$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ \hrulefill\\[0.5em] a_3=8\\ n=3\\ d=5 \end{cases} \\\\\\ a_3=a_1+(3-1)5\implies 8=a_1+(2)5 \\\\\\ 8=a_1+10\implies -2=a_1 \\\\\\ \begin{cases} a_1=-2\\ d=5 \end{cases}\implies a_n=-2+(n-1)d$

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B5%20%2B%202%20%5Cdiv%20%20%5Csqrt%7B5%20-%202%7D%20%7D%20" id="TexFormula1" title

Artemon [7]

Answer:

so yea what was the answer again

Step-by-step explanation:

6 0

3 years ago

Subtract m + 8 ​​from​ 5m + 11.

Subtract m + 8 from 5m + 11.