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Novay_Z [31]
3 years ago
5

Write an equation of the line perpendicular to y=1/6x+4 that contains (3,-3)

Mathematics
1 answer:
Vlad [161]3 years ago
8 0

Answer:

y= -6x+15

Step-by-step explanation:

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(See Attachment)
vesna_86 [32]
The solution would be the coordinates of (3a,a).
8 0
3 years ago
Evaluate 3 to the power of 3+1⋅9+12<br><br> A.24<br> B.30<br> C.48<br> D.264
Crank
24..... hope I helped
5 0
3 years ago
Read 2 more answers
A sum of money is deposited at a bank at a rate of 12½ annum simple interest. In how many years would the deposited money be dou
Dmitrij [34]

Answer:

It would take 10 years for the given sum of money be doubled at the given simple interest rate.

Step-by-step explanation:

A 10% interest would be added to the the principal amount after each year. So  the interest would reach 100% i.e. equal to the principal amount in 10 years.

7 0
3 years ago
segment MN is shown. Point M is located at (7,6). Point N is located at (7,-4) What is the midpoint segment of MN
Luda [366]

Answer:

The midpoint of MN is <u>(7,1)</u>

Add the x coordinates 7+7=14 then you divide that by 2 giving you 7.

Then you add the y coordinates together 6+(-4)=2 then divide by 2 and that gives you 1. Therefore the answer is (7,1)

6 0
2 years ago
Determine the slope of the line that passes through each pair of points.
Soloha48 [4]
1.)\\\\(5,1) \  \ and  \ \ (2,7)\\\\First \  find \ the \  slope \ of \ the \ line \ thru \  the \  points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} }   \\ \\m=\frac{ 7-1}{2-5} =  \frac{6}{-3}=-2 \\ \\   Use  \  point \  form  \ of  \ a \  line\  with \ one \ point: \\ \\ y-y_{1} =m(x-x _{1}) \\ \\m=-2 , \ \x_{1}=5, \ \ y_{1}=1 \\ \\y-1 = -2(x-5 ) \\\\y=-2x+10+1 \\ \\y=-2x+11


2.)\\\\(5,3)\ \ and \ \ (-2,3)\\\\First \  find \ the \  slope \ of \ the \ line \ thru \  the \  points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} }   \\ \\m=\frac{  3-3}{-2-5} =  \frac{0}{-7}=0\\ \\   Use  \  point \  form  \ of  \ a \  line\  with \ one \ point: \\ \\ y-y_{1} =m(x-x _{1}) \\ \\m=0 , \ \x_{1}=5, \ \ y_{1}=3 \\ \\y-3 = 0\cdot (x-5 ) \\\\y-3=0 \\ \\y=3
7 0
3 years ago
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