$1.\underline{2}\implies \cfrac{12}{1\underline{0}}\implies \cfrac{6}{5} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{\frac{6}{5}}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{6}{5}}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{10}}}\implies \cfrac{~~ \frac{6-30}{5}~~}{-6}\implies \cfrac{~~ \frac{-24}{5}~~}{-6}\implies \cfrac{~~ -\frac{24}{5}~~}{-\frac{6}{1}}$

$-\cfrac{\stackrel{4}{~~\begin{matrix} 24 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{5}\cdot -\cfrac{1}{\underset{1}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \boxed{\cfrac{4}{5}}$

4 0

3 years ago

Write an expression that can be usedto find the nth termo of each sequence 9, 17, 25, 33

ra1l [238]

Answer:

The expression used to find the nth term of each sequence 9, 17, 25, 33 will be:

$a_n=8n+1$

Step-by-step explanation:

Given the sequence

9, 17, 25, 33

a₁ = 9

<em>Determining the common difference</em>

d = 17-9 = 8

d = 25-17 = 8

d = 33-25 = 8

As the common difference between the adjacent terms is same and equal to

d = 8

Therefore, the given sequence is an Arithmetic sequence.

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

substituting a₁ = 9, d = 8 in the equation

$\:a_n=9+8n-8$

$a_n=8n+1$

Therefore, the expression used to find the nth term of each sequence 9, 17, 25, 33 will be:

$a_n=8n+1$

6 0

3 years ago