Answer:
2
Step-by-step explanation:
You add 2 subtract 3
Answer:
Denote the amount of miles she will run in 24 minutes: x
=> x/24 = 7/60 (same rate)
=> x = 7x24/60 = 2.8 (miles)
Since
, we can rewrite the integral as
![\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1f%28t%29%5C%3Bdt%20%2B%20%5Cint_1%5E4%20f%28t%29%5C%3B%20dt)
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
![\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1f%28t%29%5C%3Bdt%20%3D%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%2C%5Cquad%20%5Cint_1%5E4%20f%28t%29%5C%3B%20dt%20%3D%20%5Cint_1%5E4%202t%5C%3B%20dt)
Both integrals are quite immediate: you only need to use the power rule
![\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20x%5En%5C%3Bdx%3D%5Cdfrac%7Bx%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D)
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
No, a closed circle does include that number.
Closed circles are used for greater than or equal to or less than or equal to
Open circles are only used for greater than or less than
Answer:
Step-by-step explanation:
the answer is 3/7