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Shtirlitz [24]
3 years ago
8

Circle D was dialated to form circle D'. The diameter of circle D is 5 units and the diameter of circle D' is 15 units. What is

the relationship between the areas of circle D and circle D'?
A The area of circle D' would be 3 times the area.

B The area of circle D' would be 9 times the area.

C The area of circle D' would be 1/3 of the original area.

D The area of circle D' would be 1/9 of the original area.
​
Mathematics
1 answer:
Sphinxa [80]3 years ago
3 0

Answer:

Let A1 be original area

A2 dilated Area

A1=pi*d^2/4

=3.143*5^2/4

A1=19.63 unit square

A2= 3.143*15/4

A2=176.74 unit square

A2 is 9 times greater than A1

Hence the answer is B

Step-by-step explanation:

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b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum
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Answer:

{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

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