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svetoff [14.1K]
3 years ago
6

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(3, −6), Q(0, 9), and R(−3, 0). Triangle P'Q'R' has v

ertices P'(1, −2), Q'(0, 3), and R'(−1, 0).
Plot triangles PQR and P'Q'R' on your own coordinate grid.

Part A: What is the scale factor of the dilation that transforms triangle PQR to triangle P'Q'R'? Explain your answer. (4 points)

Part B: Write the coordinates of triangle P"Q"R" obtained after P'Q'R' is reflected about the y-axis. (4 points)

Part C: Are the two triangles PQR and P''Q''R'' congruent? Explain your answer. (2 points)
Mathematics
1 answer:
RideAnS [48]3 years ago
3 0

Answer:

I'm sorry if its not right i tried my best. have a great day!!

Step-by-step explanation:

Part A: What is the scale factor of the dilation that transforms triangle PQR to triangle P'Q'R' the scale factor dilation is 

Part B: The new coordinates of triangle P"Q"R" is p is at (1, -2), (0,-3), (1,0).

Part C: No, the two triangles are not congruent they are the same shape but not the same size. The Triangle P'Q'R' is much smaller than the triangle PQR.

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zaharov [31]
Simple...

you have: F= \frac{G m^{2} }{ d^{2} }

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d^{2} *F= \frac{G m^{2} }{ d^{2} } * d^{2}

Leaving you with...

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Keep isolating m-->>>

F* d^{2} =G m^{2}

\frac{F* d^{2} }{G} = \frac{G m^{2} }{G}

Square root to solve what just m is-->>>

\sqrt{ \frac{F* d^{2} }{G} } = \sqrt{ m^{2} }

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Thus, your answer.
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