when it comes to checking if a function is even or odd, it boils down to changing the argument, namely x = -x, and if the <u>resulting function is the same as the original</u>, then is even, if the <u>resulting function is the same as the original but negative</u>, is odd, if neither, well then neither :).
anyway, that said, let's first expand it and then plug in -x,
![\bf f(x)=(x^2-8)^2\implies f(x)=(x^2-8)(x^2-8)\implies f(x)=\stackrel{FOIL}{x^4-16x^2+64}\\\\[-0.35em]~\dotfill\\\\f(-x)=(-x)^4-16(-x)^2+64\qquad \begin{cases}(-x)(-x)(-x)(-x)=x^4\\(-x)(-x)=x^2\end{cases}\\\\\\f(-x)=x^4-16x^2+64\impliedby \stackrel{\textit{same as the original}}{Even}](https://tex.z-dn.net/?f=%20%5Cbf%20f%28x%29%3D%28x%5E2-8%29%5E2%5Cimplies%20f%28x%29%3D%28x%5E2-8%29%28x%5E2-8%29%5Cimplies%20f%28x%29%3D%5Cstackrel%7BFOIL%7D%7Bx%5E4-16x%5E2%2B64%7D%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5Cf%28-x%29%3D%28-x%29%5E4-16%28-x%29%5E2%2B64%5Cqquad%20%5Cbegin%7Bcases%7D%28-x%29%28-x%29%28-x%29%28-x%29%3Dx%5E4%5C%5C%28-x%29%28-x%29%3Dx%5E2%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5Cf%28-x%29%3Dx%5E4-16x%5E2%2B64%5Cimpliedby%20%5Cstackrel%7B%5Ctextit%7Bsame%20as%20the%20original%7D%7D%7BEven%7D%20)
Answer:
B) (3, –2)
Explanation:
The inequality is y ≤ –x + 1
There are two ways to do this. You can try the four options by seeing where they lie on the graph, or by inputting them into the inequality and seeing if they check out. I am going to do a bit of both.
I know that the solution cannot have two positive coordinates because the first quadrant is not part of the solution, so I won't guess A or C.
I'll try (3, –2) (which is option B).
On the graph, (3, –2) is on the line, which means it is part of the solution because the line is solid and the inequality is a greater than or equal to sign.
Try it in the inequality:
y ≤ –x + 1
–2 ≤ –3 + 1
–2 ≤ –2 yes this checks out.
Answer:
yes
Step-by-step explanation:
congurency property can be used for all right angles triangles
that's all ....have fun
The two factors are 2 and p=10 so I used your equations and
