Answer:
e^ (5 ln (x + 1)) = (x + 1)^5
Step-by-step explanation:
e5 In (x + 1) or did you mean e^ (5 ln (x + 1))
because then this would simplify a lot
e^ (5 ln (x + 1)) = e ^ (ln (x + 1)^5)
e^ (5 ln (x + 1)) = e ^ (ln (x + 1)^5) = (x + 1)^5
or did you mean (e^5) ( ln (x + 1)) = ln [(x+1)^(e^5)]
But I think you meant:
e^ (5 ln (x + 1)) = e ^ (ln (x + 1)^5) = (x + 1)^5
The length is (3/5 -1/5)=2/5
the height is also 2/5
area: length * height =2/5 *2/5=4/5
D is correct.
Answer:
in this form, the "-r" would cause the result to decrease
I assume that the answer is "r"
if r = .1 (10 %) then 1-.1 = .9
if you have 100 items then y = 100(.9)^1 = 90
the total decreased by 10% ... y = 100(.9)^2 after 2 time periods
Step-by-step explanation:
the answer is 9.74972315 I hope you have the right answer I hope this is helpful
Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) = 
= 
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4