It is neither.
To be even, f(x) must equal f(-x).
If you substitute -x for x, you'd get
y = (-x)^2 - 2(-x) -8
y = x^2 +2x -8
This is not the same as the original, so this is not even.
To be odd, f(x) must equal -f(-x).
If you take the -x substitution from the last step and then multiply it by -1, you'd have:
y = -1 (x^2 +2x -8)
y = -x^2 -2x +8
This is not the same as the original either.
The function is neither even nor odd.
<span>sqrt(pow(length, 2) + pow(width, 2))
</span>
Well, I bet you want your answer right away! So here it is.
<span>Given <span>f (x) = 3x + 2</span> and <span>g(x) = 4 – 5x</span>, find <span>(f + g)(x), (f – g)(x), (f × g)(x)</span>, and <span>(f / g)(x)</span>.</span>
To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
<span>\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}<span><span>(<span><span>g</span><span>f</span><span></span></span>)</span>(x)=<span><span><span>g(x)</span></span><span><span>f(x)</span></span><span></span></span></span></span><span>= \small{\dfrac{3x+2}{4-5x}}<span>=<span><span><span>4−5x</span></span><span><span>3x+2</span></span><span></span></span></span></span>
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
<span>\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}<span><span>(<span><span>g</span><span>f</span><span></span></span>)</span>(x)=<span><span><span>4−5x</span></span><span><span>3x+2</span></span><span>
Hope I helped! :) If I did not help that's okay.
-Duolingo
</span></span></span></span>
Answer:
Line BD is Congruent to Line BD due to the Reflexive property. Angle Bad is congruent to angle BCD because of the third angle theorem therefore triangle ABD is congruent to triangle CBD because all angles and sides are congruent
I don't know if that will satisfy your teacher because some prefer different things but that leaves no ground to be inferred and explains everything word for word.
