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Maru [420]
3 years ago
6

Triangle ABC is reflected across the y-axis to form the image A'B'C'. Triangle A'B'C' is then reflected across the x-axis to for

m the image A''B''C''. What type of rotation can be used to describe the relationship between triangle A"B"C" and triangle ABC
Mathematics
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

Circular!

Step-by-step explanation:

Triangle ABC is reflected across the y-axis to form the image A'B'C'. Triangle A'B'C' is then reflected across the x-axis to form the image A''B''C''. What type of rotation can be used to describe the relationship between triangle A"B"C" and triangle ABC

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Which of the following equations represents the perpendicular bisector of WX graphed below?
kotykmax [81]
The coordinates of the 2 given points are W(-5, 2), and X(5, -4).

First, we find the midpoint M using the midpoint formula:

\displaystyle{ M_{WX}= (\frac{x_1+x_2}{2},  \frac{y_1+y_2}{2} )=  (\frac{-5+5}{2},  \frac{2+(-4)}{2} )=(0, -1).

Nex, we find the slope of the line containing M, perpendicular to WX. We know that if m and n are the slopes of 2 parallel lines, then mn=-1.

The slope of WX is \displaystyle{ m= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-4)}{-5-5}= \frac{6}{-10}= -\frac{3}{5}.

Thus, the slope n of the perpendicular line is \displaystyle{  \frac{5}{3}.

The equation of the line with slope \displaystyle{ n= \frac{5}{3} containing the point M(0, -1) is given by:

\displaystyle{ y-(-1)=\frac{5}{3}(x-0)

\displaystyle{ y+1= \frac{5}{3}x

\displaystyle{ 3y+3=5x

\displaystyle{ 5x-3y-3=0

Answer: 5x-3y-3=0
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The total cost is c the you just multiply 14 by the number of hours so it would be C=$14h

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