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2 years ago

Find the area and the circumference of a circle with diameter 8 ft.

Mathematics

1 answer:

masha68 [24]2 years ago

5 0

Answer:

a. Area of circle = 50.24 ft².

b. Circumference of circle = 25.12 ft.

Step-by-step explanation:

Given the following data;

Diameter = 8ft

Pie, π = 3.14

Radius, r = diameter/2 = 8/2 = 4ft

a. To find the area of the circle;

Area of circle = πr²

Substituting into the equation, we have;

Area of circle = 3.14*(4)²

Area of circle = 3.14*16

Area of circle = 50.24 ft²

b. To find the circumference of the circle;

Circumference of a circle = 2πr

Substituting into the equation, we have;

Circumference of circle = 2*3.14*4

Circumference of circle = 25.12 ft.

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we get x > 9, as the simplified form of the inequality.

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3 years ago

Please help will give brainliest if correct

Taya2010 [7]

Answer:

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3 years ago

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The diameter of a circle is 23 inches and the circle has two chords of length 14 inches what is the distance from each chord to

Svetradugi [14.3K]

Answer:

9.12 + 9.12 = 18.24 inches

Step-by-step explanation:

Diameter = 23 inches (given)

Radius = 11.5 inches

2 Chords of length = 14 inches ( You didn't specify if the 14 inches is for both chords or for a single cord. I'll assume it's for two cords 14 and 14inches apart.

To clearly solve this, we'll make some mild assumptions.

Let the perpendicular distance of the chords from the center of the circle to represented as " x and y"

Therefore:

x^2 + 7^2 = 11.5 ^ 2

x^2 + 49 = 132.25

x^2 = 132.25 - 49

x^2 = 83.25

x = √ 83.25

x = 9.12 inches

Since the cords have thesame length (Assumed from the way the question was structured, the distance would still be thesame)

y^2 + 7^2 = 11.5 ^ 2

y^2 + 49 = 132.25

y^2 = 132.25 - 49

y^2 = 83.25

y = √ 83.25

y = 9.12 inches

Therefore, the distance will be :

9.12 + 9.12 = 18.24 inches

Have fun!

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3 years ago

Agroup of friends were working on a student film with a budget of $500, which they

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Answer:

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1 year ago

Read 2 more answers

One more time!

CaHeK987 [17]

Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

$\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}$

so we conclude that
$q(\frac{\sqrt{19} }{2} ) = 1/4$

therefore

$p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right) \right)$

plug $x=\sqrt{19}/2$ into p( q(x) ) to get answer

$p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow$

$\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}$

$p(1/4) = -\dfrac{6\sqrt{19} }{361}$

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3 years ago