Add 3 1/2 and 2 1/3,
Firstly, find the common denominator. It is is 6
3 3/6 + 2 2/6
this gets you 5 5/6.
Take this away from 8:
8 - 5 5/6 = 2 1/6
So 2 1/6 of the board is left over.
We are required to compare the magnitude for two earthquake intensity:
the larger intensity is 7.9
the smaller intensity is 3.7
thus the ratio of the intensity will be:
(larger intensity)/(smaller intensity)
=7.9/3.7
=2.135
~2
Thus we conclude that:
The larger earthquake's intensity was 2 times as great as the smaller earthquake's intensity.
First integral:
Use the rational exponent to represent roots. You have
![\displaystyle \int\sqrt[8]{x^9}\;dx = \int x^{\frac{9}{8}}\;dx](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%5Csqrt%5B8%5D%7Bx%5E9%7D%5C%3Bdx%20%3D%20%5Cint%20x%5E%7B%5Cfrac%7B9%7D%7B8%7D%7D%5C%3Bdx%20)
And from here you can use the rule
![\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}+C](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%20x%5En%5C%3Bdx%3D%5Cdfrac%7Bx%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%2BC%20)
to derive
![\displaystyle \int\sqrt[8]{x^9}\;dx = \dfrac{x^{\frac{17}{8}}}{\frac{17}{8}}=\dfrac{8x^{\frac{17}{8}}}{17}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%5Csqrt%5B8%5D%7Bx%5E9%7D%5C%3Bdx%20%3D%20%5Cdfrac%7Bx%5E%7B%5Cfrac%7B17%7D%7B8%7D%7D%7D%7B%5Cfrac%7B17%7D%7B8%7D%7D%3D%5Cdfrac%7B8x%5E%7B%5Cfrac%7B17%7D%7B8%7D%7D%7D%7B17%7D%20)
Second integral:
Simply split the fraction:
![\dfrac{3+\sqrt{x}+x}{x}=\dfrac{3}{x}+\dfrac{\sqrt{x}}{x}+\dfrac{x}{x}=\dfrac{3}{x}+\dfrac{1}{\sqrt{x}}+1](https://tex.z-dn.net/?f=%20%5Cdfrac%7B3%2B%5Csqrt%7Bx%7D%2Bx%7D%7Bx%7D%3D%5Cdfrac%7B3%7D%7Bx%7D%2B%5Cdfrac%7B%5Csqrt%7Bx%7D%7D%7Bx%7D%2B%5Cdfrac%7Bx%7D%7Bx%7D%3D%5Cdfrac%7B3%7D%7Bx%7D%2B%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%7D%7D%2B1%20)
So, the integral of the sum becomes the sum of three immediate integrals:
![\displaystyle \int \dfrac{3}{x}\;dx = 3\log(|x|)+C](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B3%7D%7Bx%7D%5C%3Bdx%20%3D%203%5Clog%28%7Cx%7C%29%2BC%20)
![\displaystyle \int \dfrac{1}{\sqrt{x}}\;dx = \int x^{-\frac{1}{2}}\;dx = 2\sqrt{x}+C](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%7D%7D%5C%3Bdx%20%3D%20%5Cint%20x%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5C%3Bdx%20%3D%202%5Csqrt%7Bx%7D%2BC%20)
![\displaystyle \int 1\;dx = x+C](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%201%5C%3Bdx%20%3D%20x%2BC%20)
So, the answer is the sum of the three pieces:
![3\log(|x|) + 2\sqrt{x} + x+C](https://tex.z-dn.net/?f=%203%5Clog%28%7Cx%7C%29%20%2B%202%5Csqrt%7Bx%7D%20%2B%20x%2BC%20)
Third integral:
Again, you can split the integral of the sum in the sum of the integrals. The antiderivative of the cosine is the sine, because
. So, you have
![\displaystyle \int \left( \cos(x)+\dfrac{1}{7}x\right)\;dx = \int \cos(x)\;dx + \dfrac{1}{7}\int x\;dx = \sin(x)+\frac{1}{14}x^2+C](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cint%20%5Cleft%28%20%5Ccos%28x%29%2B%5Cdfrac%7B1%7D%7B7%7Dx%5Cright%29%5C%3Bdx%20%3D%20%5Cint%20%5Ccos%28x%29%5C%3Bdx%20%2B%20%5Cdfrac%7B1%7D%7B7%7D%5Cint%20x%5C%3Bdx%20%3D%20%5Csin%28x%29%2B%5Cfrac%7B1%7D%7B14%7Dx%5E2%2BC%20)
Answer:
524cm^2
Step-by-step explanation:
Formula for Volume of sphere= 4/3 πr^2
We have,
r=5cm
Now,
Volume(v)=4/3 πr^2 = 4/3π 5^3= 4/3π 125 = 166.666666667π = 523.598775599
Rounding to the nearest tenth,
Volume=524cm^2
Answer:
can you list the following so I can tell you?
Step-by-step explanation: