Answer:
I think the answer is c -2
Step-by-step explanation:
but don't quote me on that and if it's wrong i truly am sorry
Answer:
ok so first we need to find the area of the circle inside which is a=Pi*raduis^2
a=pi*1(radius is half of diamater)
a=3.14
ok then the area of the sqare including the circle is 16 so
16-3.14=12.86
this is is aprox since the area of pi is infinite and i just used 3.14
Hope This Helps!!!
9514 1404 393
Answer:
15. f(x) +g(x) = 7x +2
16. f(x) -g(x) = -3x +8
17. h(x) · g(x) = 35x -21
Step-by-step explanation:
Substitute the function definition for each function expression and simplify.
__
15. f(x) + g(x) = (2x +5) +(5x -3)
f(x) +g(x) = 7x +2
__
16. f(x) - g(x) = (2x +5) -(5x -3)
f(x) -g(x) = -3x +8
__
17. h(x) · g(x) = (7) · (5x -3)
h(x) · g(x) = 35x -21
Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
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:
Both integrals are quite immediate: you only need to use the power rule
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.
