Answer:
x=12
Step-by-step explanation:
4x - 20 = -2 ( -3x + 22 )
4x-20=6x-44
24=2x
x=12
Answer:
x = 35
Step-by-step explanation:
![40 + 2x + 2x = 180](https://tex.z-dn.net/?f=40%20%2B%202x%20%2B%202x%20%3D%20180)
Subtract 40 on both sides.
![2x + 2x = 140](https://tex.z-dn.net/?f=2x%20%2B%202x%20%3D%20140)
![4x = 140](https://tex.z-dn.net/?f=4x%20%3D%20140)
Divide with 4 on both sides.
![x = 35](https://tex.z-dn.net/?f=x%20%3D%2035)
Remember that the velocity as a function of time is the derivative of the position as a function of time.
To solve this we are going to take the derivative of our position function
![s=24t+3t^2-t^3](https://tex.z-dn.net/?f=s%3D24t%2B3t%5E2-t%5E3)
. To do that er are going to apply the power rule of calculus:
![\frac{dy}{dx} x^n=nx^{n-1}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20x%5En%3Dnx%5E%7Bn-1%7D)
![\frac{dy}{dx} 24t+\frac{dy}{dx}3t^2-\frac{dy}{dx}t^3](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%2024t%2B%5Cfrac%7Bdy%7D%7Bdx%7D3t%5E2-%5Cfrac%7Bdy%7D%7Bdx%7Dt%5E3)
![24+2(3)t-3t^2](https://tex.z-dn.net/?f=24%2B2%283%29t-3t%5E2)
![v=24+6t-3t^2](https://tex.z-dn.net/?f=v%3D24%2B6t-3t%5E2)
We can conclude that the correct answer is:
<span>
B. V = 24 + 6t - 3t2</span>
Answer:
(1, 6 )
Step-by-step explanation:
5x + 2y = 17 → (1)
4x + y = 10 → (2)
Multiplying (2) by - 2 and adding to (1) will eliminate the y- term
- 8x - 2y = - 20 → (3)
Add (1) and (3) term by term to eliminate y
- 3x + 0 = - 3
- 3x = - 3 ( divide both sides by - 3 )
x = 1
Substitute x = 1 into either of the 2 equations and solve for x
Substituting into (1)
5(1) + 2y = 17
5 + 2y = 17 ( subtract 5 from both sides )
2y = 12 ( divide both sides by 2 )
y = 6
solution is (1, 6 )
Substitute
and
. Then the integral transforms to
![\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bx%5C%2Cdx%7D%7B%28x%5E2%2B4%29%5E3%7D%20%3D%20%5Cfrac12%20%5Cint%20%5Cfrac%7Bdu%7D%7Bu%5E3%7D)
Apply the power rule.
![\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bdu%7D%7Bu%5E3%7D%20%3D%20-%5Cdfrac1%7B2u%5E2%7D%20%2B%20C)
Now put the result back in terms of
.
![\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bx%5C%2Cdx%7D%7B%28x%5E2%2B4%29%5E3%7D%20%3D%20%5Cfrac12%20%5Cleft%28-%5Cdfrac1%7B2u%5E2%7D%20%2B%20C%5Cright%29%20%3D%20-%5Cdfrac1%7B4u%5E2%7D%20%2B%20C%20%3D%20%5Cboxed%7B-%5Cdfrac1%7B4%28x%5E2%2B4%29%5E2%7D%20%2B%20C%7D)