Answer:
65
Step-by-step explanation:
(-6.5)(-5) = +32.5, and then
(32.5)(2) = 65 (answer)
Answer:
The maximum volume of cone is 138.25 m³
m
m
Step-by-step explanation:
A right circular cone whose hypotenuse is
m
It is revolved about one leg to generate a right circular cone.
Let radius be r m and height be h m
For right angle triangle,
![r^2+h^2=7^2](https://tex.z-dn.net/?f=r%5E2%2Bh%5E2%3D7%5E2)
![r^2=49-h^2](https://tex.z-dn.net/?f=r%5E2%3D49-h%5E2)
Volume of generated cone
Differentiate w.r.t h
![\dfrac{dV}{dh}=\dfrac{1}{3}\pi (49-3h^2)](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdh%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%20%2849-3h%5E2%29)
For maximum/minimum ![\dfrac{dV}{dh}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdh%7D%3D0)
![\dfrac{1}{3}\pi (49-3h^2)=0](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B3%7D%5Cpi%20%2849-3h%5E2%29%3D0)
![r^2=49-\dfrac{49}{3}](https://tex.z-dn.net/?f=r%5E2%3D49-%5Cdfrac%7B49%7D%7B3%7D)
![r=7\sqrt{\dfrac{2}{3}}](https://tex.z-dn.net/?f=r%3D7%5Csqrt%7B%5Cdfrac%7B2%7D%7B3%7D%7D)
<em>Using double derivative test </em>
![\dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi (-6h)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2V%7D%7Bdh%5E2%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%20%28-6h%29)
At
so get maximum volume.
Dimension of cone:
m
m
![V=\frac{1}{3}\pi\cdot\left(7\sqrt{\frac{2}{3}}\right)^{2}\cdot\frac{7}{\sqrt{3}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Ccdot%5Cleft%287%5Csqrt%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cright%29%5E%7B2%7D%5Ccdot%5Cfrac%7B7%7D%7B%5Csqrt%7B3%7D%7D)
The maximum volume of cone is 138.25 m³
Answer: Don’t listen to the other comment it’s a scam. It’s 15 percent. Or 0.15
Step-by-step explanation:
Z=8+6x-px
-8 -8
------------------------
Z-8=6x-px
-------------------------
Z-8 = x(6-p)
/(6-p) /(6-p)
-----------------------------
Z-8
X= ---------
6-p