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Rudiy27
3 years ago
10

How to find the circumference of a circle

Mathematics
1 answer:
Alborosie3 years ago
5 0

Answer:

Step-by-step explanation:

The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference (c).

circle diameter radius

A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d)

Half of the diameter, or the distance from the midpoint to the circle border, is called the radius of the circle (r).

The circumference of a circle is found using this formula:

C=π⋅dorC=2π⋅r

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Answer:

Lower Quartile: 62

Upper Quartile: 81

Interquartile Range: 19

Step-by-step explanation:

To find the lower quartile, you want to find the median from the minimum to the median.

49, 55, 62, 64, 67

The median of this is 62. Therefore, 62 is the lower quartile.

To find the upper quartile, you want to find the median from the median to the maximum.

76, 79, 81, 82, 83

The median of this is 81. Therefore, 81 is the upper quartile.

To find the interquartile range, you subtract the upper and lower quartile.

81-62=19

The difference is 19. Therefore, the interquartile range is 19.

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Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
  • sin (270°-θ)  =  -cos θ
  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
  • cot (270°+θ)  =  -tan θ
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