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mrfishyyyy
01.02.2021 •
Mathematics
F(6)=3x6+2= i need help solving this
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Ответ:
Part A: The scale factor is 2
Part B:
P''(-4, 0)
Q''(-3, 1)
R''(1, -2)
Part C: No
Step-by-step explanation:
Part A:
The coordinates of the triangle PQR are P(8, 0), Q(6, 2), and R(-2, -4)
The coordinates of the triangle P'Q'R' are P'(4, 0), Q'(3, 1), and R'(-1, -2)
Please find attached the plot of triangles PQR and P'Q'R'
The length of the sides are given by the following formula;
Using an online length
For PQ. the length =![l_{PQ} = \sqrt{\left (2-0 \right )^{2}+\left (6-8 \right )^{2}} = 2 \cdot \sqrt{2}](/tpl/images/0839/5402/24eb3.png)
For PR. the length =![l_{PR} = \sqrt{\left (-4-0 \right )^{2}+\left (-2-8 \right )^{2}} = 2 \cdot \sqrt{29}](/tpl/images/0839/5402/ac827.png)
For RQ. the length =![l_{RQ} = \sqrt{\left (-4-2 \right )^{2}+\left (-2-6 \right )^{2}} = 10](/tpl/images/0839/5402/55a3e.png)
While for triangle P'Q'R', we have;
For P'Q'. the length =![l_{P'Q'} = \sqrt{\left (1-0 \right )^{2}+\left (3-4 \right )^{2}} = \sqrt{2}](/tpl/images/0839/5402/29c9c.png)
For P'R'. the length =![l_{P'R'} = \sqrt{\left (-2-0 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}](/tpl/images/0839/5402/6ff6e.png)
For R'Q'. the length =![l_{R'Q'} = \sqrt{\left (-2-1 \right )^{2}+\left (-1-3 \right )^{2}} = 5](/tpl/images/0839/5402/bb9d4.png)
∴ PQ = 2 × P'Q'
PR = 2 × P'R'
QR = 2 × Q'R'
Therefore, the scale factor = 2
Part B
By reflecting about the y-axis we have;
Coordinates of the preimage (x, y) after reflection across the y=axis becomes (-x, y)
Therefore, the coordinates of the triangle P''Q''R'' are;
P'(4, 0) after reflection = P''(-4, 0)
Q'(3, 1) after reflection about the y-axis is Q''(-3, 1)
R'(-1, -2) after reflection about the y-axis becomes R''(1, -2)
Part C: No
Given that the triangle P''Q''R'' is congruent to triangle P'Q'R', and triangle P'Q'R' is similar to but not congruent to triangle PQR, we have that triangle P''Q''R'' is similar to but not congruent to triangle PQR.