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Sergeeva-Olga [200]
3 years ago
5

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)

4 0
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


5 0
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.
8 0
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}

3 0
3 years ago
Read 2 more answers
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)

5 0
3 years ago
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