Answer:
-4
Step-by-step explanation:
1. Always start with the parenthesis. (26/13) = 2
2. Then multiply. 2*-7 = -14
3. After, you add 14 to your precious answer (-14) then subtract 4.
-14 + 14 = 0 - 4 = -4
Check the picture below.
since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.
1)
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%5C%5C%20h%3D4%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%283%29%5E2%284%29%7D%7B3%7D%5Cimplies%20V%3D12%5Cpi%20%5Cimplies%20V%5Capprox%2037.7)
2)
now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%283%29%5E3%7D%7B3%7D%5Cimplies%20V%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semisphere%7D%7D%7BV%3D18%5Cpi%20%7D%5Cimplies%20V%5Capprox%2056.55)
3)
well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.
4)
pretty much the same thing, we get the volume of the cone and its top, add them up.

We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704
![\sqrt[3]{x + 1} = 2](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%20%2B%201%7D%20%20%3D%202)
Let us cube both sides.
![= > (\sqrt[3]{x + 1} ) ^{3} = {2}^{3}](https://tex.z-dn.net/?f=%20%3D%20%20%3E%20%20%28%5Csqrt%5B3%5D%7Bx%20%2B%201%7D%20%29%20%5E%7B3%7D%20%20%3D%20%20%7B2%7D%5E%7B3%7D%20)
Now, it will be only x + 1.

Transpose 1 to the Right Hand Side.

Now, subtract.

<u>Answer</u><u>:</u>
x = 7
Hope you could understand.
If you have any query, feel free to ask.
They would have to win 12.6 games. You need to find 60% of 21