Perhaps I need some clarity of the origial problem.

Well then, your book is WRONG; should be y = k / sqrt(x) : as apm says

these work like this:
if a = b
then c = b / a * c

with yours:
15 = 1/3
y = 1/5

(1/3) / 15 * y = 1/5
y / 45 = 1/5
5y = 45
y = 9