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2 years ago

What is the approximate 47? Give your answer and explain why it is correct. HELP

Mathematics

2 answers:

valentina_108 [34]2 years ago

5 0

Answer:

Step-by-step explanation:

$6^{2} =36\\7^2=49\\therefore,6< \sqrt{47}$

a_sh-v [17]2 years ago

3 0

Answer:

Step-by-step explanation:

7 is approximate to $\sqrt{47}$ because 7 is $\sqrt{49}$ , and 49 is close to 47

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Read 2 more answers

Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

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.

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

$8x^3-1=0$

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

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

$\mathrm{Simplify}$

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

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

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

$\mathrm{Simplify}$

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

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

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

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

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