3 years ago

write in slope intercept form an equation of the line that passes through the points (-15, -1) and (10,4)

1 answer:

4 0

Find the slope first
m=y2-y1/x2-x1
m=4+1/10+15
m=5/25 = 1/5
use point slope formula and plug in any points
y-y1=m(x-x1)
y-4=1/5(x-10)
y-4=1/5x-2
y=1/5x+2

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Read 2 more answers

What is the difference in simplest form? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%5E2%2B3n%2B2%7D%7Bn%5E2%2B6n%2B8%7D-%5

Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$