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Arte-miy333 [17]
3 years ago
6

While keeping the height of the cone constant, move the vertex of the cone so that it is not directly above the center of the ba

se. Check the Allow
oblique and Freeze height boxes to assist you. Does changing the position of the vertex affect the volume if the height remains the same? Why or
Why not? (Hint: Think of the cone as a stack of disks of decreasing radius.)
Mathematics
2 answers:
podryga [215]3 years ago
7 0

Answer:

No, it doesn't.

Step-by-step explanation:

If we think of the cone as a stack of disks of decreasing radius, and we move the vertex of the cone while keeping the height of the cone, the number of disks don't change nor their radius, it only changes their position respect the original disposition. Then, the volume of the cone remains the same.

Dmitry [639]3 years ago
4 0

Answer:

ANSWER option 1:

Changing the position of the vertex doesn't change the volume because the height remains the same, all that changes is the position of hypothetical disks. Changing the vertex without changing the height has no effect on the volume of said cone.

ANSWER option 2:

The volume does not change. Cavalieri's principle states that if two three-dimensional figures have the same height and have the same cross-sectional area at every level, then they have the same volume. Cavalieri's principle is applicable to the cone before and after the vertex is changed, assuming that the height is fixed.

Step-by-step explanation:        (both answers above are correct)

option 1:

I got that answer correct on edmentum (PLATO)

option 2:

sample answer on edmentum (PLATO)

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Step-by-step explanation:

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3 0
3 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
3 years ago
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