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Keith_Richards [23]
3 years ago
12

What’s the answer pls I can’t figure it out

Mathematics
2 answers:
marissa [1.9K]3 years ago
6 0

Answer:

(0,-1)

Step-by-step explanation:

The current cooridnates for T are (5,6)

If you translate anything 7 units down and 5 units to the left. (x,y) --> (x-5,y-7)

So, the coordinates for T' are (5-5, 6-7) --> (0,-1)

Ira Lisetskai [31]3 years ago
4 0

Answer:

(0,-1)

Step-by-step explanation:

Basically, the T coordinate in the second shape, move it 7 units down on the graph, and after you have done so, move it 5 units to the left.

The coordinate for t will be (0,-1)

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That gives you 20 * 19 * 18 * 17 * 16 * 15 * 14.

That number would allow you to write the students in different order. Since order here does not matter, any group with the same students in any order is the same group, you need to divide by the number of way you can order 7 items. Divide by 7 * 6 * 5 * 4 * 3 * 2 * 1

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What is the linear function equation represented by the graph?
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Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
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Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

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Answer:

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