Answer: 1/30
Step-by-step explanation:
∫[0,4] arcsin(x/4) dx = 2π-4
x = 4sin(u)
arcsin(x/4) = arcsin(sin(u)) = u
dx = 4cos(u) du
∫[0,4] 4u cos(u) du
∫[0,4] f(x) dx = ∫[0,π/2] g(u) du
v = ∫[1,e] π(R^2-r^2) dx
where R=2 and r=lnx+1
v = ∫[1,e] π(4-(lnx + 1)^2) dx
Using shells dy
v = ∫[0,1] 2πrh dy
where r = y+1 and h=x-1=e^y-1
v = ∫[0,1] 2π(y+1)(e^y-1) dy
v = ∫[0,1] (x-x^2)^2 dx = 1/30
Answer:
10000000000000000000000,
the next number: 1-934028049747284
that's all
pls report
thanks!))))
Step-by-step explanation:
f(x) = x + 3
when x = 0, f(x) = 3
when x = 1, f(x) = 1 + 3 = 4
when x = 2, f(x) = 2 + 3 = 5
when x = 3, f(x) = 3 + 3 = 6
answer is D
f(x) = x + 3
Answer is C! Hope it helps
Alright, here goes:
For this question, you are asked about the median and Interquartile Range(IQR), so first you need to understand what that means.
The IQR is the box in the middle of the box-and-whisker-plot. The question asks for the place to have a small IQR, so the most likely places are Pensacola and Jacksonville.
The question is also asking for a high median temperature. The median is the line in the middle of the plot.
The median of Pensacola's plot is higher than the median of Jacksonville, so Pensacola Beach is your answer :)
