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.
Answer: Point of intersection (0, -2)
Step-by-step explanation:
Answer:
31.4 in³
Step-by-step explanation:
The box is just big enough to hold the 3 balls, so it must have a length 6 times the radius of each ball, a width 2 times the radius, and a height 2 times the radius.
The volume of the box is:
V = (6r)(2r)(2r)
V = 24r³
The volume of the 3 balls is:
V = 3 (4/3 π r³)
V = 4πr³
So the volume of the air is:
V = 24r³ − 4πr³
V = (24 − 4π) r³
Since r = 1.4 inches:
V = (24 − 4π) (1.4 in)³
V ≈ 31.4 in³
Answer:
Y= -5.5 There is no X intercept.
Step-by-step explanation:
Please mark brainliest.
Answer:
33x2=66
66-3=62
62x4=248
Is this a question for a test cause if it is i'm gonna feel embarrassed