Answer:
(x, y) = (77/240, -3/10)
Step-by-step explanation:
It is convenient to write the equations in standard form.
Multiplying the first equation by 21 gives ...
21y = 24x -14
Multiplying the second equation by 8 gives ...
24x +9y = 5
Then the system of equations in standard form is ...
Subtracting the first from the second, we get ...
(24x +9y) -(24x -21y) = (5) -(14)
30y = -9
y = -9/30 = -3/10
Substituting this into the second equation, we have ...
24x +9(-3/10) = 5
24x = 7.7 . . . . . . . add 27/10
x = 7.7/24 = 77/240
The solution is (x, y) = (77/240, -3/10).
Answer:
She will need 10.5 quarts
Step-by-step explanation:
ratio of yellow to blue paint = 3:2
7 ÷ 2 = 3.5
3 x 3.5 = 10.5 quarts
H = 3.
FIRST STEP:
<span>Add 1 to both sides to get rid of the -1 on the left side.
4h-1 = 3h+2
</span><span>4h-1 (+1) = 3h+2 (+1)
</span><span>4h = 3h+3
SECOND (FINAL) STEP:
Subtract 3h from both sides to get rid of the 3h on the right side.
</span>4h(-3h) = 3h+3 (-3h)
h = 3
Hope this helps, sorry if it's hard to understand :)
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.