Mya should put 3/5 in each serving.
It’s 8^6, you’re correct :)
Step-by-step explanation:
1. when x is 0
y will be
4x+2y=10
4(0)+2y=10
0+2y=10
collect like terms
2y=10-0
2y=10
divide both sides by 2
2y/2=10/2
y=5
so when x =0 y will equal to 5
2. when x is 1
y will be
4x+2y=10
4(1)+2y=10
4+2y=10
collect like terms
2y=10-4
2y=6
divide both sides by two
2y/2=6/2
y=3
so when x is 1, y is 3
3. when x is 2
4x+2y=10
4(2)+2y=10
8+2y=10
(CLT) 2y=10-8
2y =2
y= 1
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)
1] y - 3x = -8
[2] y + 9x = 4
-3x + y = -8 9x + y = 4
Solve equation [2] for the variable y
[2] y = -9x + 4
// Plug this in for variable y in equation [1]
[1] (-9x+4) - 3x = -8
[1] - 12x = -12
// Solve equation [1] for the variable x
[1] 12x = 12
[1] x = 1
// By now we know this much :
y = -9x+4
x = 1
// Use the x value to solve for y
y = -9(1)+4 = -5
{y,x} = {-5,1}
