To make a dollar u u can u use 2 dimes,1 nickel,2 quarters,25 pennies
Here's a shocking idea:
Nobody needs the answer. Figuring out the pattern is much more important
than getting the answer, and it's the reason they gave you the question.
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.
To find the number of ways to make 75 cents using pennies, nickels, dimes, and quarters, you can use a combinatorial approach. By considering the different combinations of each coin type, you can set up an equation based on their values. This problem can also be solved using generating functions or dynamic programming methods to count the possibilities efficiently. The exact number of combinations requires detailed calculations, but there are numerous ways to achieve this with the given coin denominations.
To determine the number of ways to make a dollar using quarters (25 cents), dimes (10 cents), and nickels (5 cents), we can represent the problem using a combination of these coins. By systematically counting combinations for different quantities of quarters (0 to 4), and then varying the number of dimes and nickels to reach the total of 100 cents, we find there are 292 different combinations to make a dollar with these coins.
One solution is 10 nickels, because if you put in a dollar it can be broken into: 2 quarters (50 cents) so 50 cents are left, and you have 10 coins left to use If these 10 coins are all nickels, (total: 50 cents), the problem is solved! .5 +.5 =1 dollar
110 nickels and 40 dimes my little brother had the same problem
2 quarters, 3 dimes 3 nickles, 5 pennies 7 dimes, 6 nickles 1 quarter, 7 dimes, 5 pennies 1 quarter, 3 dimes, 9 nickles It;s only four but maybe you van think of the fith
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.
Oh, isn't that a lovely question! Let's see, to make 36 cents, you can use different combinations of coins like quarters, dimes, nickels, and pennies. There are several ways to do this, and it's like creating a beautiful painting with different colors and textures. Just remember, there's no right or wrong way to make 36 cents with coins, so have fun exploring all the possibilities!
My third period amazing math class did this as their problem of the day today, and we discovered that there are 50 ways to make 50 cents. We included a 50-cent piece as well!
Call the number of nickels n and the number of quarters q. From the problem statement, q = n + 78 and 25q + 5n = 3240. Substituting the first equation into the second yields 25(n + 78) + 5n = 3240, or 30n = 3240 - 1950 or n = 1290/30 = 43 nickels.
The question is incomplete. Please post a new version with the rest of the problem.
There are two possible solutions: You could have one quarter, two dimes, two nickels and forty-five pennies, or you could have two dimes, eight nickels and forty pennies. An easy way of approaching this problem is to start by imagining that you have fifty pennies. You have the right number of coins, but are fifty cents short. Instead of adding other coins, you replace pennies with them: replacing a penny with a nickel gains four cents, a dime gains nine, and a quarter gains twenty-four. You can't possibly use more than two quarters, so there are few cases to consider there: If you replace two pennies with quarters, you've gained forty-eight cents, so you only need two more; but any further replacement will give you too much. If you use one quarter, you need to make up twenty-six more cents in steps of four or nine; it's easy to see that two of each works. Finally, with no quarters, you need to gain fifty cents using increments of four or nine; this yields the second solution.
One solution is 10 nickels, because if you put in a dollar it can be broken into: 2 quarters (50 cents) so 50 cents are left, and you have 10 coins left to use If these 10 coins are all nickels, (total: 50 cents), the problem is solved! .5 +.5 =1 dollar
Call the unknown number of cents c. From the problem statement, c = (8 X 25) + (25 X 5) = 325 cents, which equals $3.25.
Well honey, you can make 26 cents with 26 pennies, 5 nickels and 1 penny, 2 dimes and 6 pennies, or just 1 quarter and 1 penny. That's a total of 4 different ways to make 26 cents using those coins. Math can be a real hoot sometimes, can't it?
110 nickels and 40 dimes my little brother had the same problem
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.