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11 months ago

Drag and drop an answer to each box to correctly complete the explanation for deriving the formula for the volume of a sphere

Mathematics

1 answer:

denis-greek [22]11 months ago

5 0

The answers that would fill in the blanks are

2r
a circle
an annulus
1/3πr³
4/3πr³

<h3>What is the Cavalier's principle?</h3>

This principle states that if two solids are of equal altitude then the sections that the planes would make would have to be parallel and also be at the same distances from their bases which are equal such that the volumes of the solids would be equal.

Now we have to fill in the blanks with the solution.

For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius r and height 2r minus the volume of two cones, each with a radius and height of r. A cross section of the sphere is a circle base of cylinder, is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is an annulus_ .The volume of the cylinder with radius r and height 2r is 2πr³, and the volume of each cone with radius r and height r is 1/3πr³. So the volume of the cylinder minus the two cones is 4/3πr³. Therefore, the volume of the sphere is by Cavalieri's principle

Read more on Cavalieri's principle here

brainly.com/question/22431955

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3 years ago

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liq [111]

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3 years ago

1. Find the Least Common Multiple of these two monomials: See picture

gtnhenbr [62]

Answer:

The last choice is correct

$LCM=120a^4b^7c^5$

Step-by-step explanation:

Least Common Multiple (LCM)

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

$12a^4b^2c^5$

$40a^3b^7c^1$

The prime factors of the first monomial are:

$2^2,3,a^4,b^2,c^5$

The prime factors of the second monomial are:

$2^3,5,a^3b^7c^1$

LCM = Multiply $2^3*3*5*a^4*b^7*c^5$

These are all the factors the greatest number of times they occur.

Operating:

$LCM=8*15*a^4*b^7*c^5$

$\boxed{LCM=120a^4b^7c^5}$

The last choice is correct

3 0

3 years ago

Mean Value theorem help! BRAINLIEST if you can explain it clearly!!!

Korolek [52]

$\bf f(x)=ln(x-5)\qquad [6,8]\qquad \cfrac{df}{dx}=\cfrac{1}{x-5}\\\\
-----------------------------\\\\
\textit{mean value theorem}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\
-----------------------------\\\\
f'(c)=\cfrac{1}{c-5}\qquad thus\quad \cfrac{1}{c-5}=\cfrac{f(8)-f(6)}{8-6}
\\\\\\
\cfrac{1}{c-5}=\cfrac{ln(3)-ln(1)}{2}\implies \cfrac{1}{c-5}=\cfrac{ln(3)-0}{2}
\\\\\\
\cfrac{2}{ln(3)}=c-5\implies \boxed{\cfrac{2}{ln(3)}+5=c}$

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3 years ago

Lisa says that 43 is a 2 - digit odd number that is a composite. is she correct? explain

klemol [59]

No, she is not correct.

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43 has no other numbers that can divide into it and leave it a whole number.

8 0

3 years ago

Read 2 more answers