V=5*5+10*x, where V is the value in cents.
40 Dimes and 8 Nickles
Let's represent the number of nickels as "N" and the number of dimes as "D." Since you have $2.35 in total, we can set up the equation 0.05N + 0.10D = 2.35. Additionally, we know that there is one more dime than nickels, so we have the equation D = N + 1. By substituting the second equation into the first and solving simultaneously, we find that N = 9 and D = 10. Therefore, you have 9 nickels and 10 dimes.
In order to eliminate the decimal point, let us work with cents. So, let $ 4.50 = 450 cents. If the number of nickels is x, then the value of x nickels is 5x cents. Since the number of dimes is twice the number of nickels, then we can write the number of dimes as 2x, and the value of 2x dimes as 10(2x) or 20x cents. So we have: 5x + 20x = 450 25x = 450 divide by 25 to both sides x = 18 the number of nickels 2x = 2 x 18 2x = 36 the number of dimes Thus, Cameron has 18 nickels and 36 dimes. Check: 18 nickels = 18 x 5 cents = 90 cents or $ 0.90 36 dimes = 36 x 10 cents = 360 cents or $ 3.60 0.90 + 3.60 = 4.50? 4.50 = 4.50 True
slope
The equation of line in terms of y , x coordinates and intercept can be written as : y= m x+ c where m is the slope of the equation and the c is value of intercept.
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.
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 are no dimes in nickels. But 25 dimes have the same monetary value as 50 nickels have.
Let N be the number of nickels. The value of nickels is 0.05N and the value of dimes is 0.10(2N-12). We can form the equation 0.05N + 0.10(2N-12) = 2.55 and solve for N. After solving, we find that there are 9 nickels.
Nickels are worth five cents and dimes are worth ten.
11 dimes and 5 nickels.
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?
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.
Two nickels equal the value of one dime, therefore fourteen (14) nickels would equal the value of seven (7) dimes.
he has 27 dimes. to add on, he has 8 nickels to make your total of 3.10
To find the number of dimes, nickels, and pennies in $1.10, we can set up a system of equations. Let x represent the number of dimes, y represent the number of nickels, and z represent the number of pennies. The total value equation is 10x + 5y + z = 110 cents. Since there are no constraints on the number of coins, there are multiple solutions to this problem. One possible solution is 8 dimes, 0 nickels, and 10 pennies.
Nickels, dimes are only slightly smaller in height than nickels.