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Snowcat [4.5K]
3 years ago
13

Polygon MNOPQ is dilated by a scale factor of 0.8 with the origin as the center of dilation, resulting in the image M′N′O′P′Q′.

If M = (2, 4) and N = (3, 5), what is the slope of line M'N'?
A.√2
B.1
C.√3
D.√1/2
Mathematics
1 answer:
Goryan [66]3 years ago
4 0
M` = ( 0.8 * 2,  0.8 *4 ) = 0.8 * ( 2, 4 );
N` = ( 0.8 * 3, 0.8 * 5 ) = 0.8 * ( 3 , 5 ).  
The slope of line M`N` ( after dilation ):
m = Rise / Run = ( y 2 - y 1 ) / ( x 2 - x 1 ) =
= 0.8 * ( 5 - 4 ) / ( 0.8 ( 3 - 2 ) = 0.8 / 0.8 = 1
m = 1
Answer:
The slope of line M`N` is : B ) 1
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7 0
2 years ago
Solve for x: x^2 + 28x = 60
Alisiya [41]

For this case we have a quadratic equation, x ^ 2 + 28x-60 = 0, of the form ax ^ 2 + bx + c = 0

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a = 1\\b = 28\\c = -60

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Then, the factorization is given by:

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The roots are:

x_ {1} = 2\\x_ {2} = - 30

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x_ {1} = 2\\x_ {2} = - 30

8 0
3 years ago
Please help me ASAP!!!
Mama L [17]
9/20 

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7 0
3 years ago
To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)
Pavlova-9 [17]

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\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}


\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill

5 0
3 years ago
Who can answer this question
lyudmila [28]
The answer is: c
Explination 30-9=21
8 0
2 years ago
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