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Q: How many 50c coins make 2 dollars?
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Related questions

What 20 coins make up 6?

One combination anyway: American: 3.5 dollars in 50c is 7 coins 2 dollars in quarters = 8 coins 5 dimes = 5 coins Euro: 3.5 euro in 50c is 7 coins 2.40 euro in 20c is 12 coins and 1 10 cent coin


How many 50c coins make up $3.00?

大变


If you have 20.40 Of that 2.40 are in 20c Coins and 10.00 in 1.00 coins The rest are in 50c Coins how many 50c Coins do you have?

16 $0.50 coins


If you have 50 cents 2 dollars and 1 dollar coins in how many ways can you make up 10 dollars?

it's 36. Only use 50c or $1 or $2, 3 ways only use 50c and $1, 9 ways, only use 50c and $2, 4 ways, only use $1 and $2, 4 ways. use 50c, $1 and $2, 16 ways. total is $36


Ann has 20 coins in her purse They are 10c 20c and 50c coins and the total value of the coins is 5 dollars If she has more 50c than 10c coins- how many 10c coins does she have?

Well, isn't that a happy little math problem we have here! If Ann has more 50c coins than 10c coins and the total value is $5, we can figure out that she must have 4 10c coins. Let's give Ann a little encouragement as she counts her coins and solves this puzzle.


How many ways can you make a dollar with 21 coins only using 50c 25c 10c 5c and 1c coins?

3 dimes 13 nickles 5 pennies


Which 5 coins would you need to make 3.90?

$2, $1, 50c, 20c, 20c.


What is the fewest number of bills and coins the make 11.50?

Three. 1x $10, 1x $1 and 1x 50c


What 3 different coins will make 75 cents?

If in america, 3 quarters make 75c. With the euro, a 50c, a 20c, and a 5c make 75c


What four coins make 1.75 euro?

umm so 1 euro+50c+20c+5c=1.75 euro:)


Can you make 74 cents with 7 coins?

Indeed. 74 cents can be made with the following set of 7 coins: 50c 10c 10c 1c 1c 1c 1c


What is the largest possible value of coins is it possible to have without having any coins which will add to 4 dollars coins are 5c 10c 20c 50c 1 dollar 2 dollar?

Tentatively I will say $4.35, with for instance: 19 x 20c 1 x 50c 1 x 5c It's possible that a rigorous method exists, but this is just a fudged guess. Any advances on it?