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10 dimes

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โˆ™ 2011-03-17 22:00:51
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A polynomial of degree zero is a constant term

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Q: Cash register has 4.00 in dimes nickels and quarters how many are dimes?
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The coins in the store's cash register total 12.50 The cash register contains only nickels dimes and quarters There are twice as many dimes as nickels There are also twice as many quarters as di?

The coins in the store's cash register total $12.50. The cash register contains only nickels, dimes, and quarters. There are twice as many dimes as nickels. There are also twice as many quarters as dimes. How many quarters are in the cash register?


The coins in the stores cash register total 12.50?

The coins in a cash register amount to $12.50. One coin combination that would produce this total is 40 quarters, 19 dimes, 2 nickels, and 50 pennies. Another combination is 20 quarters, 50 dimes, 45 nickels, and 25 pennies.


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.


The coins in the store's cash register total 4.00 The cash register contains only nickels dimes and quarters There is an equal number of each coin How many dimes are in the cash register?

There are 10 of each coin in the cash register. 10 quarters (2.50) plus 10 dimes (1.00) plus 10 nickels (.50) = 4.00 The formula to solve this problem is: (X x .25) + (X x .10) + (X x .05) = 4.00 X (.25+.10+.05) = 4.00 X = 4.00/.4 = 10Woah that is really hard math


The coins in a cash register total 12.50 and there are only nickels dimes and quarters. There are twice as many dimes as nickels and there are twice as many quarters as dimes. How many many dimes are?

You need to define variables for the different types of coins, write the corresponding equations, then solve them. One equation for each fact. Here are the equations:5N + 10D + 25Q = 1250 D = 2N Q = 2D


The coins in the store's cash register total 12.50. The cash register contains quarters nickels and dimes how many quarters are in the register?

Assuming the plural is used because there are more than 1 of each coin, then there are at least 2 quarters and at most 48 quarters. If the plural is being used in a generic way, meaning at least 1 of each coin must be present, then there is at least 1 quarter and at most 49 quarters. If you want a more specific answer, then more specific data is needed in the question.


If you have 2.50 in cash how many nickels and dimes are there?

There's no specific answer, except to say that the number of nickels must be even (otherwise the total would end in 5 rather than 0). You could have combinations like50 nickels and no dimes40 nickels and 10 dimes25 dimes and no nickels22 dimes and 6 nickelsand so on. Each one, and many other possibilities, all add up to $2.50


The coins in a cash register total 12.50 and there are only nickels dimes and quarters. There are twice as many dimes as nickels and there are twice as many quarters as dimes. How many of each coin?

There are 10 nickels, 20 dimes and 40 quarters in the cash register. The 10 nickels is 10 x 5 cents or 50 cents. The 20 dimes is 20 x 10 cents or 200 cents. The 40 quarters is 40 x 25 cents or 1000 cents. Converting and adding these, we get $0.50 + $2.00 + $10.00 = $12.50, which is the sum given in the question. Let's work through it. The number of nickels is N, the number of dimes is D and the number of quarters is Q. These are our variables in this problem. We don't know how many of them there are, and their numbers could vary. That's why we call them variables. We might also call them unknowns, too. A nickel is 5 cents, so the value of the nickels is the number of nickels, which is N, times the value of the nickel, which is 5 cents. That's 5N here. A dime is 10 cents, so the value of the dimes is the number of dimes, which is D, times the value of the dime, which is 10 cents. That's 10D here. A quarter is 25 cents, so the value of the quarters is the number of quarters, which is Q, times the value of the quarter, which is 25 cents. That's 25Q here. The sum of the values of the coins was given as $12.50, or 1250 cents, because we are working with coins, whose values are measured in cents. Further, we can write this expression as 5N + 10D + 25Q = 1250 on our way to the answer. Of the last two facts, the first was that there were twice as many dimes as nickels. We could write that as D = 2N because said another way, there are twice the number of dimes as nickels. We might also say that for every nickel, there are 2 dimes, so doubling the number of nickels will give us the number of dimes. The last fact is that there were twice as many quarters as dimes. We could write that as Q = 2D because said another way, thre are twice the number of quarters as dimes. We might also say that for every dime, there are 2 quarters, so doubling the number of dimes will give us the number of quarters. The last two bits of data we have allow us to solve the problem, because the do something special for us. Each bit of data expresses one variable in terms of another. That means we can make substitutions in our expressions for the sum of the values of the coins. Let's put up or original expression, and then do some substitutions. 5N + 10D + 25Q = 1250 This is the original expression. We know that D = 2N, so lets put the 2N in where we see D and expand things a bit. 5N + 10(2N) + 25Q = 1250 5N + 20N + 25Q = 1250 We changed the "look" of the expression, but we didn't change its value. Let's go on. We know that Q = 2D, so lets put that in. 5N + 20N + 25Q = 1250 5N + 20N + 25(2D) = 1250 5N + 20N + 50D = 1250 We're almost there. Remember that D = 2N, and we can substitute that in here. 5N + 20N + 50D = 1250 5N + 20N + 50(2N) = 1250 5N + 20N + 100N = 1250 Groovy! We have substituted variables and now have an expression with only one variable in it! Let's proceed. 5N + 20N + 100N = 1250 125N = 1250 We're close! N = 1250/125 = 10 N = 10 The number of nickels is 10, and because the nickel is 5 cents, the value of these coins is their number times their value, or 10 x 5 cents = 50 cents = $0.50 We were told the number of dimes was twice the number of nickels. This means that since there are 10 nickels, there will 2 x 10 or 20 dimes. And 20 x 10 cents = 200 cents = $2.00 We were also told the number of quarters was twice the number of dimes. This means that since there are 20 dimes, there will be 2 x 20 or 40 quarters. And 40 x 25 cents = 1,000 cents = $10.00 If we add the values of the coins, we should get the $12.50 that we were told was in the register. $0.50 + $2.00 + $10.00 = $12.50 We're in business. The value of each denomination of coins adds up to the given value of all the coins in the register. Piece of cake.


How much in coins is 1000 in USA cash?

This is a very vague question. Are you asking "how many coins equal $1,000.00 in cash?" Depending on the denomination of coins it can be 1,000 One dollar coins = $1,000.00 or 100,000 One cent coins = $1,000.00 or any combination of cents, nickels, dimes, quarters, half dollars or dollar coins.


Customer buys a six pack of soft drinks for 2.79 and gives the cashier a 10.00 bill The cash register is out of 1.00 bills The customer's change is?

The answer is $7.21. The customer gets a 5 dollar bill, 8 quarters ,2 dimes and 1 penny.


The cash register says to give 47cent in change to the customer What are the correct coins to give this customer?

one quarter two dimes two pennies


How do you program royal cash register?

i want to program my cash register

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