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3 years ago

How to find area of polygon please explain

Mathematics

1 answer:

Nat2105 [25]3 years ago

5 0

To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothegm. Here is what it means: Perimeter = the sum of the lengths of all the sides. Apothegm = a segment that joins the polygon's center to the midpoint of any side that is perpendicular to that side.

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Write the percent as a mixed number in simplest form.<br> 168%

vladimir2022 [97]

Answer:

1 and 17/25

Step-by-step explanation:

First, you make it into a fraction. So because percents are out of 100, it would be 168/100. Then, you take 100 out of it to get a mixed number, 1 and 68/100, and then simplify 68/100 by dividing both sides by the greatest common factor, which is 4. 68/100÷4=17/25. The answer would be 1 and 17/25.

(hope this helps :P)

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2 years ago

Janelle earned 90% on a test and got 63 points. How many total points were possible on the test? Draw a diagram to organize your

Slav-nsk [51]

Answer:

63 / 0.9 = 70

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2 years ago

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How can you tell whether a solution to a rational equation is extraneous?

BARSIC [14]

An extraneous solution doesn't belong to the domain.

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3 years ago

Find the slope between these two points: (3, -10) and (-2, -30). Show all of your work. what the answer

Harlamova29_29 [7]

$\huge\text{$m=\boxed{4}$}$

Hey there! Start with the slope formula, where $(x_1,y_1)$ and $(x_2,y_2)$ are the two known points.

$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\&=\frac{-30-(-10)}{-2-3}\end{aligned}$

Simplify.

$\begin{array}{c|l}\textbf{Solving}&\textbf{Reason}\\\cline{1-2}\\m=\dfrac{-30+10}{-2-3}&x-(-y)=x+y\\\\m=\dfrac{-20}{-5}&\text{Addition and subtraction}\\\\m=\dfrac{20}{5}&\text{The negatives cancel out}\\\\m=\dfrac{4}{1}&\text{Divide the numerator and denominator by $5$}\\\\m=\boxed{4}&\dfrac{x}{1}=x\end{aligned}$

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3 years ago

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Solve the equation by completing the square.Round to the nearest hundredth if necessary. X^2–x–7=0

lesya [120]

NOTE: $\left(x-\dfrac{1}{2}\right)^2=x^2-x+\dfrac{1}{4}$

$x^2-x-7=0 \\ \\ \underbrace{x^2-x+\frac{1}{4}}_{\left(x-\frac{1}{2}\right)^2}-7\dfrac{1}{4}=0 \\ \\ \\ \left(x-\frac{1}{2}\right)^2=7\frac{1}{4}=\frac{29}{4} \qquad /\sqrt{}\\ \\ \left|x-\frac{1}{2}\right|=\dfrac{\sqrt{29}}{2} \\ \\ x_1=\dfrac{1}{2}-\dfrac{\sqrt{29}}{2} \approx -2.19 \\ \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{29}}{2} \approx 3.19$

ANSWER: c)

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3 years ago