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Q: The total value in cents of n nickels?

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answer: 5n

Set up your equations based upon the following: Let N = number of nickles. Let N + 1 = number of dimes. Let 5N = the value (in cents) of all the nickles. Let 10(N + 1) = the value (in cents) of all the dimes. Let 235 = the total value (in cents) of all the coins (nickles and dimes). In English, the value of the dimes plus the value of the nickles equals the total value of the coins. Algebraically, we write: 5N + 10(N + 1) = 235 The distributive law of multiplication lets us rewrite that this way: 5N + 10N + 10 = 235 Grouping lets us write that this way: 15N + 10 = 235 Subtracting 10 from both sides of the equation yields: 15N = 225 Dividing both sides by 15 yields: N = 15 So, we have 15 nickles and 15 + 1, or 16, dimes.

Nickles - 10 $0.50dimes - 20 $2.00quarters - 40 $10.000.50+2.00+10.00 = $12.50 containing 40 quarters---Here's how to solve this with Algebra :Let N be the number of nickels, so that2N is the number of dimes, and2(2N) or 4N is the number of quarters.A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents,and the total in the cash register is $12.50Multiplying by their cents values, we have5(N) plus 10(2N) plus 25 (4N) = 12505N + 20N + 100N = 1250125 N = 1250N= 10So the number of nickels is 10, dimes 20, and quarters 40.

This problem can be solved using a system of equations.If we call the number of nickles n and the number of dimes d, we can write the following two equations:n + d = 42 (The number of nickels and dimes is 42).05n + .1d = 3.35 (5 cents for every nickel plus 10 cents for every dime is $3.35)Then:n + d = 425n + 10d = 335 (Switch to cents instead of dollars, gets rid of decimals)n = 42 - dWe have now just defined n in terms of d. Now every time you see n in the second equation, you can replace it. This is called substitution.5n + 10d = 335 (from above)5(42 - d) + 10d = 335 (substituting for n)Follow through with the algebra, and d (# of dimes) = 25, then there must be 17 nickels.25 dimes, 17 nickels

The idea is to write two equations, one for the number of coins, one for the amount of money. Then solve the equations.Assuming "n" is the number of nickels, and "q" the number of quarters, the equations for the coins, of course, is quite simply: n + q = 64 And the equation for the money (I'll use cents; you can just as well use dollars if you prefer): 5n + 25q = 740 You can solve the first equation for "n", then replace that in the second equation.

Related questions

A nickel is worth 5 cents so n nickels have a value of 5n cents.

A nickel is worth 5 cents so if you have N nickels their value in cents is 5*N

5n

answer: 5n

A nickel is 5 cents and a quarter is 25 cents. If T is the total number of cents, the expression would beT = 5n + 25y

If you let N represent how many of each coin you have, their values are:N pennies = N centsN nickels = 5N centsN dimes = 10N centsso the total value is N + 5N + 10N = 16N cents. That must be equal to $7.68, or 768 cents, so the final equation is 16N = 768Solving for N gives you 768/16, or 48 of each coin.

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.

Set up your equations based upon the following: Let N = number of nickles. Let N + 1 = number of dimes. Let 5N = the value (in cents) of all the nickles. Let 10(N + 1) = the value (in cents) of all the dimes. Let 235 = the total value (in cents) of all the coins (nickles and dimes). In English, the value of the dimes plus the value of the nickles equals the total value of the coins. Algebraically, we write: 5N + 10(N + 1) = 235 The distributive law of multiplication lets us rewrite that this way: 5N + 10N + 10 = 235 Grouping lets us write that this way: 15N + 10 = 235 Subtracting 10 from both sides of the equation yields: 15N = 225 Dividing both sides by 15 yields: N = 15 So, we have 15 nickles and 15 + 1, or 16, dimes.

11 dimes and 5 nickels.

Nickles - 10 $0.50dimes - 20 $2.00quarters - 40 $10.000.50+2.00+10.00 = $12.50 containing 40 quarters---Here's how to solve this with Algebra :Let N be the number of nickels, so that2N is the number of dimes, and2(2N) or 4N is the number of quarters.A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents,and the total in the cash register is $12.50Multiplying by their cents values, we have5(N) plus 10(2N) plus 25 (4N) = 12505N + 20N + 100N = 1250125 N = 1250N= 10So the number of nickels is 10, dimes 20, and quarters 40.

This problem can be solved using a system of equations.If we call the number of nickles n and the number of dimes d, we can write the following two equations:n + d = 42 (The number of nickels and dimes is 42).05n + .1d = 3.35 (5 cents for every nickel plus 10 cents for every dime is $3.35)Then:n + d = 425n + 10d = 335 (Switch to cents instead of dollars, gets rid of decimals)n = 42 - dWe have now just defined n in terms of d. Now every time you see n in the second equation, you can replace it. This is called substitution.5n + 10d = 335 (from above)5(42 - d) + 10d = 335 (substituting for n)Follow through with the algebra, and d (# of dimes) = 25, then there must be 17 nickels.25 dimes, 17 nickels

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.

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