how many ridges in a dime?
613444
The dime has an edge with 118 ridges. This coin is 17.91 mm in diameter and is 1.35 mm thick - it is the smallest, thinnest and lightest US coin.
1092149
One quarter, one nickel, and 7 dimes. :)
462376
20\n\n 12 edges and 8 vertices
210388
10 nickels =  .50      - 1.00\n2 quarters =  .50      -\n\n\n4 dimes       =.40\n1 quarters    =.25\n7 nickels     =.35\n\n2 quarters 10 nickels\n1 quarter 4 dimes 7 nickels\n8 dimes 4 nickels
942574
i think it's either called Dime or Bojangles
1124992
Dimes:\n\n2268 grams  79.8336 ounces  4.9896 pounds\n\nQuarters:\n\n2268 grams  79.8336 ounces  4.9896 pounds\n\nIn other words, exactly the same.
412128
71. 7 continents on the earth\n72. \n73.25 statistics on a baseball card\n74.\n75.\n76.10 little indians\n77.13 original colonies\n78. 10 dimes in a dollar
263065
40 pennies, 8 nickels, 2 dimes  \n\nor\n\n45 pennies, 1 quarter, 2 dimes, 2 nickels
331834
1:1,000,000,000\n\n(One in a Billion)
380874
Or 7:3 if the number of coins
882485
1 Nickel = 5 c\n1 Dime = 10 c\n1 Quarter = 25 c\n5.50$ = 550 c\n\nAssume number of nickels = X\n\nThen number of quarters = X + 4\n\nThen number of dimes = total number of coins – (number of nickels + number of quarters) = 37 – (X + X + 4) = 33 – 2X\n\nnumber of dimes * dime value + number of nickels * nickel value + number of quarters * quarter value = Total value\n\n(33 – 2X) * 10 + X * 5 + (X + 4) * 25 = 550\n330 – 20X + 5X + 25X + 100 = 550\n10X + 430 = 550\n10X = 550 – 430\n10X = 120\nX = 120 / 10\nX = 12\n\nNumber of nickels = X = 12\nNumber of dimes = 33 – 2X = 33 – 2 * 12 = 9\nNumber of quarters = X + 4 = 16\n\n12 * 5c + 9 * 10c + 16 * 25c = 60c + 90c + 400c = 550 c = 5.50 $\n\n12 + 9 + 16 = 37 coins\n\nAnswer: Jo has 9 dimes :-)
547564
Estimate it=use about 70 sq meters, as supplied by Mars, and figure 1.5 candies per centimeter long, and 2 per centimeter high-so 6 candies per cubic cm. I would estimate about 5, though, because I dont think they are as small as I estimated.  So at 1 million cub cm/cub M, that would get you about 5 X 1 million, X 70, or about 350 million.  Because they are oblate spheroids, they would pack slightly tighter than the flat rectangular boxes, so I'd add a fudge factor of about 20%.  Say anything from a low end of about 200 million to a high end of about 500 million.
347016
Normally 336, but anywhere between 300 and 500 is acceptable. Different balls have different numbers of dimples.
669176
p=pennies, n=nickels, d=dimes\n3p= 3,  2p+1n=7,  2p+1d=12\n\n3n=15,  2n+1p=11,  2n+1d=20\n\n3d=30,  2d+1p=21,  2d+1n=25\n\n1p+1n+1d=16\nSorry, I could only come up with 10.  :)
668507
You would use three dimes and a nickel. The dime being one coin the nickel being the other coin.
957112
The answer was A..       dime, nickel and quarter or something like that.
259226
here's one with coins that was on one of the worksheets i have:\n\nlet's say, for example, that you have $8.75 in dimes and quarters and there are 65 coins. how many dimes and quarters are there?\n\nso there would be 2 equations:\nd+q=65 (which is how many coins there are)\n\nand\n\n10d+25q=875 (how much money there is in dimes and quarters)\n\nfrom there, you'd solve it up and down like\n\n 10d+25q=875\n+  d+  q= 65 (Multiply by 10)\n---------------------------\n 10d+25q  =875\n+(-10d-10q=-650)\n-----------------------\n       15q= 225\n         q=15\n\nthen you plug in 15 for q for one of the equations.\n\n15+d=65\n65-15=50\nd=50\n\nyou'd have 50 dimes and 15 quarters. (i hope)\n\ni also hoped this helped ^^;