Mathematics, 06.06.2020 02:58, Dalmcdan01

A cone with radius 5 and height 12 has its radius doubled. How many times greater is the volume of the larger cone than the smaller cone? Use a pencil and paper. Explain how the volume of the cone would change if the radius were halved.

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answered: jshu1998

$V = \frac{1}{3} \pi (5)^2 (12)= 314.159$

Now if we increase the radius by a factor of 2 the new volume would be:

$V_f = \frac{1}{3} \pi (2*5)^2 (12)= 1256.637$

And we can find the increase factor for the volume like this:

$\frac{V_f}{V}= \frac{1256.637}{314.159}= 4$

Then if we increase the radius by 2 the volume increase by a factor of 4

If we reduce the radius by a factor of 2 then we will have that the volume would be reduced by a factor of 4.

On the figure attached we have an illustration for the cases analyzed we see that when we increase the radius the volume increase and in the other case decrease.

Step-by-step explanation:

For this case we have the following info given:

r = 5 , h =12

and we can find the initial volume:

$V = \frac{1}{3} \pi r^2 h$

And replacing we got:

$V = \frac{1}{3} \pi (5)^2 (12)= 314.159$

Now if we increase the radius by a factor of 2 the new volume would be:

$V_f = \frac{1}{3} \pi (2*5)^2 (12)= 1256.637$

And we can find the increase factor for the volume like this:

$\frac{V_f}{V}= \frac{1256.637}{314.159}= 4$

Then if we increase the radius by 2 the volume increase by a factor of 4

If we reduce the radius by a factor of 2 then we will have that the volume would be reduced by a factor of 4.

On the figure attached we have an illustration for the cases analyzed we see that when we increase the radius the volume increase and in the other case decrease.

A cone with radius 5 and height 12 has its radius doubled. How many times greater is the volume of t

answered: Guest

$\frac{a^{2} }{4}$

answered: Guest

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt