20 times with a remainder of 3 or 20.75 times
There are 83 coins. If there are N nickels then there are (83 - N) dimes. Davida has nickels worth 5N and dimes worth 10(83 - N) but. 5N + 10(83 - N) = 695 5N + 830 - 10N = 695 5N = 830 -695 = 135 therefore N = 135/5 = 27 : D = 83 - N = 83 - 27 = 56 Davida has 27 nickels and 56 dimes.
83-n
83
83 - 28 = 5555- 19 = 36.
The difference is 83 minus 56 = 27
one, remainder 27
56+27 = 83
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56Factors of 83: 1, 83GCF (56, 83) = 1
20 times with a remainder of 3 or 20.75 times
There are 83 coins. If there are N nickels then there are (83 - N) dimes. Davida has nickels worth 5N and dimes worth 10(83 - N) but. 5N + 10(83 - N) = 695 5N + 830 - 10N = 695 5N = 830 -695 = 135 therefore N = 135/5 = 27 : D = 83 - N = 83 - 27 = 56 Davida has 27 nickels and 56 dimes.
The GCF is 7.
3/83/821/56 = 3/83/821/56 = 3/8
there are 9 prime numbers between 56 and 101 these are 59, 61, 67, 71, 73, 79, 83, 89, 97
101 to 95 =-6 95 to 83 =-12 83 to 74 =-9 6 to 12 is x2 12 to 9 is -3 Therefore,-9x2 is -18 74-18=56 Ans: 56
9. They are: 59, 61, 67, 71, 73, 79, 83, 89 & 97.
There are nine: 59 61 67 71 73 79 83 89 97.