Plug in -2 for a, 3 for b, and -5 for c. Then it'd be | (-3)^2 - 2(-2)(-5) +5 (3) |. Then, multiply everything to get | 9 -20 +15 |. Then, add/subtract to get | 4 |. Finally, take the absolute value and get 4.
Answer: Choice A -- f(x) = (1/3)*|x|
============================================
The input of a function is x and the output is y = f(x)
To vertically compress a function, we will multiply the y value by some fraction smaller than 1. This is so that the y coordinates are a fraction of what they once were.
In this case, we multiply y by 1/3 so that something that has a y coordinate of y = 81 becomes y = 27 (divide by 3)
So we have y = f(x) become (1/3)*y = (1/3)*f(x) = (1/3)*|x|
Answer:
he must burn 294.6
Step-by-step explanation:
Just minus 305.5 from 600
600
305.5
is equal to 294.6
Answer:
a) So, this integral is convergent.
b) So, this integral is divergent.
c) So, this integral is divergent.
Step-by-step explanation:
We calculate the next integrals:
a)
![\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D%5Cleft%5B-%5Cfrac%7Be%5E%7B-2x%7D%7D%7B2%7D%5Cright%5D_1%5E%7B%5Cinfty%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D-%5Cfrac%7Be%5E%7B-%5Cinfty%7D%7D%7B2%7D%2B%5Cfrac%7Be%5E%7B-2%7D%7D%7B2%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D%5Cfrac%7Be%5E%7B-2%7D%7D%7B2%7D%5C%5C)
So, this integral is convergent.
b)
![\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D%5Cleft%5B-%5Cfrac%7B1%7D%7Bz-1%7D%5Cright%5D_1%5E2%5C%5C%5C%5C%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D-%5Cfrac%7B1%7D%7B1-1%7D%2B%5Cfrac%7B1%7D%7B2-1%7D%5C%5C%5C%5C%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D-%5Cinfty%5C%5C)
So, this integral is divergent.
c)
![\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D%5Cleft%5B2%5Csqrt%7Bx%7D%5Cright%5D_1%5E%7B%5Cinfty%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D2%5Csqrt%7B%5Cinfty%7D-2%5Csqrt%7B1%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D%5Cinfty%5C%5C)
So, this integral is divergent.
Answer:
the motion of the object will stay the same
Step-by-step explanation:
