The answer would be B. because a unit rate is usually equal to one for the sake of conversion
Step-by-step explanation:
Given:
and ![f(0) = 0](https://tex.z-dn.net/?f=f%280%29%20%3D%200)
We can solve for f(x) by writing
![\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cint%20f%27%28x%29dx%3D%5Cint%20x%5E2e%5E%7B2x%5E3%7Ddx)
Let ![u = 2x^3](https://tex.z-dn.net/?f=u%20%3D%202x%5E3)
![\:\:\:\:du=6x^2dx](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3Adu%3D6x%5E2dx)
Then
![\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cint%20x%5E2e%5E%7B2x%5E3%7Ddx%20%3D%20%5Cdfrac%7B1%7D%7B6%7D%5Cint%20e%5Eu%20du)
![\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%3D%5Cfrac%7B1%7D%7B6%7De%5E%7B2x%5E3%7D%20%2B%20k)
We know that f(0) = 0 so we can find the value for k:
![f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}](https://tex.z-dn.net/?f=f%280%29%20%3D%20%5Cfrac%7B1%7D%7B6%7D%281%29%20%2B%20k%20%5CRightarrow%20k%20%3D%20-%5Cfrac%7B1%7D%7B6%7D)
Therefore,
![\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%5Cleft%28e%5E%7B2x%5E3%7D%20-%201%20%5Cright%29)
The answer is A. or B. because it states that both the central and inscribed angle have the same endpoints.
Answer:
13. m∠MN + m∠NMP = m∠MP.
14. m∠NMP - m∠OMP = m∠NMO.
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15. x = 3 + y and -2y + 3 < -1 : Given
16. x > y : x = 3 + y
17. -2y < -4 : -2y < -4
18. y > 2 : y > 2
19. x > 2 : x = 3 + y and y > 2
<u>------------------------------------------</u>
Answer:
D. -2a∧2-5a+10
Step-by-step explanation: