Answer:
p is 16
Step-by-step explanation:
[tex]p \: \alpha {(q + 2)}^{2} [/tex]
introduce a constant on a right hand side:
[tex]p = k {(q + 2)}^{2} - - - eqn \: (a)[/tex]
[ when q is 1, also p is 1 ]
[tex]1 = k(1 + 2) {}^{2} \\ 1 = 9k \\ k = \frac{1}{9} [/tex]
substitute for k in eqn (a):
[tex]p = \frac{1}{9} (q + 2) {}^{2} [/tex]
so, when q is 10:
[tex]p = \frac{1}{9} (10 + 2) {}^{2} \\ \\ p = \frac{1}{9} (144) \\ \\ p = 16[/tex]
