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3 years ago

Find the cube roots of 8i. Write the answer in a + bi form.

Mathematics

2 answers:

Murrr4er [49]3 years ago

8 0

Answer:

a0 = ∛8 [ cos [ (pi/2)/3] + i *sin [(pi/2)/3 ] = ∛8 [ cos [ (pi/6)] + i *sin [(pi/6) ] =

2 [ √3/2 + 1/2 * i ] = √3 + 1i

a1 = ∛8 [ cos ( (pi/2)/3 + 2pi/3 ) + i sin ( (pi/2 )/3 + 2pi/3 ) ] =

2 [ cos (5pi/6) + i sin (5pi/6) ] = 2 [ - √3/2 + (1/2)*i ] = - √3 +1i

a2 = ∛8 [ cos ( (pi/2)/3 + 4pi/3 ) + i sin ( (pi/2 )/3 + 4pi/3 ) ] =

2 [ cos(3pi/2) + i sin (3pi/2) ] = 2 [ -1 i ] = -2i

like this

AnnZ [28]3 years ago

3 0

Answer:

It's B

Step-by-step explanation:

Just took the quiz

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3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

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