83-63=20
So your answer is 20
Simplify the integrands by polynomial division.
![\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7Bt%5E2%7D%7B1%20-%203t%7D%20%3D%20-%5Cdfrac19%20%5Cleft%283t%20%2B%201%20-%20%5Cdfrac1%7B1%20-%203t%7D%5Cright%29)
![\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)](https://tex.z-dn.net/?f=%5Cdfrac%20t%7B1%20%2B%204t%7D%20%3D%20%5Cdfrac14%20%5Cleft%281%20-%20%5Cdfrac1%7B1%20%2B%204t%7D%5Cright%29)
Now computing the integrals is trivial.
5.
![\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bt%5E2%7D%7B1%20-%203t%7D%20%5C%2C%20dt%20%3D%20-%5Cfrac19%20%5Cint%20%5Cleft%283t%20%2B%201%20-%20%5Cfrac1%7B1-3t%7D%5Cright%29%20%5C%2C%20dt%20%5C%5C%5C%5C%20%3D%20-%5Cfrac19%20%5Cleft%28%5Cfrac32%20t%5E2%20%2B%20t%20%2B%20%5Cfrac13%20%5Cln%7C1%20-%203t%7C%5Cright%29%20%2B%20C%20%5C%5C%5C%5C%20%3D%20%5Cboxed%7B-%5Cfrac%7Bt%5E2%7D6%20-%20%5Cfrac%20t9%20-%20%5Cfrac%7B%5Cln%7C1-3t%7C%7D%7B27%7D%20%2B%20C%7D)
where we use the power rule,
![\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20x%5En%20%5C%2C%20dx%20%3D%20%5Cfrac%7Bx%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%20%2B%20C%20~~~~%20%28n%5Cneq-1%29)
and a substitution to integrate the last term,
![\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bdt%7D%7B1-3t%7D%20%3D%20-%5Cfrac13%20%5Cint%20%5Cfrac%7Bdu%7Du%20%5C%5C%5C%5C%20%3D%20-%5Cfrac13%20%5Cln%7Cu%7C%20%2B%20C%20%5C%5C%5C%5C%20%3D%20-%5Cfrac13%20%5Cln%7C1-3t%7C%20%2B%20C%20~~~%20%28u%3D1-3t%20%5Ctext%7B%20and%20%7D%20du%20%3D%20-3%5C%2Cdt%29)
8.
![\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%20t%7B1%2B4t%7D%20%5C%2C%20dt%20%3D%20%5Cfrac14%20%5Cint%20%5Cleft%281%20-%20%5Cfrac1%7B1%2B4t%7D%5Cright%29%20%5C%2C%20dt%20%5C%5C%5C%5C%20%3D%20%5Cfrac14%20%5Cleft%28t%20-%20%5Cfrac14%20%5Cln%7C1%20%2B%204t%7C%5Cright%29%20%2B%20C%20%5C%5C%5C%5C%20%3D%20%5Cboxed%7B%5Cfrac%20t4%20-%20%5Cfrac%7B%5Cln%7C1%2B4t%7C%7D%7B16%7D%20%2B%20C%7D)
using the same approach as above.
The answer is -8 because the opposite of 4 is -4 so 32 divided by -4 = -8
The formula of a slope:
![m=\dfrac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
We have the points (-2, 2) and (-4, -2). Substitute:
![m=\dfrac{-2-2}{-4-(-2)}=\dfrac{-4}{-2}=2](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B-2-2%7D%7B-4-%28-2%29%7D%3D%5Cdfrac%7B-4%7D%7B-2%7D%3D2)
<h3>Answer: the slope = 2</h3>
Answer:
27 logs
Step-by-step explanation:
<u><em>4hrs = 6 logs</em></u>
<u><em></em></u>
<u><em>18hrs= ?x</em></u>
<u><em></em></u>
<u><em>Then you cross multiply.</em></u>
<u><em>so,18x6= 4x</em></u>
<u><em /></u>
<u><em>4x/4=108/4</em></u>
x=27 logs