well, the sequence goes, 1100, to 1135, to 1170.... notice, is simply adding 35 to get the next term, so the common difference is 35, and the first term is 1100 of course.
![\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}\\[-0.5em] \hrulefill\\ a_1=1100\\ d=35\\ a_n=3725 \end{cases} \\\\\\ 3725=1100+(n-1)35\implies 3725=1100+35n-35 \\\\\\ 3725=1065+35n\implies 2660=35n\implies \cfrac{2660}{35}=n\implies 76=n](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%0A%5C%5C%5C%5C%0Aa_n%3Da_1%2B%28n-1%29d%5Cqquad%0A%5Cbegin%7Bcases%7D%0An%3Dn%5E%7Bth%7D%5C%20term%5C%5C%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ad%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa_1%3D1100%5C%5C%0Ad%3D35%5C%5C%0Aa_n%3D3725%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A3725%3D1100%2B%28n-1%2935%5Cimplies%203725%3D1100%2B35n-35%0A%5C%5C%5C%5C%5C%5C%0A3725%3D1065%2B35n%5Cimplies%202660%3D35n%5Cimplies%20%5Ccfrac%7B2660%7D%7B35%7D%3Dn%5Cimplies%2076%3Dn)
Answer: Well if the number has no end it must be irrational same as sqrt of 2 or pi
Step-by-step explanation:
Answer:
0.1
Step-by-step explanation:
20.15625 is the answer to your question if you would like the method you move the percentage two decimal places and multiply.
D. 1/4 + 1/5 = 5/20 + 4/20 = 9/20 = 45/100 = 0.45