Question

How many questions I have to ask?​

Answer 1

Answer:

p is 16

Step-by-step explanation:

$p \: \alpha {(q + 2)}^{2}$

introduce a constant on a right hand side:

$p = k {(q + 2)}^{2} - - - eqn \: (a)$

[ when q is 1, also p is 1 ]

$1 = k(1 + 2) {}^{2} \\ 1 = 9k \\ k = \frac{1}{9}$

substitute for k in eqn (a):

$p = \frac{1}{9} (q + 2) {}^{2}$

so, when q is 10:

$p = \frac{1}{9} (10 + 2) {}^{2} \\ \\ p = \frac{1}{9} (144) \\ \\ p = 16$