If p(x)=2(x^2+1)+16, what is the value of p(12)?

Question

asked Aug 20, 2021 32.0k views

2 Answers

Tim Raynor · Answer 1 · 2021-08-21T11:48:17+0000

Answer: P(12)=306

Explanation:

Replace the ‘x’ with 12 in the equation. 2(12^2+1)+16

2•144+1+16 (distribute the 2)

288+2+16

306

Steve McLeod · Answer 2 · 2021-08-24T19:52:14+0000

Explanation:

Hey there!

Given;

p(x)= 2(x^2+1)+16

And p(x)= 12

Now,

Putting value of p(12) in p(x)= 2(x^2+1)+16.

$p(12) = 2( {12}^(2) + 1) + 16$

p(12) = 290 + 16

Therefore p(12)= 306.

Hope it helps...