Question

y varies directly as the square of x. When x = 3, y = 54. Find y when x = 8
1) y = 96
2) y = 144
3) y = 256
4) y = 384

y varies inversely as twice x. When x = 3, y = 8. Find y when x = 2
1) y = 4
2) y = 6
3) y = 9
4) y = 12

y varies jointly with p and q. When p = 7 and q = 2, y = 70. Find y when p = 8 and q = 4
1) y = 120
2) y = 160
3) y = 200
4) y = 240

Answer 1

The first sentence could be rewritten in maths terms as \(y=kx^2\). Your job is to find k when x=3, and y=54.

Answer 2

i know but i dont know how to do that

Answer 3

can anyone teach me how to do this?

Answer 4

i think the first one the anser is 384

Answer 5

\[y=kx^2 \Leftrightarrow k=\frac{y}{x^2} \]
Use any values x,y and find k, then just plug in the equation.

Answer 6

First question:
\[k=\frac{y}{x^2}=\frac{54}{9}=6 \Rightarrow y=6x^2 \Rightarrow 6 \times 8^2=6 \times 64=384\]
For instance. try doing the same with the others