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2 years ago

What is the circumference if the diameter of a circle is 23 centimeters?

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

4 0

The equation is 2(3.14)(11.5) because 11.5 is half of 23
they multiply and you get: 72.22
Hope this helps! :D

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Examine the work shown. Explain the error and find the correct result.

Zepler [3.9K]

2(-12) isn't 24.

It is -24.

So the answer would be 6.

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2 years ago

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What is 33 over 16 in a decimal

TiliK225 [7]

Answer:

2.0625

Step-by-step explanation:

Convert the fraction to a decimal by dividing the numerator by the denominator.

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3 years ago

The area of triangle LMN is 45 in². What is the area of the parallelogram LMNO? A.45 in² B. 90 in² C. 135 in² D.180 in²

krok68 [10]

Answer:its A.

Step-by-step explanation:

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2 years ago

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

$8x^3-1=0$

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

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

$\mathrm{Simplify}$

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

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

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

$\mathrm{Simplify}$

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

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

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

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

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2 years ago

Whats the measure of

maxonik [38]

Acute angle i think about 60

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2 years ago