if it has a diameter of 8, that means its radius is half that, or 4.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=4\\ h=5 \end{cases}\implies V=\cfrac{\pi (4)^2(5)}{3}\implies V=\cfrac{80\pi }{3} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{using~\pi =3.14}{V= 83.7\overline{3}}~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D4%5C%5C%0Ah%3D5%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%284%29%5E2%285%29%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B80%5Cpi%20%7D%7B3%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%5Cstackrel%7Busing~%5Cpi%20%3D3.14%7D%7BV%3D%2083.7%5Coverline%7B3%7D%7D~%5Chfill%20)
Answer:
A. 7/11
Step-by-step explanation:
Answer:
X 1 = -4,X 2 =12
EXPLANATION:
Determine the defined range
X+6/x=6/x-8,x#0,x#8
Simplify the equation using cross-multiplication
(X+6) x (x-8)=6x
Move variable to the left-hand side and change its sign
(X-6)x(x-8)-6x=0
Multiply the parentheses
X^2-8x+6x-48-6x=0
Since two opposites add up to zero, remove them from the expression
X^2-8x-48=0
Write -8x as a difference
X^2+4x-12x-48=0
Factor out x from the expression
Xx(x+4)-12x-48=0
Factor out -12 from the expression
Xx(x+4)-12(x+4)=0
Factor out from the expression
(x+4)x(x-12)=0
When the product of factors equals 0, at least one factor is 0
x+4=0
x-12=0
Solve the equation for x
X=-4
X-12=0
X=-4,x#0,x#8
Check if the solution is in the defined range
X=-4
x=12
The equation has 2 solutions
X 1 = -4,X 2 =12
Hope this helps
Answer:
the answer is B
Step-by-step explanation:
6w + 5 - 3w + 7
<h3>i) collect like terms</h3>
= 6w - 3w + 5 + 7
= 3w + 5 + 7
<h3>ii) add the numbers</h3>
= 3 w + 5 + 7
= 3w + 12
