Yes I think it is because
First solve x³=9x

We can ignore x = -3 because it is not in the first quadrant.
<span>So our integral is going to go from x=0 to x=3.
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<span>Now we can use the formula
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![\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\pi\int\limits_{0}^{3}(81x^2-x^6)dx \\2\times \int\limits_{0}^{3}[\pi(9x)^2-\pi(x^3)^2]dx \\\pi\left[81\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{3} \\ \\\frac{2916\pi}{7}](https://tex.z-dn.net/?f=%5Cint%5Climits_%7B0%7D%5E%7B3%7D%28%5Cpi%20%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%29dx%0A%5C%5C%5Cint%5Climits_%7B0%7D%5E%7B3%7D%28%5Cpi%20%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%29dx%0A%5C%5C%5Cpi%5Cint%5Climits_%7B0%7D%5E%7B3%7D%2881x%5E2-x%5E6%29dx%0A%5C%5C2%5Ctimes%20%5Cint%5Climits_%7B0%7D%5E%7B3%7D%5B%5Cpi%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%5Ddx%0A%5C%5C%5Cpi%5Cleft%5B81%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Cfrac%7Bx%5E7%7D%7B7%7D%5Cright%5D_%7B0%7D%5E%7B3%7D%20%5C%5C%20%5C%5C%5Cfrac%7B2916%5Cpi%7D%7B7%7D)
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Answer:
m∠AVF = 158°
Step-by-step explanation:
In the diagram attached,
m∠AVD = m∠BVE = m∠CVF = 90°
m∠CVD = 22°
∠AVB = ∠EVF
Since ∠AVD = ∠AVC + ∠CVD
Therefore, 90° = m∠AVC + 22°
m∠AVC = 90° - 22°
= 68°
Similarly, m∠CVF = m∠CVD + m∠DVF
90° = 22° + m∠DVF
m∠DVF = 68°
m∠AVC = m∠DVF = 68°
Now ∠AVF = ∠AVD + ∠DVF
= 90° + 68°
= 158°
3 times a # (# isn't specified, so you should use a variable): 3x
result is 5: =5
Put it all together: 3x=5
Answer:
The seth has to be worked for 10.89 hours for earn the rest of the money.
Step-by-step explanation:
Given data
Cost of the skateboard = $ 167
Amount that seth has in the bank = $ 88
Rest of the amount needed = 167 - 88 = $ 79
Earning of the seth per hour = $ 7.25
No. of hours to be worked by seth for earning rest of the $ 79 amount is
10.89 hours.
Therefore the seth has to be worked for 10.89 hours for earn the rest of the money.