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choli [55]
3 years ago
5

The length around the outside of semicircle O from point L through point M to point N is 14 centimeters. The perimeter of semici

rcle O is about 22.92 centimeters. What is the approximate area of semicircle O? Use 3.14 for . A. 41.83 square centimeters B. 83.67 square centimeters C. 62.46 square centimeters D. 31.23 square centimeters Reset
Mathematics
1 answer:
Luden [163]3 years ago
3 0
The area would be 83.67 cm.

A semicircle is half of a circle.  The perimeter of the semicircle would be half of the perimeter (circumference) of the entire circle.  The formula for circumference is:

C=πd

Using our information, we have  

22.92 = 0.5(3.14)d
22.92 = 1.57d

Divide both sides by 1.57:

22.92/1.57 = 1.57d/1.57
14.6≈d

Since the diameter is 14.6, the radius is 14.6/2 = 7.3.

We use the radius for the area of the semicircle:

A=0.5πr²
=0.5(3.14)(7.3)²
=83.67
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The four vertices of the new quadrilateral is the midpoints of the four sides. (5,3) and (5,-1) is ((5+5)/2,(3-1)/2)=(5,1). Midpoint between (-1,3) and (5,3)=(2,3). Midpoint between (-1,3) and (-1,-1) is (-1,1). Midpoint between (-1,-1) and (5,-1) is (2,-1). So the four vertices are (5,1), (2,3), (-1,1), and (2,-1). The resulting figure is a rhombus, centered at (2,1).
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4 years ago
Solve the exponential equation for<br> 5^6x+5/125^3x−1=5^−5x+2
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3 years ago
In a bag of m&amp;m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&amp;m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
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There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=&#10;\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
4 0
3 years ago
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