Q: If Andy has 16 coins. Some of them are dimes and some are nickels. The combined value of his dimes and nickels is 1.35. How many dimes and nickels does he have?

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Helen has twice as many dimes as nickels and five more quarters than nickels the value of her coins is 4.75 how many dimes does she have?

he has 27 dimes. to add on, he has 8 nickels to make your total of 3.10

11 dimes.

ms lynch has 21 coins in nickels and dimes. their total value is 1.65. how many of each coin does she have

Ten it each group.

1/35 or 2.86%

There are no dimes in nickels. But 25 dimes have the same monetary value as 50 nickels have.

There are no dimes in nickels, but 20 dimes have the same purchasing value as 40 nickels have.

Tjats a pretty simple question. Anthony has. 2 pennies, 4 dimes, and 4 nickels

56 pennies, 11 nickels, 16 dimes, 5 quarters = $3.96

All coins such as quarters, nickels, dimes, and half dollars

2 quarters 5 dimes 10 nickels 50 pennies

Nickels are worth five cents and dimes are worth ten.

The denominations for 1959 were: Half dollars, Quarters, Dimes, Nickels and Cents.

47 Quarters 83 Nickels

7

40 Dimes and 8 Nickles

You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.

You have 40 pennies, 8 nickels and 2 dimes. 40+40+20=100

Two nickels equal the value of one dime, therefore fourteen (14) nickels would equal the value of seven (7) dimes.

Let the Number of Dimes be D, then the number of nickels = 73 - DBy value, 5(73 - D) + 10D = 580 : 365 - 5D + 10D = 580 : 5D = 215 : D = 43There are 43 dimes.

there r 40 nikels

Pennies, nickels, dimes, or what?

Nickels, dimes are only slightly smaller in height than nickels.

We have 3 times as many nickles as dimes. Let X equal our number of dimes then 3X equals our number of nickles. Now multiply these values by the value of the coins they represent. So we get : 10(X) + 5(3X)=150 cents or $1.50 next : 10X + 15X = 150 25X = 150 Divide both sides of the equation by 25 X = 6 so we have six dimes = 60 cents we have 3(X) nickels or 3(6) nickels or 18 nickels and 18 times 5 cents = 90 cents 60 cents in dimes plus 90 cents in nickels = $1.50