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TEA [102]
3 years ago
11

Use De Moivre’s Theorem to compute the following:

29%2B8isin%284%5Cpi%2F5%29%29%7D%20" id="TexFormula1" title="\sqrt[3]{(8cos(4 \pi /5)+8isin(4\pi/5))} " alt="\sqrt[3]{(8cos(4 \pi /5)+8isin(4\pi/5))} " align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Kipish [7]3 years ago
6 0
Hello here is a solution : 

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Line A goes through the points (1, 5) and (-1, 9). Line B goes through the points (0, 2) and (4, -6). Which of the following sta
Delvig [45]
B is correct. Explanation Below:
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Line A never intersects Line B, as they are parallel lines-the slopes are the same-and parallel lines never intersect. 
Line A's Slope: (9-5)/(-1-1) = -2
Line B's Slope: (-6-2)/(4-0) = -2
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Hope this helps!
7 0
3 years ago
PLEASE HELP 30 POINTS
Gelneren [198K]

Answer:

3 + 1(3) +3 (the dollar amounts)

3+3+3

9

0.5 + 0.1 +0.1 +0.1 +0.5 (the cents)

1.3

9 +1.3 = 10.30 worth of groceries using front end estimation

3 0
3 years ago
−3x^ 2 +6x+9=<br> factor
GrogVix [38]

Answer:

the answer in factor form would be

-3(

7 0
3 years ago
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midni
Alex17521 [72]

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, \hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13

p = 0.20

So, hypothesis:

H_0:p=\hat{p}\\\\H_a:\hat{p}

so, test statistic value would be

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

4 0
3 years ago
Read 2 more answers
Consider a deck of cards:
Ganezh [65]

Ohhhh nasty !  What a delightful little problem !

The first card can be any one of the 52 in the deck.  For each one ...
The second card can be any one of the 39 in the other 3 suits. For each one ...
The third card can be any one of the 26 in the other 2 suits.  For each one ...
The fourth card can be any one of the 13 in the last suit.

Total possible ways to draw them = (52 x 39 x 26 x 13) = 685,464 ways.

But wait !  That's not the answer yet.

Once you have the 4 cards in your hand, you can arrange them
in (4 x 3 x 2 x 1) = 24 different arrangements.  That tells you that
the same hand could have been drawn in 24 different ways.  So
the number of different 4-card hands is only ...

                     (685,464) / (24) = <em>28,561 hands</em>.

I love it !


3 0
3 years ago
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