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zvonat [6]
2 years ago
8

27x39 Plz help jk jk

Mathematics
1 answer:
lesya [120]2 years ago
8 0

Answer: 1053

Step-by-step explanation: ...............................

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<h3>stay safe healthy and happy<u>.</u></h3>

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Which value is NOT a solution of 8x3 – 1 = 0?
Tpy6a [65]

<u><em>Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.</em></u>

Answer:

Any value other than the values x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4} will not be a solution of 8x^3\:-\:1\:=\:0.

Step-by-step explanation:

Considering the equation

8x^3\:-\:1\:=\:0

Steps to solve the equation

8x^3-1=0

\mathrm{Add\:}1\mathrm{\:to\:both\:sides}

8x^3-1+1=0+1

\mathrm{Simplify}

x^3=\frac{1}{8}

\mathrm{Divide\:both\:sides\:by\:}8

\frac{8x^3}{8}=\frac{1}{8}

\mathrm{Simplify}

x^3=\frac{1}{8}

As

\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}

x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}

So,

x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}

Therefore,

Any value other than the values x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4} will not be a solution of 8x^3\:-\:1\:=\:0.

Keywords: solution, value

Learn more about equation solution from  brainly.com/question/1679491

#learnwithBrainly

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If th<span>e equation y=1/3x is the boundary line for the inequality y>1/3 x, then the solution set is illustrated by shading the area ABOVE the line y=(1/3)x.</span>
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98.75 + 5% sales tax
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103.69

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5 0
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