The coins in the stores register total $12.50. the register only contains 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 register?

First, identify the variables:n = amount of nickelsd = amount of dimes Next, setup the equations based on what you know.  The first equation is:n + d = 28 For the second equation, we know that a dime is worth 10¢ and a nickel is 5¢, so it should be:0.05n + 0.10d = 1.95 This a three-step In one formula (you can use any of them; most people use the simplest one), single out the variable on one sideApply the first formula into the second formula, and solve it to get the value of one variableApply the answer from the second formula into the first formula, and solve it to get the value of the other variable  Step One:n + d = 28n + d - d = 28 - dn = 28 - d Step Two:0.05n + 0.10d = 1.95
(0.05 * (28 - d)) + 0.10d = 1.951.40 - 0.05d + 0.10d = 1.951.40 + 0.05d = 1.951.40 - 1.40 + 0.05d = 1.95 - 1.400.05d = 0.55d = 11 Step Three:n = 28 - dn = 28 - 11n = 17 Your answer should be 17 nickels and 11 dimes. You can double check by applying the variables into both formulas. n + d = 2817 + 11 = 2828 = 28
0.05n + 0.10d = 1.95(0.05 * 17) + (0.10 * 11) = 1.950.85 + 1.10 = 1.951.95 = 1.95 I hope this helps.

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.