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OlgaM077 [116]
3 years ago
13

The figure shows a planetarium’s viewing room, which is in the shape of a hemisphere. The diameter of the hemisphere is 50 feet.

What is the volume of the viewing room of the planetarium? (Use 3.14 for π. Round the answer to the nearest tenth, if necessary. Recall that the formula for the volume of a sphere is .) 10,416.7 cu. ft. 16,354.2 cu. ft. 32,708.3 cu. ft. 65,416.7 cu. ft.
Mathematics
2 answers:
Nimfa-mama [501]3 years ago
8 0
You did not include the figure, but according to the statement the viewing room of the planetarium is in the shape of a hemisphere.

A hemisphere is half a sphere.

So, the volume of a hemisphere is the half of the volume of the complete sphere.

The formula for the volume of a sphere is:

V = (4/3) π (radius)^3

Using π = 3.14 and radius = diameter / 2 = 50 feet / 2 = 25 feet.

You get:

V = (4/3) (3.14) (25 feet)^3 = 65,416.7 feet^3

Answer: 65,416.7 cu. ft.
Jlenok [28]3 years ago
5 0

Answer:

32,708.3 cu. ft

Step-by-step explanation:

First we find the volume of a sphere with the same dimensions

V=\frac{4}{3} \pi r^{3} is the formula for the volume of a sphere

V=\frac{4}{3} \((3.14)25^{3} substitute the radius with 25 because radius is half the diameter and use 3.14 for pie. The answer to that formula is 65,416.6

Since you're finding the volume of a hemisphere, divide the volume of the sphere in half

65,416.6/2= 32,708.3

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IRINA_888 [86]

Answer:

A=1.53\ m^2

Step-by-step explanation:

The length of a chalkboard, l = 1.7

The width of a chalkboard, b = 0.9 m

We need to find the area of the chalkboard. The formula for the area of a rectangle is given by :

A=L\times B\\\\A=1.7\times 0.9\\\\A=1.53\ m^2

So, the area of the chalkboard is equal to 1.53\ m^2.

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Alan needs to manufacture a circular metal plate with a perimeter of `10pi` centimeters. If the allowed error tolerance in the p
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The ideal radius Alan must control is \frac{1}{2\pi } cm.

<h3>Define perimeter of circle.</h3>

The measurement of the circle's perimeter, also known as its circumference, is called the circle's boundary. The area of a circle determines the space it takes up. A circle's diameter is equal to the length of a straight line traced through its center. Usually, it is stated in terms of units like cm or m.

Given data -

Perimeter of circular plate = 10\pi cm

We know that perimeter of a circle is 2\pir

Therefore   10\pi = 2\pir

r = 5 cm

The given error Alan can make is +-1 cm.

Minimum radius is given by

2\pir = 10\pi - 1

r = \frac{10\pi - 1 }{2\pi }

r = 5 - \frac{1}{2\pi }

Maximum radius is given by

2\pir = 10\pi + 1

r = \frac{10\pi + 1 }{2\pi }

r = 5 + \frac{1}{2\pi }

The ideal radius Alan must control is \frac{1}{2\pi } cm.

To know more about perimeter of circle, visit:

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Step-by-step explanation:

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Natalka [10]
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You could have ribbons the size of 1/4 our yard and 1/6 yard, but you would have different amounts of each.
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