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Blizzard [7]
3 years ago
14

If the domain of the coordinate transformation (, ) = ( + 2,− − 4) is (1, -4), (3, -2), (0, −1), what is the range?

Mathematics
1 answer:
timama [110]3 years ago
8 0

Answer:

A. (-2, -5), (0, -7), (1, -4)

Step-by-step explanation:

The following transformation is applied:

(x,y) \rightarrow (y + 2, -x - 4)

To find the range:

We apply the transformation to the points in the domain. Thus:

(1,-4) \rightarrow (-4 + 2, -1 - 4) = (-2, -5)

(3,-2) \rightarrow (-2 + 2, -3 - 4) = (0, -7)

(0,-1) \rightarrow (-1 + 2, 0 - 4) = (1, -4)

Thus the correct answer is given by option a.

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\large\boxed{x=\dfrac{3\pi}{2}\ \vee\ x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}}

Step-by-step explanation:

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\sin x=-1\to x=-\dfrac{\pi}{2}+2k\pi,\ k\in\mathbb{Z}\\\\\sin x=\dfrac{1}{2}\to x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}\\\\x\in[0,\ 2\pi)

x=-\dfrac{\pi}{2}\notin[0,\ 2\pi)\\\\x=-\dfrac{\pi}{2}+2\pi=\dfrac{3\pi}{2}\in[0,\ 2\pi)\\\\x=-\dfrac{\pi}{2}+4\pi=\dfrac{7\pi}{2}\notin[0,\ 2\pi)\\\\x=\dfrac{\pi}{6}\in[0,\ 2\pi)\\\\x=\dfrac{\pi}{6}+2\pi=\dfrac{13\pi}{6}\notin[0,\ 2\pi)\\\\x=\dfrac{5\pi}{6}\in[0,\ 2\pi)\\\\x=\dfrac{5\pi}{6}+2\pi=\dfrac{17\pi}{6}\notin[0,\ 2\pi)

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