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fredd [130]
3 years ago
5

A person is using a new tennis ball launching machine that is 15 inches by 15 inches by 14. Assuming the machine uses tennis bal

ls that are spherical and 2.7 inches in diameter, how many tennis should fit in one of the top of the machine.
Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

305 tennis balls should fit

Step-by-step explanation:

The volume of the machine is length * width * height

So Volume of Machine = 15 * 15 * 14 = 3150

Volume of Sphere is  V=\frac{4}{3}\pi r^3

Where

V is volume and r is radius (half of diameter)

Since 2.7 is diameter, 2.7/2 = 1.35 inches is radius

Volume of 1 tennis ball = V=\frac{4}{3}\pi r^3\\V=\frac{4}{3}\pi (1.35)^3\\V=10.3 (rounded to 1 decimal)

We simply divide the volume of the machine by volume of 1 tennis ball to get number of tennis balls:

3150/10.3 = 305.82

the fractional amount (.82) won't be possible so the max number of balls possible is 305 balls

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Roman55 [17]

Given:

A line through the points (7,1,-5) and (3,4,-2).

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The parametric equations of the line.

Solution:

Direction vector for the points (7,1,-5) and (3,4,-2) is

\vec {v}=\left

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y=1+3t

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Therefore, the required perimetric equations are x=7-4t, y=1+3t and z=-5+3t.

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3 years ago
In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
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Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}
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Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
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Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

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d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

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Answer:

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I hope I helped and have a wonderful day!

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