Question

please please answer :​

Answer 1

Answer:

ajskskslsossoeokekemdmdmdkdkkdkdkdidodod

Answer 2

Step-by-step explanation:

$let \: \frac{x}{y} = {t}^{2} \\ now \: \: t + \frac{1}{t} = \frac{10}{3} \\ \frac{ {t}^{2} + 1 }{t} = \frac{10}{3} \\ 3 {t}^{2} + 3 = 10t \\ 3 {t}^{2} - 10t + 3 = 0 \\ 3 {t}^{2} - 9t - t + 3 = 0 \\ 3t(t - 3) - 1(t - 3) = 0 \\ (t - 3)(3t - 1) = 0 \\ t = 3 \: \: or \: \: \frac{1}{3} \\ i.e \: \: \: \frac{x}{y} = {t}^{2} = 9 \: \: or \: \: \frac{1}{9}$

but with these information, we can't find xy.