Question

z varies directly with x and inversely with y2. When x = 2 and y = 5, z = 8.

What is the value of z when x = 4 and y = 9?

Answer 1

\[z=\frac{kx}{y^2}\]

Answer 2

Plug in the given values and find the constant, k.

Answer 3

What did you get for k?

Answer 4

Im looking at my notes but I cant figure out which number goes where? @Mertsj

Answer 5

Why do you have to look at your notes? Why don't you read the problem? It clearly says x=2. That means replace x with 2

Answer 6

Then it says y = 5. That means replace y with 5

Answer 7

Then it says z = 8. That means replace z with 8

Answer 8

ok, so I plugged in the numbers, 8=2k/5, and I made 2/5 a reciprocal and multiplied to 8 so it would even out right? and I got k= 1525.87, then I plugged that number in for the other side of the equation, z=1525.87(4)/9 and I got 678.16, which is not even close to the options I have.

Answer 9

\[z = k\frac{x}{y^2}\]\[8= k\frac{2}{5^2}\]\[25*8 = 2k\]\[k=100\]
You failed to square \(y\) in the denominator, plus some other mistake I'm not going to try to figure out.

Answer 10

lol^,I got it already, thanks

Answer 11

if you already have the answer, close the question so no one wastes time on it...