Answer:
Step-by-step explanation:
yay a branliest :o
Part A:
For the first figure it can be seen that the orientation of the pre-image (<span>∆RST) is the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by shifting the pre-image some units down.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a translation.
Part B:</span>
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For the second figure it can be seen that the orientation of the pre-image (<span>∆RST) is the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by shifting the pre-image some units down and some units to the right.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a translation.
Part C:
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For the third figure it can be seen that the orientation of the pre-image (<span>∆RST) is not the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by rotating the pre-image some 180 degrees.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a rotation.
Part D:
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For the fourth figure it can be seen that the orientation of the pre-image (<span>∆RST) is not the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by refrecting the pre-image.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a reflection.</span></span>
Answer:
c g(x)=2x^2
Step-by-step explanation:
hi