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How do you write a numerical or variable expression for the value in cents of n nickels?
answer: 5n
If you have 2.35 in dimes and nickels and you have one more dime than nickels how many of each do you have?
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.
A cash register contains 12.50 and there are only nickels dimes and quarters There are twice as many dimes as nickels and twice as many quarters as dimes How many quarters are in the cash register?
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.
You save 42 nickels and dimes that total 3.35 how many nickels and dimes do you save?
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
You piggy bank has all nickels and quarters. You count up the coins and find out that there are 64 coins and you have a total of 7.40. How many of each type of coin do you have?
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.
What is the value in cents of nickels?
A nickel is worth 5 cents so if you have N nickels their value in cents is 5*N
What is the value in cents of n nickels?
A nickel is worth 5 cents so n nickels have a value of 5n cents.
The value of cents in n Nickels?
5n
How do you write a numerical or variable expression for the value in cents of n nickels?
answer: 5n
You have 7.68 with equal amounts of dimes nickels and pennies. How many dimes do you have?
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.
How would you write an algebraic expression for the number of cents in n nickels and y quarters?
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
10 coins consistinig of nickels and dimes equal 80 cents how many coins 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.
If you have 2.35 in dimes and nickels and you have one more dime than nickels how many of each do you have?
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.
A stack of nickels and dimes is worth 2.55 Find the number of nickels if the number of dimes is12 less than twice the number of nickels?
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.
A cash register contains 12.50 and there are only nickels dimes and quarters There are twice as many dimes as nickels and twice as many quarters as dimes How many quarters are in the cash register?
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.
You save 42 nickels and dimes that total 3.35 how many nickels and dimes do you save?
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 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?
11 dimes and 5 nickels.
