Question

The value of y varies directly with x2, and y = 64 when x = 4.



What is the value of y when x = 6?

Answer 1

if that x2 stands for \(x^2\), then:
\[y=Ax^2\]
where \(A\) is some constant, if you plug \(y=64\) for \(x=4\) you can find the value of \(A\) and then plug \(x=6\) and see what you get for \(y\)...

Answer 2

6^2 would make it 36 right?

Answer 3

@Greg_D

Answer 4

that woul be the case \(A=1\)...
you can et the value for \(A\) from:
\[64=A\times 4^2\]
find that and then calculate \(y=A\times 6^2\)

Answer 5

what is a

Answer 6

the problem says:
"The value of y varies directly with x^2, and y = 64 when x = 4"
so y must equal "something" times x^2, i called that something A

Answer 7

im sorry im the worst XD

Answer 8

576

Answer 9

no problem, just keep asking until you understand !
what did you get for A ?

Answer 10

well 64/4 is 16 so i multiplied 36 by 16 to get 576

Answer 11

note that \(A=\frac{64}{4^2}\) dont forget the square!

Answer 12

oh yeah

Answer 13

i dont remember how to do that

Answer 14

well \(4^2=16\) so \(A=\frac{64}{16}=4\)

Answer 15

so 4 times 36 ?

Answer 16

for 144?

Answer 17

:)
that is what i get... good work!