Answer:
The answer to your question is 600 cups
Step-by-step explanation:
Data
Cylinder Cup
diameter = 30 in diameter = 3 in
height = 24 in height = 4 in
Process
1.- Calculate the volume of the cylinder
Volume = πr²h
-Substitution
Volume = (3.14)(30/2)²(24)
-Simplification
Volume = (3.14)(15)²(24)
Volume = (3.14)(225)(24)
-Result
Volume = 16959 in³
2.- Calculate the volume of the cup
Volume = (3.14)(3/2)²(4)
-Simplification
Volume = (3.14)(1.5)²(4)
Volume = (3.14)(2.25)(4)
-Result
Volume = 28.26 in³
3.- Divide the volume of the cylinder by the volume of the cup
Number of full cups = 16959 in³ / 28.26 in³
Number of full cups = 600
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20a_n%3D2-5%28n-1%29%5Cimplies%20a_n%3D%5Cstackrel%7B%5Cstackrel%7Ba_1%7D%7B%5Cdownarrow%20%7D%7D%7B2%7D%2B%28n-1%29%28%5Cstackrel%7B%5Cstackrel%7Bd%7D%7B%5Cdownarrow%20%7D%7D%7B-5%7D%29)
so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.
just a quick note on notation:
![\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_n%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bthe%20term%20before%20it%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_%7Bn-1%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_5%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B5-1%7D%5Cimplies%20a_4%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_%7B12%7D%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B12-1%7D%5Cimplies%20a_%7B11%7D%7D)
The answer to this question is C<span>. $611.20 hope this helps :)</span>
The answer is what that guys said
(Also what grade is this?)
If you get 0 as the last value in the bottom row, then the binomial is a factor of the dividend.
Let's say the binomial is of the form (x-k) and it multiplies with some other polynomial q(x) to get p(x), so,
p(x) = (x-k)*q(x)
If you plug in x = k, then,
p(k) = (k-k)*q(k)
p(k) = 0
The input x = k leads to the output y = 0. Therefore, if (x-k) is a factor of p(x), then x = k is a root of p(x).
It turns out that the last value in the bottom row of a synthetic division table is the remainder after long division. By the remainder theorem, p(k) = r where r is the remainder after dividing p(x) by (x-k). If r = 0, then (x-k) is a factor, p(k) = 0, and x = k is a root.
