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inessss [21]
4 years ago
5

Find the cube roots of 27(cos 279° + i sin 279°).

Mathematics
1 answer:
UNO [17]4 years ago
7 0

The solution would be like this for this specific problem:

<span>27(cos 279° + i sin 279°).
</span><span>(cosx +i sinx) = cox(nx)) + i sin(nx)
</span><span>(27×(cos 279+i sin 279)<span>)1/3</span>=<span>27 1/3</span>×(cos 279+i sin 279<span>)<span>1/3
</span></span></span><span><span>2713</span>=<span>27−−√3</span>=?
</span><span>3×(cos279+i sin279<span>)<span>1/3
</span></span></span><span>3×(cos 279+i sin 279<span>)13</span>=3(cos 279/3+i sin 279/3)

</span><span>279/3 = 93
cube root = </span>3(cos 93 + i sin 93)

<span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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Please answer i need help quick
Serjik [45]

Answer:

(-2, 6)

Step-by-step explanation:

Since you want a 1 to 7 ratio, you want to divide the line into 2 parts, where one part has a length of 1 and the other has a length of 7. So the total length of the line is 8.

Start by looking at the difference in the X and Y coordinates.

X = | -4 - 12 | = | -16 | = 16

Y = | 7 - -1 | = | 8 | = 8

You could calculate the length of the line using pythagorian's theorem, but that's not needed. Simply use similar triangles. We have a right triangle with legs of length 16 and length 8. We want a similar triangle that is 1/8th as large (to get the desired 1 to 7 ratio). So divide both legs by 8, getting lengths of 16/8 = 2, and 8/8 = 1.

Now add those calculated offsets to point A.

A has an X coordinate of -4 and B has an X coordinate of 12 and the X coordinate for C must be between those limits. So calculate -4 + 2 = -2 to get the X coordinate for C.

The Y coordinate of A is 7 and the Y coordinate of B is -1. And since the Y coordinate must be between then, you have 7 - 1 = 6.

So the coordinates for C is (-2, 6)

8 0
4 years ago
If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?
Tju [1.3M]
Not necessarily. \mathbf u and \mathbf v may be linearly dependent, so that their span forms a subspace of \mathbb R^2 that does not contain every vector in \mathbb R^2.

For example, we could have \mathbf u=(0,1) and \mathbf v=(0,-1). Any vector \mathbf w of the form (r,0), where r\neq0, is impossible to obtain as a linear combination of these \mathbf u and \mathbf v, since

c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)

unless r=0 and c_1=c_2.
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3 years ago
Help me please I’d really appreciate it
Semenov [28]

Answer:

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7 0
3 years ago
What integer best represents that you lost $5.00 on the way to school? ​
gogolik [260]
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3 0
3 years ago
Read 2 more answers
Write the equation of the line parallel to 5x + y = -2 that goes through the point (2, -3).​
gladu [14]

Answer:

y = -5x + 7

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

Step 1: Define

5x + y = -2

Random point (2, -3)

Step 2: Rewrite

y = -5x - 2

m = -5

Step 3: Write parallel line

y = -5x + b

-3 = -5(2) + b

-3 = -10 + b

7 = b

y = -5x + 7

8 0
3 years ago
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