Free thanks and crown thing.
2 answers:
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Answer:
Correct integral, third graph
Step-by-step explanation:
Assuming that your answer was 'tan³(θ)/3 + C,' you have the right integral. We would have to solve for the integral using u-substitution. Let's start.
Given : ∫ tan²(θ)sec²(θ)dθ
Applying u-substitution : u = tan(θ),
=> ∫ u²du
Apply the power rule ' ∫ xᵃdx = x^(a+1)/a+1 ' : u^(2+1)/ 2+1
Substitute back u = tan(θ) : tan^2+1(θ)/2+1
Simplify : 1/3tan³(θ)
Hence the integral ' ∫ tan²(θ)sec²(θ)dθ ' = ' 1/3tan³(θ). ' Your solution was rewritten in a different format, but it was the same answer. Now let's move on to the graphing portion. The attachment represents F(θ). f(θ) is an upward facing parabola, so your graph will be the third one.
4 minutes 34 seconds will takes to empty the tank, if the starts completely full and oil drained at a rate of 2.5 per minute.
Step-by-step explanation:
The given is,
Tank is shaped like a cylinder that is 3 ft long with a diameter of 2.2 ft.
Oil drained at a rate of 2.5 per minute.
Step:1
Time taken to dry the oil tank is,
T = ....................................(1)
Step:2
Volume of the oil is,
.................................................(2)
Where, r - Radius of Cylinder
r = 1.1 ft
From eqn (1),
V = ×
× 3
= 11.40398
Step:3
From equation (1)
=
= 4.56
= 4.56 minutes
T = 4 minutes 34 seconds
Result:
Time taken to dry the oil tank is 4 minutes 34 seconds, if a cylinder is 3 ft long with a diameter of 2.2 ft and oil is drained at a rate of 2.5ft^3 per minute.