Question

y varies directly as the square of x. When x = 5, y = 5. Find y when x = 10.

Answer 1

PLEASE HELP ME

Answer 2

First, try to find 'k'
\(\large
y=kx^2\)

Answer 3

did you get 20?

Answer 4

yes? no?

Answer 5

Let's substitute x and y into our
\(\large \begin{align}

y&=kx^2\\
5&=k*5^2 \\
5 &= 25k\\
k& =5\div 25\\
k&=\frac{1}{5}
\end{align}\)

Answer 6

y = 10

y = 20

y = 50

y = 100

are my choices

Answer 7

im pretty sure im right

Answer 8

Okay, now you found k, Find y when x = 10.

\(\large y=\frac{1}{5}10^2\)
y=20

Yeah, you are right :)