Question

13. Given that y varies as x^2 and that y = 36 when x=3, find:

a) k
b) the value of y when x=2
c) the value of x when y= 64​

Answer 1

K=4

Y=16

X=4

Explanation:

`y = kx {}^(2) \\ finding \: k \: when \: y= 36 \: and \: x = 3 \\ 36 = k(3) {}^(2) \\ 36 = 9k \\ dividing \: through \: by \: 9 \\ (36)/(9) = (9k)/(9) \\ 4 = k \\ k = 4 \\ findnig \: y \: when \: x = 2 \\ y = 4(2) {}^(2) \\ y = 4 * 4 = 16 \\ finding \: x \: when \: y = 64 \\ 64 = 4 * x \\ 64 = 4x \\ dividing \: through \: by \: 4 \\ (64)/(4) = \frac{4x {}^(2) }{4} \\ 16 = x {}^(2) \\ square \: root \: bothsides \\ √(16) = x {}^(2) \\ 4 = x \\ x = 4`