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ivanzaharov [21]
3 years ago
12

Use de moivre's theorem to write [2(cos 12 degrees + i sin 12 degrees)]^5 in standard form

Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0

DeMoivre's theorem says

(2 (cos(12°) + <em>i</em> sin(12°)))⁵ = 2⁵ (cos(5×12°) + <em>i</em> sin(5×12°))

… = 32 (cos(60°) + <em>i</em> sin(60°))

… = 32 (1/2 + √3/2 <em>i</em> )

… = 16 + 16√3 <em>i</em>

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nasty-shy [4]

Answer:

2239grams

Step-by-step explanation:

4 0
3 years ago
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m_a_m_a [10]

74048 is the answer i think


3 0
3 years ago
Please help me this is very important its a test due in 30 minutes
Mashcka [7]
A
Please mark as a brainliest
5 0
3 years ago
Pls answe quick and correctly
andriy [413]

Answer:

9890

Step-by-step explanation

7×43×30+5×4×43=9890

4 0
2 years ago
I don’t know how to do this
stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as x approaches the edges. For example,

g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}

has two pieces, 2x+5 and x^2-10, both of which are continuous by themselves on the provided intervals. In order for g to be continuous everywhere, we need to have

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)

By definition of g, we have g(-3)=2(-3)+5=-1, and the limits are

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1

\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1

The limits match, so g is continuous.

For the others: Each of the individual pieces of f,h are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.

4 0
3 years ago
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