20 liters I looked up the answer and did research on how to find it.
[x-1]=5x+10
We have two equations:
1) x-1=5x+10
x-1=5x+10
x-5x=10+1
-4x=11
x=-11/4
2) x-1=-(5x+10)
x-1=-(5x+10)
x-1=-5x-10
x+5x=-10+1
6x=-9
x=-9/6=-3/2
we have two possible solutions:
solution₁; x=-11/4
solution₂: x=-3/2
we check it out:
1) x=-11/4
[x-1]=5x+10
[-11/4 - 1]=5(-11/4)+10
[(-11-4)/4]=-55/4 + 10
[-15/4]=(-55+40) /4
15/4≠-15/4 This solution don´t work.
2) x=-3/2
[x-1]=5x+10
[-3/2 - 1]=5(-3/2)+10
[(-3-2)/2]=-15/2 + 10
[-5/2]=(-15+20)/2
5/2=5/2; this solution works.
Therefore:
Answer: x=-3/2.
Answer:
8 points????????? TYYYYYYYYYYYYYY
Step-by-step explanaI DONT KNOOOOOOOOOOOONIGAAAAAAAAAAAAAAxD
Answer:
The volume of the solid is 
Step-by-step explanation:
In this case, the washer method seems to be easier and thus, it is the one I will use.
Since the rotation is around the y-axis we need to change de dependency of our variables to have
. Thus, our functions with
as independent variable are:
For the washer method, we need to find the area function, which is given by:
![A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ccdot%20%5B%28%5Crm%7Bouter%5C%20radius%29%5E2%20-%28%5Crm%7Binner%5C%20radius%29%5E2%20%5D)
By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function
and the inner one by
. Thus, the area function is:
![A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)](https://tex.z-dn.net/?f=A%28y%29%3D%5Cpi%5Ccdot%20%5B%28%5Csqrt%7By%7D%20%29%5E2-%28y%5E2%29%5E2%5D%5C%5CA%28y%29%3D%5Cpi%5Ccdot%20%28y-y%5E4%29)
Now we just need to integrate. The integration limits are easy to find by just solving the equation
, which has two solutions
and
. These are then, our integration limits.
