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3 years ago

Find the circumference of the circle if the radius is 4 inches

Mathematics

2 answers:

AlladinOne [14]3 years ago

5 0

The answer is 25.13

Kitty [74]3 years ago

3 0

Answer:

25.13

Step-by-step explanation:

C = d( $\pi$ )

d = r(2) in this case 4(2) = 8 = d

so 8( $\pi$ ) = C

C ≈ 25.13

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2 years ago

The length of a rectangle is 3 more inches than the width. Find the length and width of the rectangle is the perimeter of the re

emmainna [20.7K]

Answer:

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2 years ago

What is the prime factorization of 110?

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Answer:

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2 years ago

How to solve 8 = 2^x + 4?

bearhunter [10]

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3 0

3 years ago

Read 2 more answers

Please help me with geometry

spayn [35]

For each sub-problem shown below, I'm using the distance formula to compute the distance between the two points

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For the points (-10,2) and (-2,2), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((-2-(-10))^2+(2-2)^2)

d = sqrt((-2+10)^2+(2-2)^2)

d = sqrt((8)^2+(0)^2)

d = sqrt(64+0)

d = sqrt(64)

d = 8

------------------------------------------------------

For the points (-3,-1) and (5,1), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-(-3))^2+(1-(-1))^2)

d = sqrt((5+3)^2+(1+1)^2)

d = sqrt((8)^2+(2)^2)

d = sqrt(64+4)

d = sqrt(68)

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For the points (-3,-2) and (1,3), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((1-(-3))^2+(3-(-2))^2)

d = sqrt((1+3)^2+(3+2)^2)

d = sqrt((4)^2+(5)^2)

d = sqrt(16+25)

d = sqrt(41)

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For the points (-3,-5) and (-2,-4), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((-2-(-3))^2+(-4-(-5))^2)

d = sqrt((-2+3)^2+(-4+5)^2)

d = sqrt((1)^2+(1)^2)

d = sqrt(1+1)

d = sqrt(2)

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For the points (0,0) and (5,5), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-0)^2+(5-0)^2)

d = sqrt((5)^2+(5)^2)

d = sqrt(25+25)

d = sqrt(50)

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For the points (1,2) and (1,-10), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((1-1)^2+(-10-2)^2)

d = sqrt((0)^2+(-12)^2)

d = sqrt(0+144)

d = sqrt(144)

d = 12

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For the points (1,2) and (5,2), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-1)^2+(2-2)^2)

d = sqrt((4)^2+(0)^2)

d = sqrt(16+0)

d = sqrt(16)

d = 4

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For the points (2,3) and (10,9), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((10-2)^2+(9-3)^2)

d = sqrt((8)^2+(6)^2)

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d = 10

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See the attached image for how the answers match up (take note of the letter labels)

3 0

3 years ago