Enter the coordinates of the vertices of r(180,0) and t(3,-2) (QRST) will give crown
1 answer:
Answer:
Step-by-step explanation:
From the graph attached,
Coordinates of the vertices are,
Q(1, 3), R(3, -3), S(0, -2) and T(-2, 1)
Following the rule of translation by 3 units to the right and 2 units down
(x, y) → (x+3, y-2)
Q(1, 3) → Q''(4, 1)
R(3, -3) → R"(6, -5)
S(0, -2) → S"(3, -4)
T(-2, 1) → T"(1, -1)
Following rule (rotation of a point by 180° about the origin) will give the image points,
(x, y) → (-x, -y)
Q"(4, 1) → Q'(-4, -1)
R"(6, -5) → R'(-6, 5)
S"(3, -4) → S'(-3, 4)
T"(1, -1) → T'(-1, 1)
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Step-by-step explanation:
this can be found by determination of the highest and lowest x values
Answer:
When Ø = 300°, Ø = 60 degrees.
When Ø = 225°, Ø = 45 degrees.
When Ø = 480°, Ø = 60 degrees.
When Ø = -210°, Ø = 30 degrees.
Step-by-step explanation:
Reference angles are in Quadrant I (0° to 90°).
1. Find 300° (Quadrant IV) on the unit circle. Since it's in Quadrant IV, you use 360 - 300 = 60° to get your answer.
2. Find 225° (Quadrant III) on the unit circle. Since it's in Quadrant III, you use 225 - 180 = 45° to get your answer.
3. The angle 480° is not on the unit circle. To find its corresponding angle between 0° and 360°, use 480 - 360 = 120°. Then, find 120° (Quadrant II) on the unit circle. Since it's in Quadrant II, you use 180 - 120 = 60° to get your answer.
4. The angle -210° is not on the unit circle. To find its corresponding angle between 0° and 360°, use -210 + 360 = 150°. Then, find 150° (Quadrant II) on the unit circle. Since it's in Quadrant II, you use 180 - 150 = 30° to get your answer.
