The purifier filters 1/3 water
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
12(x-5)=45 This is the answer I hope so
Assuming the question marks are minus signs
to find max, take derivitive and test 0's and endpoints
take derivitive
f'(x)=18x²-18x-108
it equal 0 at x=-2 and 3
if we make a sign chart to find the change of signs
the sign changes from (+) to (-) at x=-2 and from (-) to (+) at x=3
so a reletive max at x=-2 and a reletive min at x=3
test entpoints
f(-3)=83
f(-2)=134
f(3)=-241
f(4)=-190
the min is at x=3 and max is at x=-2
Answer:
Step-by-step explanation:
We need to find the value of
Solving:
We know,
a^0 = 1
so,
So, the value of