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diamong [38]
4 years ago
6

Find all solutions of the equation. leave answers in trigonometric form x^3-8=0 2cis

Mathematics
1 answer:
Lera25 [3.4K]4 years ago
7 0

x^3-8=0

x^3=8=8\mathrm{cis}\,0^\circ

By DeMoivre's theorem, we have

x=8^{1/3}\mathrm{cis}\left(\dfrac{0^\circ+360^\circ k}3\right)

with k=0,1,2. Then we have three solutions,

x=2\mathrm{cis}\,0^\circ=2

x=2\mathrm{cis}\,120^\circ

x=2\mathrm{cis}\,240^\circ

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Z=y/xm , solve for x
fgiga [73]

Answer:

y/mz

Step-by-step explanation:

7 0
3 years ago
Help ASAP! And explain!
stich3 [128]

Answer:

Option (2)

Step-by-step explanation:

Given:

AC is an angle bisector of ∠DAB and ∠DAB

m∠BCA ≅ m∠DCA

m∠BAC ≅ m∠DAC

To Prove:

ΔABC ≅ ΔADC

Solution:

               Statements                                  Reasons

1). m∠BCA ≅ m∠DCA                            1). Given

2). m∠BAC ≅ m∠DAC                           2). Given

3). AC ≅ AC                                            3). Reflexive property

4). ΔABC ≅ ΔADC                                 4). ASA property of congruence

Therefore, Option (2) will be the correct option.

4 0
3 years ago
(Solve and write addition Equations) Zacarias and Paz together have $756.80. If Zacarias has $489.50, how much does Paz have? Wr
serious [3.7K]
$489.50+x=$756.80 

x=amount paz has.

756.80-489.50 = 267.3

so Paz has $267.30
4 0
3 years ago
Read 2 more answers
What is 2/5 divided by 3/4
Triss [41]
It’s 8/15 decimal form it’ll be 0.53 hope this helped !
4 0
3 years ago
Write the equation of a possible rational
kondor19780726 [428]

Answer:

The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)

Step-by-step explanation:

A rational equation is a equation where

\frac{p(x)}{q(x)}

where both are polynomials and q(x) can't equal zero.

1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of

(x - 2)

in our denomiator.

So right now we have

\frac{p(x)}{(x - 2)}

2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.

We can model -2.5 as

(2x + 5)

So we have as of right now.

\frac{(2x + 5)}{(x - 2)(2x + 5)}

Now let see if this passes throught point (6,-3).

\frac{(2x + 5)}{(x - 2)(2x + 5)}  = y

\frac{(17)}{68}  =  \frac{1}{4}

So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.

If we have a variable r, in the numerator that will make this applicable, we would get

\frac{(2x + 5)r}{(2x + 5)(x - 2)}  =  - 3

Plug in 6 for the x values.

\frac{17r}{4(17)}  =  - 3

\frac{r}{4}  =  - 3

r =  - 12

So our rational equation will be

\frac{ - 12(2x + 5)}{(2x + 5)(x - 2)}

or

\frac{ - 24x - 60}{2 {x}^{2}  + x - 10}

We can prove this by graphing

5 0
3 years ago
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