Answer:
-x-3
Step-by-step explanation:
(x+3) + the additive inverse = 0
the additive inverse = -(x+3)
5.a) Steve - 60=5×12
= <u>5×3×2×2</u>
5.b) Ian - 60=6×10
= <u>3×2×5×2.</u>
<u>Discussion</u> - The prime factors reduce to the same numbers in both Steve's and Ian's case.
5.c) Case 1 of 48 = 6×8
= <u>3×2×2×2×2.</u>
Case 2 of 48 = 12×4
= <u>2×2×3×2×2.</u>
If I am reading the question correctly it would come out to -6x+42 First do all the addition and squaring, then factor in that negative. You come out to -2x^3 +2x^3-4x-2x+25+5+9+3. The x^3's cancel, the -x's add to -6x, then all that other adds to 42.
Answer:
-3
Step-by-step explanation:
3 x 6 - 42/2 =
18 - 42/2 =
18 - 21 =
-3
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.