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?
11 dimes and 5 nickels.
Let's represent the number of nickels as 'n' and the number of dimes as 'd'. The value of n nickels is 0.05n, and the value of d dimes is 0.10d. We can set up the equation 0.05n + 0.10d = 1.65 to represent the total value of the coins. Solving this equation will give us the values of n and d.
If n is the number of nickels and d the number of dimes, then the equations are:n + d = 160 (total number of coins) 5n + 10d = 1050 (total value). And I have thought through to the answer.
there r 40 nikels
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
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?
Tjats a pretty simple question. Anthony has. 2 pennies, 4 dimes, and 4 nickels
11 dimes and 5 nickels.
7
Let's represent the number of nickels as 'n' and the number of dimes as 'd'. The value of n nickels is 0.05n, and the value of d dimes is 0.10d. We can set up the equation 0.05n + 0.10d = 1.65 to represent the total value of the coins. Solving this equation will give us the values of n and d.
Ten it each group.
If n is the number of nickels and d the number of dimes, then the equations are:n + d = 160 (total number of coins) 5n + 10d = 1050 (total value). And I have thought through to the answer.
there r 40 nikels
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.