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Let line \(l_1 \) be the graph of \(5x+8y=-9\). Line \(l_2 \) is perpendicular to line \(l_1 \) and passes through the point \((10,10)\). If line \(l_2\) is the graph of the equation \(y=mx+b\), then find \(m+b \)

 Aug 11, 2019
 #1
avatar+18961 
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l1  is re-written as   y = -5/8 x -9/8     slope m is  -5/8

l2 perpindicular slope is     -1/(-5/8) = 8/5

y = 8/5 x + b     sub in point   10,10

10 = 8/5 (10) + b

b = 10 - 80/5 = -30/5

 

-30/5 +8/5=  -22/5

 Aug 11, 2019
 #2
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+1

The answer is \(-\frac{22}{5}\) but if you want the full explantion you get get that here 

 

https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.986526.html  

 

I don't know if this is the correct way to post a answer because I have not done it. 

Hope it Helps!!laugh

 Aug 11, 2019

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