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Aneli [31]
3 years ago
11

HELPPPPPPPPP!!!!!!!!!!!!

Mathematics
1 answer:
WARRIOR [948]3 years ago
3 0

Answer:

(1) commutative property of the addition.

(2) distributive property of the multiplication and associative property of the addition.

(3) addition property of the equality.

(4) multiplication property of the equality.

Step-by-step explanation:

First we start with:

4x + 1 + x + 3 = -11

Here we use the commutative property of the sum, which says that:

A + B = B + A

So we can reorder the terms in a sum, and we can rewrite our equation as:

4*x + x + 3 + 1 = -11

Here we also used the associative property of the sum, that says that:

A + B + C = (A + B) + C = A + ( B + C)

wich means that we can perform the sum in any order we want, then we can add the 3 and the 1

4*x + x + 4 = -11

Now we use the distributive property of the multiplication, which says that:

(A + B)*C = A*C + B*C

Then we can rewrite our equation as:

(4 + 1)*x + 4 = -11

Now we solve the sum in the parentheses to get:

5*x + 4 = -11

Now we can use the addition property of the equality, we can add (-4) in both sides to get:

5*x + 4 - 4 = -11  - 4

5*x = -15

Now we can use the multiplication property of the equality, and multiply both sides by (1/5) to get:

5*x*(1/5) = -15*(1/5)

x = -5

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4 years ago
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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4 years ago
Can anyone please teach me or send me the answer if thus​
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thw answer is c if you count the blocks

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