Two adjacent sides of a rhombus are represented by 5x + 7 and 6x -1. find the value of x

A group of friends decided to buy bulk chocolate from a candy store. If some friends each want 5 over 7 of a kilogram of a choco

AnnZ [28]

Each person = 5/7 kg

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Number of people that can share the 10kg of chocolate:
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$10 \div \dfrac{5}{7}$

$= 10 \times \dfrac{7}{5}$

$= 14$

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Answer: 14 can share the chocolate.
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7 0

3 years ago

What is 3/5 divided by 1/3 ??

zhannawk [14.2K]

1 4/5 would be the answer

3 0

3 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

Mariana [72]

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).

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

7 0

3 years ago

PLEASE HELP!!!! Find the value of -7 - 8 - (-8) - 7. 0 14 -14 -30

miskamm [114]

-7 - 8 - (-8) - 7 × 0(14) - 14 - 30

(I think that's what that says)

Do PEMDAS

0(14) = 0

-7 × 0 = 0

You're left with -7 - 8 - (-8) - 14 - 30

Go from left to right

-7 - 8 = -15

-15 - (-8) = -7

-7 - 14 = -21

-21 - 30 = -51

7 0

3 years ago

Read 2 more answers

The Powerball lottery is open to participants across several states. When entering the powerball lottery, a participant selects

mr Goodwill [35]

Answer and Step-by-step explanation:

Number of ways = 59C5 × 35= 59!/(54!*5!)* 35= 175223510

Odds are thus 1 in 175223510

b) 1 million - Number of ways = 59!/(54!*5!) =5006386*35/34 = 5153632

Odds are thus 1 in 5153632

6 0

3 years ago