Please Explain!

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!!!

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.

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.

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))

krisolaw said:

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.

Click to expand...

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