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.
A would be equal to 1 since a is equal to one already
Answer:
20 miles with an error margin of ± 8 miles
Step-by-step explanation:
The margin of error of a result is the range in which an error can vary. To find the margin of error between both distances we have to
28-12 = 16, that is, the variation of the result has a range of 16 miles. So we will look for the midpoint of both distances
(X2-X1)/2+X1=(28-12)/2+12=16/2+12=8+12=20
So from this midpoint the value can vary between 8 points below and 8 points above that would cover the difference of 16 miles that we observed at the beginning
In this way, the correct answer is 20 miles with an error margin of ± 8 miles
Done
Answer:
n=36
Step-by-step explanation:
36 divided by 2 is 16 then subtract 10 and you will get 6 hope this helps
