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Lesechka [4]
4 years ago
12

Your lab develops a synthetic compound that is extremely light but very durable. You test it by making a solid ball with a radiu

s of 20.0 millimeters. Which of the following is closest to the volume of the synthetic compound ball, in cubic millimeters?  a.) 1,260  b.) 1,670 c.) 25,100 d.) 33,500 e.) 50,200
Mathematics
1 answer:
Maksim231197 [3]4 years ago
6 0
The volume of a spherical shaped object is given by

V= \frac{4}{3} \pi r^3

Given that the <span>solid ball has a radius of 20.0 millimeters, the volume of the solid ball is given by

V= \frac{4}{3} \pi(20)^3\approx33,510mm^3

Therefore, the </span><span>closest to the volume of the synthetic compound ball, in cubic millimeters is 33,500 cubic milimeters.</span>
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If you would like to solve 1/3 / 6, you can do this using the following steps:

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8 0
3 years ago
Triangle ABL is an isosceles triangle in circle A with a radius of 11, PL = 16, and ∠PAL = 93°. Find the area of the circle encl
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Answer:

The area of the circle enclosed by line PL and arc PL is approximately 37.62 square units

Step-by-step explanation:

The given parameters in the question are;

The radius of the circle, r = 11

The length of the chord PL = 16

The measure of angle ∠PAL = 93°

The segment of the circle for which the area is required = Minor segment PL

The shaded area of the given circle is the minor segment of the circle enclosed by line PL and arc PL

The area of a segment of a circle is given by the following formula;

Area of segment = Area of the sector - Area of the triangle

In detail, we have;

Area of segment = Area of the sector of the circle that contains the segment) - (Area of the isosceles triangle in the sector)

Area of a sector = (θ/360)×π·r²

Where;

r = The radius of the circle

θ = The angle of the sector of the circle

Plugging in the the values of <em>r</em> and <em>θ</em>, we get;

The area of the sector enclosed by arc PL and radii AP and AL = (93°/360°) × π × 11² ≈ 98.2 square units

Area of a triangle = (1/2) × Base length × Height

Therefore;

The area of ΔAPL = (1/2) × 16 × 11 × cos(93°/2) ≈ 60.58 square units

∴ The area of the segment PL ≈ (98.2 - 60.58) square units = 37.62 square units

Therefore, the area of the shaded segment PL ≈ 37.62 square units

More examples on area of a shaded segment can be found here:

brainly.com/question/22599425

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hope this helps :)

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Answer: the answer is 0.435

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