![\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)
Answer:
R / 5 = -9
is that what u want if not i'll change it. :D
The answer is 0.833333333
9514 1404 393
Answer:
62.5 oz of 100%
Step-by-step explanation:
Let p represent the number of ounces of pure syrup. Then (100-p) is the number of ounces of 60% syrup. The amount of maple syrup in the desired mix is ...
p +0.60(100 -p) = 0.85(100)
0.40p +60 = 85 . . . . . . . . . . . simplify
0.40p = 25 . . . . . . . . . . subtract 60
p = 62.5 . . . . . . . . . divide by 0.4
62.5 ounces of pure syrup should be mixed with the diluted syrup to make 100 ounces of 85% maple syrup.
_____
The other 37.5 ounces will be 60% syrup.
Answer:
Step-by-step explanation:
Set the cost equation C(l) = l3 – l2 + l + 2.5 equal to $11.00 and solve for l:
C(l) = 2l + 2.5 = $11, or
2l = 8.5, or l = length = 4.25 inches