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TEA [102]
2 years ago
10

A sphere has a radius of 12 inches. A second sphere has a radius of 9 inches. What is the difference of the volumes of the spher

es?
The volume of the larger sphere is π
cubic inches greater than the volume of the smaller sphere.
Mathematics
1 answer:
just olya [345]2 years ago
3 0

Answer:

Solution given;

for sphere

radius [r]:12inches

another

radius [R]:9inches

Now

difference in volume of sphere:

volume of sphere of having radius [r-R]

=\frac{4}{3} πr³-\frac{4}{3} πR³

=\frac{4}{3} π(12³-9³)

= 1332π cubic inches

The volume of the larger sphere is <u>___1332</u>__π cubic inches greater than the volume of the smaller sphere.

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A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
2 years ago
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UkoKoshka [18]

Answer:

I think 87.2% is the correct answer.

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2 years ago
I need the answer to this problem please help me
LekaFEV [45]
I'm pretty sure both B and C equal 12 as well
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Cynthia Besch wants to buy a rug for a room that is 19 ft wide and 32 ft long. She wants to leave a uniform strip of floor aroun
Rainbow [258]

Answer:

The rug should be 15 ft wide and 28 ft long.

Step-by-step explanation:

I have attached a figure that represents the situation.

The the rug is l by h, the width of the strip of floor is w.

We are told that Cynthia can only afford 420 square feet of carpeting; therefore, it must be that

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From the figure we see that

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We expand this equation and get:

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using the quadratic equation we get two solutions:

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since the second solution, namely w=23.5, is larger than one of the dimensions of the room (is greater than 19 ft) it cannot be the width of the strip; therefore, we take w=2 to be our solution.

Now we find the dimensions of the rug:

l=32-2(2)=28\\\\h=19-2(2)=15

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3 years ago
A window washer clean 38 windows in two hours at this rate how many windows did he clean seven hours
madreJ [45]

Answer:

133 windows

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