Question

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Answer 1

15. $\boxed{\sf{Pythagorean\:Theorem}}$

$\sf{(\sqrt{5})^2=1^2+2^2}$

Therefore √5 is a hypotenuse of a triangle having 1 and 2 as side lengths.

16. $\boxed{\sf{Splitting\:the\:middle\:Term}}$

(1) $\sf{6x^2+17x+5}$

$\sf{=6x^2+15x+2x+5}$

$\sf{=3x(2x+5)+(2x+5)}$

$\sf{=(3x+1)(2x+5)}$ (ANSWER)

(2) $\sf{15t^2-2t-1}$

$\sf{=15t^2+3t-5t-1}$

$\sf{=3t(5t+1)-(5t+1)}$

$\sf{=(3t-1)(5t+1)}$ (ANSWER)

Answer 2

Step-by-step explanation:

this will be the answer