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sp2606 [1]
2 years ago
7

Let y in the form of a + ib, where a and b are real numbers, be the cubic roots of a complex number z 20, where z =2 4+i3. Find

a + b.​
Mathematics
1 answer:
GalinKa [24]2 years ago
6 0

The possible values of a + b are 0.89, 3.01 and -3.9

<h3>How to determine the value of a + b?</h3>

The given parameters are:

y = a + ib

z = 24 + i3

Where:

y = ∛z

Take the cube of both sides

y³ = z

Substitute the values for y and z

(a + ib)³ = 24 + i3

Expand

a³ + 3a²(ib) + 3a(ib)² + (ib)³ = 24 + i3

Further, expand

a³ + i3a²b + i²3ab² + i³b³ = 24 + i3

In complex numbers;

i² = -1 and i³= -i

So, we have:

a³ + i3a²b + (-1)3ab² -ib³ = 24 + i3

Further expand

a³ + i3a²b - 3ab² - ib³ = 24 + i3

By comparing both sides of the equation, we have:

a³ - 3ab² = 24

i3a²b - ib³ = i3

Divide through by i

3a²b - b³ = 3

So, we have:

a³ - 3ab² = 24

3a²b - b³ = 3

Using a graphing tool, we have:

(a,b) = (-1.55, 2.44), (2.89,0.12) and (-1.34,-2.56)

Add these values

a + b = 0.89, 3.01 and -3.9

Hence, the possible values of a + b are 0.89, 3.01 and -3.9

Read more about complex numbers at:

brainly.com/question/10662770

#SPJ1

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If you have $65 to spend at a concert and you already spend $32.25 on a ticket and fee. What is the maximum number of shirts he
Marrrta [24]
The maximum number of shirts I can buy is 2.

14.50 x 2  = 29
14.50 x 3 = 43.5

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7 0
3 years ago
The function h(x)=x² +5 maps the domain given by the set {-2, -1, 0, 1, 2}. Which of the following sets
frozen [14]

For the given function h(x) = x² +5 ,the domain of the function h(x) maps by the set { -2, -1, 0, 1, 2 } then the range of the function h(x) is given by the set { 5, 6, 9 } .

As given in the question,

Given function is equal to :

h(x) = x² +5

Domain of the function h(x) = x² +5 maps by the set { -2, -1, 0, 1, 2 }

Range of the function h(x) = x² +5 is represented by the set as follow :

Substitute the value x = -2, -1, 0, 1, 2 in the function h(x) = x² +5 we get,

h(-2) = (-2)² + 5

       = 9

h(-1) = (-1)² + 5

       = 6

h(0) = (0)² + 5

       = 5

h(1) = (1)² + 5

       = 6

h(2) = (2)² + 5

       = 9

Range of the function is represented by the set : { 5, 6, 9 }

Therefore, for the given function h(x) = x² +5 ,the domain of the function h(x) maps by the set { -2, -1, 0, 1, 2 } then the range of the function h(x) is given by the set { 5, 6, 9 }.

Learn more about function here

brainly.com/question/12426369

#SPJ1

7 0
1 year ago
Can someone tell me how
gulaghasi [49]

the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.

multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.

so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5   <----- notice, current coefficient times 3 divided by 4+1.

anyhow, with that out of the way, lemme proceed in this one.

\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ \qquad \qquad (1+ax)^n\implies \begin{array}{llll} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^n(ax)^0\\\\ 2&+\frac{(1)(n)}{1}\to n&(1)^{n-1}(ax)^1\\\\ 3&+\frac{n\cdot (n-1)}{2}&(1)^{n-2}(ax)^2 \end{array}

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.

so then, we know that the expanded 2nd term is 24x therefore

\bf n(1)^{n-1}(ax)1 = 24x\implies n(1)(ax)=24x\implies nax=24x\implies n=\cfrac{24}{a}

we also know that the expanded 3rd term is 240x², therefore we can say that

\bf \cfrac{n(n-1)}{2}~~(1)^{n-2}(ax)^2 = 240x^2\implies \cfrac{n(n-1)}{2}(1)(a^2x^2) = 240x^2 \\\\\\ \cfrac{(n^2-n)(a^2x^2)}{2}=240x^2\implies \cfrac{(n^2-n)(a^2)}{2}=\cfrac{240x^2}{x^2}\implies \cfrac{a^2n^2-a^2n}{2}=240 \\\\\\ a^2n^2-a^2n=480

but but but, we know what "n" equals to, recall above, so let's do some quick substitution

\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft

7 0
3 years ago
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