Question

The value of y varies directly with x^2, and y = 64 when x = 4.

What is the value of y when x = 6?

Answer 1

y=kx^2
y=k(6 * 6)
y=k(36)
k=36 I believe
now i took 36 * 6 to find y and it came out as 216. To test that i did 216 divided by 6 and it came out 36.
Would y be 216?

Answer 2

@YoungStudier can you help?

Answer 3

@rebeccaxhawaii

Answer 4

I believe itd be 216 cause when doing the problem they gave y=64 x=4. 4*4 = 16 and 64 divided by 4 was 16... .3.

Answer 5

Take a look at this

\[y = kx^{2}\]

\[\frac{ y }{ x^{2} } = k \]

\[\frac{ y }{ x^{2} } = \frac{ 64 }{ 4^{2} } = 4 = k \]

\[y = 4*x^{2}\]

Answer 6

@VesperSaviour you need to find what k is first

Answer 7

4^2 is 16 which should be K @Photon336

Answer 8

well, you know what y is and you know what x is right?

so you need to solve for k

\[\frac{ y }{x^{2} } = k \]

Answer 9

if its 64 divided by 4^2 (16) then its 4...

Answer 10

Yes, make sure you re-arrange the formula for the unknown variable FIRST. then plug in the numbers.

Answer 11

For the onewe gotta figure out...
y=kx^2
x=6 so 6^2 = 36....from what you are saying we take the y divide it by the x for the k but we need to find y...i took 216 divided it by 36 and that came out as 6....i still think 216.. @Photon336

Answer 12

okay
k =4
and
x = 6
right?

Answer 13

\[4*(6)^{2} = y \]

Answer 14

144?

Answer 15

so for the y i have to figure out its 144? @Photon336

Answer 16

yep that's your y value

Answer 17

thank you