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torisob [31]
3 years ago
6

Can you help me find all the seventh roots of unity? what do they look like graphed?

Mathematics
1 answer:
alex41 [277]3 years ago
8 0

Answer:

There are seven seventh roots of unity, e2πki7 , all on the unit circle, r=1 above.

The first one is at θ=2π7=360∘7=5137 ∘ , and there are others at 4π7,6π7,8π7,10π7,12π7 and of course at 0 radians, i.e. unity itself.

How to find?

There are 4 fourth roots of unity and they are 1, i,−1 and−i

Each of the roots of unity can be found by changing the value of k k k in the expression e 2 k π i / n e^{2k\pi i/n} e2kπi/n. By Euler's formula, e 2 π i = cos ⁡ ( 2 π ) + i sin ⁡ ( 2 π ) = 1.

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Answer:

2.8

Step-by-step explanation:

we do 56% times 5= 2.8

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ValentinkaMS [17]

Answer:

The vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5). The graph of image and preimage is shown below.

Step-by-step explanation:

From the given figure it is noticed that the vertices of  triangle ABC are A(0,0), B(1,5) and C(-4,5).

If a figure rotated at 180º about the origin, then

P(x,y)\rightarrow P'(-x,-y)

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A(0,0)\rightarrow A'(0,0)

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Therefore the vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5).

The graph of image and preimage is shown below.

3 0
3 years ago
Safie can write a minimum of five questions per hour and a maximum of eleven questions per hour. What is the difference betwwen
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Answer:

<em>The difference is 30 hours</em>

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<u>Proportions</u>

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7 0
3 years ago
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