Question

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

Answer 1

Answer:

K=4

Y=16

X=4

Step-by-step explanation:

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