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:
wergeyhrfed
Step-by-step explanation:
Answer:
x = 7.5
Step-by-step explanation:
4 = 2/3 (2x - 9)
4 = 4/3x - 6
4 + 6 = 4/3x - 6 + 6
10 = 4/3x
10 x 3/4 = x
x = 7.5
Answer:
A = (100 - y)y
Maximum area = 2500 sq. feet.
Step-by-step explanation:
Let the length of the combined rectangular area is y feet and the shared width is x feet.
So, perimeter of the two rectangle together is (3x + 2y) = 200 {Given}
⇒ 2y = 200 - 3x
⇒
...... (1).
a) Now, area of the total plot in sq. feet is
........ (2)
So, this is the expression for area A in terms of length of shared side x.
b) For area to be maximum the condition is 
Now, differentiating equation (2) on both sides with respect to x we get

⇒ x = 33.33 feet.
So, from equation (1) we get 
⇒ x = 50 feet.
So, the value of maximum area = 50 × 33.33 = 1666.5 sq. feet. (Answer)
Answer:
FE= 20
Step-by-step explanation:
…......................
SAS