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AlekseyPX
3 years ago
14

Henry constructed circle A with a radius of 4 units. He then created a sector

Mathematics
1 answer:
salantis [7]3 years ago
4 0

Answer:

C

Step-by-step explanation:

area of circle is found by \pir^{2}

r=4 so 4*4=16

so A=16\pi

we are finding area of shaded part so 45/360(16\pi)

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umka21 [38]
The answer is 3/2,-6
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3 years ago
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What is the volume of the composite figure? Explain your work. A complete answer should include how you broke up the figure, whi
Sphinxa [80]

Answer:

15,000\:\mathrm{mm^3}

Step-by-step explanation:

The composite figure consists of a square prism and a trapezoidal prism. By adding the volume of each, we obtain the volume of the composite figure.

The volume of the square prism is given by V=s^2\cdot h, where s is the base length and h is the height. Substituting given values, we have: V=14^2\cdot 30=196\cdot 30=5,880\:\mathrm{mm^3}

The volume of a trapezoidal prism is given by V=\frac{b_1+b_2}{2}\cdot l\cdot h, where b_1 and b_2 are bases of the trapezoid, l is the length of the height of the trapezoid and h is the height. This may look very confusing, but to break it down, we're finding the area of the trapezoid (base) and multiplying it by the height. The area of a trapezoid is given by the average of the bases (\frac{b_1+b_2}{2}) multiplied by the trapezoid's height (l).

Substituting given values, we get:

V=\frac{14+24}{2}\cdot (30-14)\cdot 30,\\V=19\cdot 16\cdot 30=9,120\:\mathrm{mm^3}}

Therefore, the total volume of the composite figure is 5,880+9,120=\boxed{15,000\:\mathrm{mm^3}} (ah, perfect)

Alternatively, we can break the figure into a larger square prism and a triangular prism to verify the same answer:

V=30^2\cdot 14+\frac{1}{2}\cdot10\cdot 16\cdot 30=\boxed{15,000\:\mathrm{mm^3}}\checkmark

8 0
2 years ago
Suppose that angle P = 3x + 14 and angle Q = 3x - 27. If these angles are supplementary, angle Q would need to be ___ degrees.
adelina 88 [10]

Answer:

Q = 69.5 degrees

Step-by-step explanation:

3x + 14 + 3x - 27 = 180

6x - 13 = 180

6x = 193

x = 32.1667

Q = 3(32.1667) - 27

Q = 69.5

5 0
3 years ago
A kilometer is larger than a meter True or false
stira [4]

A Kilometer is 1,000 times larger than a meter

5 0
3 years ago
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(1 point) A bucket that weighs 3.6 pounds and a rope of negligible weight are used to draw water from a well that is 78 feet dee
Deffense [45]

Answer:

The total work done in pulling the bucket to the top of the well is approximately 3,139.1 ft·lb

Step-by-step explanation:

The given parameters are;

The mass of the bucket, W = 3.6 pounds

The depth of the well, h = 78 feet deep

The mass of water in the bucket = 38 ponds

The rate at which the water is pulled up = 2.9 feet per second

The rate at which water is leaking from the bucket, \dot m = 0.1 pounds per second

We separate and find the work done for lifting the bucket and the water individually, then we add the answers to get the solution to the question as follows;

The work done in lifting bucket empty from the well bottom, W_b = W × h

∴  W_b = 3.6 pounds × 78 feet = 280.8 ft-lb

The work done in lifting bucket empty from the well bottom, W_b = 280.8 ft-lb

The time it takes to lift the bucket from the well bottom to the top, 't', is given as follows;

Time, t = Distance/Velocity

The time it takes to pull the bucket from the well bottom is therefore;

t = 78 ft./(2.9 ft./s) ≈ 26.897

The time it takes to pull the bucket from the well bottom to the top, t ≈ 26.897 s

The mass of water that leaks out from the bucket before it gets to the top, m₂, is therefore;

m₂ = \dot m × t

∴ m₂ = 0.1 lbs/s × 26.897 s = 2.6897

The mass of the water that leaks, m₂ = 2.6897 lbs

The mass of water that gets to the surface m₃ = m - m₂

∴ m₃ = 38 lbs  - 2.6897 lbs ≈ 35.3103 lbs

Given that the water leaks at a constant rate the equation representing the mass of the water as it is lifted can b represented by a straight line with slope, 'm' given as follows;

The slope of the linear equation m = (38 lbs - 35.3103 lbs)/(78 ft. - 0 ft.) = 0.03448\overline 3 lbs/ft.

Therefore, the equation for the weight of the water 'w' can be expressed as follows;

w = 0.03448\overline 3·y + c

At the top of the well, y = 0 and w = 38

∴ 35.3103 = 0..03448\overline 3 × 0 + c

c = 35.3103

∴ w = 0.03448\overline 3·y + 35.3103

The work done in lifting the water through a small distance, dy is given as follows;

(0.03448\overline 3·y + 38) × dy

The work done in lifting the water from the bottom to the top of the well, W_{water}, is given as follows;

W_{water} = \int\limits^{78}_0 {0.03448\overline 3 \cdot y + 35.3103 } \, dy

\therefore W_{water} = \left [ {\dfrac{0.03448\overline 3 \cdot y^2}{2}   + 35.3103 \cdot y\right ]^{78}_0

W_{water} = (0.034483/2 × 78^2 + 35.3103 × 78) - (0.034483 × 0 + 38 × 0)  ≈ 2,859.1

The work done in lifting only the water, W_{water} ≈ 2,859.1 ft-lb

The total work done, in pulling the bucket to the top of the well, W = W_b + W_{water}

∴ W = 2,859.1 ft.·lb + 280.8 ft.·lb ≈ 3,139.1 ft·lb

The total work done, in pulling the bucket to the top of the well, W ≈ 3,139.1 ft·lb.

6 0
3 years ago
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