It is 5/3 !! 8^(5/3)=....32!
The required equation is y = -9
Step-by-step explanation:
Step 1 :
Given the line l is perpendicular to the y axis
The equation of the y axis is x = 0
So any line perpendicular to the y axis will have equation as y = k , where k is a constant value for any value of x
Step 2 :
Its given that the line is passing through the point (0.-9). Here the y co ordinate is -9. Hence the perpendicular line has a constant y co ordinate of y = -9 for any value of x
So the required equation is y = -9
Step 3 :
Answer :
The required equation is y = -9
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 awnswer is a = 7 + c
Step-by-step explanation:
First, set the equation in exponential form.
x=logbY to b^x=Y
x=log44. so x=x, b=10, Y=44
10^x=44
I know that 10^1=10, 10^2=100, and 10^1.5=31.6 so x should be between 1.5 and 2.
The answer is x=1.64.