Answer:
Step-by-step explanation:
There are many such polynomials, but the one with leading coefficient of 1 is
y = (x+1)(x-1)(x-6) = x^3 - 6x^2 - x + 6
Not sure what you want to do with the 5, If that is the y-intercept, then since
(1)(-1)(-6) = 6, we will need
y = 5/6 (x^3 - 6x^2 - x + 6)
<span>3x^2 = 7x - 3
Rearranging,
</span><span>3x^2 - 7x + 3 = 0
</span>Using quadratic formula :
x= [7 + sqrt (49 - 36)]/6
x= [7 + sqrt 13]
or
x= [7 - sqrt (49 - 36)]/6
x= [7 - sqrt 13]/6
<span>d. x - 7 plus or minus the square root of 13 divided by 6</span>
Answer:
![h=5in, r=1in](https://tex.z-dn.net/?f=h%3D5in%2C%20r%3D1in)
Step-by-step explanation:
The volume of a cylinder is:
![V=\pi*r^2*h](https://tex.z-dn.net/?f=V%3D%5Cpi%2Ar%5E2%2Ah)
where
is the radius and
is the height.
The height is 4 inches greater that the radius, so:
![h=r+4in](https://tex.z-dn.net/?f=h%3Dr%2B4in)
Substituting this in the formula for volume:
![V=\pi*r^2*(r+4)](https://tex.z-dn.net/?f=V%3D%5Cpi%2Ar%5E2%2A%28r%2B4%29)
the volume is ![V=5\pi in^3](https://tex.z-dn.net/?f=V%3D5%5Cpi%20in%5E3)
thus:
![5\pi in^3=\pi*r^2*(r+4in)](https://tex.z-dn.net/?f=5%5Cpi%20in%5E3%3D%5Cpi%2Ar%5E2%2A%28r%2B4in%29)
Dividing everything between pi:
![5in^3=r^2*(r+4in)](https://tex.z-dn.net/?f=5in%5E3%3Dr%5E2%2A%28r%2B4in%29)
![5in^3=r^3*+4r^2](https://tex.z-dn.net/?f=5in%5E3%3Dr%5E3%2A%2B4r%5E2)
We can see that the solution for this is ![r=1in](https://tex.z-dn.net/?f=r%3D1in)
since ![(1)^3*+4(1)^2=1+4=5](https://tex.z-dn.net/?f=%281%29%5E3%2A%2B4%281%29%5E2%3D1%2B4%3D5)
We have the radius, now we find the height:
![h=r+4in=1in + 4in = 5in](https://tex.z-dn.net/?f=h%3Dr%2B4in%3D1in%20%2B%204in%20%3D%205in)
![h=5in, r=1in](https://tex.z-dn.net/?f=h%3D5in%2C%20r%3D1in)
SAD is the correct answer, hope this help :3
I believe the sequence here is that you double the 1+ digits after the first 1, and then put it in front of the 1. Unfortunately, I can not think of a sequence. Try working this one out, as you know how to work it out. Tell me if you need more help!