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wolverine [178]
3 years ago
10

ABCD is a rectangle. A, E and B are points on the straight line L with equation x + 3y = 18. A and D are points on the straight

line M. AE = EB. Find the equation for M in the form y = ax + b where a and b are integers
Mathematics
1 answer:
lawyer [7]3 years ago
8 0

Answer:

If  B  is the  x-intercept and  E  is the  y-intercept of  x + 3y  =  18 , then

point B  =  (18, 0)       and       point E  =  (0, 6)

Then...

point A  =  (-18, 6+6)  =  (-18, 12)

Line M  is perpendicular to line L.

slope of line L  =  -6/18  =  -1/3   so..

slope of line M  =  3

So line M passes through the point  (-18, 12)  with a slope of  3 . So the equation of line M is...

y - 12  =  3(x - -18)

y - 12  =  3(x + 18)

y - 12  =  3x + 54

y  =  3x + 66

Step-by-step explanation:

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Examine the following table of points, which are all on a certain line.
Andrew [12]

9514 1404 393

Answer:

  -1

Step-by-step explanation:

The slope formula is useful:

  m = (y2 -y1)/(x2 -x1)

  m = (2 -4)/(0 -(-2)) = -2/2 = -1

The slope of this line is -1.

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7 0
3 years ago
-1/2x+2=-x+7 i need to know
ludmilkaskok [199]

=  \frac{ - 1}{ \:  \: 2} x + 2 = x + 7

=  \frac{ - 1}{ \:  \:  \: 2} x + 2 - x = 7

=  \frac{ - 3}{ \:  \:  \: 2}x  + 2 = 7

=  \frac{ - 3}{ \:  \ \:  2}x = 7 - 2

=  \frac{ - 3}{ \:  \: \:  2} x = 5

x = 5 \:  \div  \frac{ - 3}{ \:  \: \:  2}

x = 5 \times  \frac{ \:  \:  \: 2}{ - 3}

x =  \frac{ - 10}{ \:  \:  \: 3}

∴  \frac{ - 1}{ \:  \:  \: 2}  \times  \frac{ - 10}{ \:  \:  \: 3}  =  \frac{ - 10}{ \:  \:  \: 3}  + 7

6 0
4 years ago
Please please please help
scoundrel [369]
First option: p< -8 or p>5
-6+\left|2p+3\right|>7
-6+\left|2p+3\right|+6>7+6
\left|2p+3\right|>13
2p+3<-13\quad \quad \mathrm{or}\quad \:\quad \:2p+3>13
Combine:
p< -8 or p>5
Hope this helps!
3 0
3 years ago
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