Four hundred twenty-three billion, ninety million, seven hundred nine thousand
Hope this helped:))
Question: <span>70% of 120 = ?
Work:
To find what </span><span>70% of 120 is. Do .70 times 120.
.70 x 120 = 84
Final Answer:
84
</span>
Answer:
Yes
Step-by-step explanation:
If you add 87.84, 85, and 100 you get 272.84. Once you divide that by 3 you get 90.946, which would be your test average if you score 100 on your next test.
Answer:
105 km
Step-by-step explanation:
3.75 x 28 = 105 km
Computing an integral by substitution is the reverse of the chain rule for computing the derivative. Substitution is intended to rewrite a complicated-looking integral involving the derivative of some component expression as another much simpler integral. For example, if
, then
and
![\displaystyle\int f(x)f'(x)\,\mathrm dx=\int u\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%20f%28x%29f%27%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint%20u%5C%2C%5Cmathrm%20du)
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Integration by parts is the reverse of the product rule for derivatives:
![(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\implies f(x)g'(x)=(f(x)g(x))'-f'(x)g(x)](https://tex.z-dn.net/?f=%28f%28x%29g%28x%29%29%27%3Df%27%28x%29g%28x%29%2Bf%28x%29g%27%28x%29%5Cimplies%20f%28x%29g%27%28x%29%3D%28f%28x%29g%28x%29%29%27-f%27%28x%29g%28x%29)
Integrating both sides with respect to
gives
![\displaystyle\int f(x)g'(x)\,\mathrm dx=\int(f(x)g(x))'\,\mathrm dx-\int f'(x)g(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%20f%28x%29g%27%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint%28f%28x%29g%28x%29%29%27%5C%2C%5Cmathrm%20dx-%5Cint%20f%27%28x%29g%28x%29%5C%2C%5Cmathrm%20dx)
![\displaystyle\int f(x)g'(x)\,\mathrm dx=f(x)g(x)-\int f'(x)g(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%20f%28x%29g%27%28x%29%5C%2C%5Cmathrm%20dx%3Df%28x%29g%28x%29-%5Cint%20f%27%28x%29g%28x%29%5C%2C%5Cmathrm%20dx)
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Personally, I think the best way to grasp the idea behind the two methods is to practice. You start to notice patterns to the point where knowing which is the "right" method to use becomes second nature.