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Mathematics

1 answer:

Klio2033 [76]3 years ago

7 0

Check the picture below

if that red segment, GJ, is parallel to the AE base segment of the triangle, then, the segment GJ is the midsegment of the triangle, and by the side-splitter theorem, those two triangles are similar.

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What are the correct answers and why? Simple and concise explanation please!!!

dezoksy [38]

Answer:

$\large\boxed{-\bigg[(x-3)^2+2x\bigg]+1}\\\\\boxed{-x^2+4x-8}$

Step-by-step explanation:

$(g\ \circ\ f)(x)=g\bigg(f(x)\bigg)$

$f(x)=(x-3)^2+2x\\\\g(x)=-x+1\\\\g\ \circ\ f\to\text{put f(x) instead of x in the function g(x)}:\\\\(g\ \circ\ f)(x)=-\bigg[\underbrace{(x-3)^2+2x}_{x}\bigg]+1$

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$-\bigg[(x-3)^2+2x\bigg]+1=-(x-3)^2-2x+1\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=-(x^2-(2)(x)(3)+3^2)-2x+1=-(x^2-6x+9)-2x+1\\\\=-x^2-(-6x)-9-2x+1=-x^2+6x-9-2x+1\\\\\text{combine like terms}\\\\=-x^2+(6x-2x)+(-9+1)=-x^2+4x-8$