Answer is B hope this helps
I'm assuming you mean
, not
, like your prompt suggests.
First, let's figure out what rule we can use. A likely noticeable one is the Power Rule, which says the following:
![\dfrac{d}{dx} [u^a] = a(u)^{a-1} du](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdx%7D%20%5Bu%5Ea%5D%20%3D%20a%28u%29%5E%7Ba-1%7D%20du)
Applying this, we can solve for the derivative:

While you can simplify the expression to your liking, I believe that this form is not overly complex and will thus leave it as is.
Thus, our answer is:

Answer:
BADC
Step-by-step explanation:
-4x + 7
27 Subtract 7
-4x
20 Divide both sides by -4
x
-5 B
3x - 10 < -25 Add 10
3x < -15 Divide by 3
x < -5 A
5(x - 2) > -15 Use distributive property
5x - 10 > -15 Add 10
5x > -5 Divide by 5
x > -1 D
x - 9 < -10 Add 9
x < -1 C
Answer:
well I don't know if there is a typo but the factors of 4 are going to be 1, 2, and 4 so they could possibly be multiplied by any of those numbers
Hope this helped : )