Let's see if this is true.
Assume that their sum is 48. So,
x + (x +1) + (x + 2) = 48
3x + 3 = 48 Subtract 3 at both sides;
3x = 45 Divide 3 to both sides;
x = 15
x + 1 = 15 + 1 = 16
x + 2 = 15 + 2 = 17 So,
15 + 16 + 17 = 48
Answer: The sum of three consecutive integers such as 15, 16, and 17 is 48.
There are three quarters in 12...
3x+5y=48 5y=48-3x-3x+5y=12 -3x+(48-3x)=12-6x=-36x=65y=48-3(6)5y=30y=6(6,6)
3x - 5 = 48 So 3x = 53 and x = 53/3 = 17.66...
c = 4, c squared = 16, 3c squared = 48, 48 + 9 = 57.
48
14,16.18
The integers are 44, 46 and 48.
They are 14, 16 and 18.
The numbers are -18, -16 and -14.
45, 46, 47, 48 = 186
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15, 16 and 17
There are no such answers. The sum of even numbers is even. The sum of odd numbers could be even, but the sum of three consecutive odd numbers is always odd. One could swap the questions to get potentially answerable questions: Sum of 3 consecutive EVEN integers = 150 48 + 50 + 52 = 150 unfortunately, we can already see that if we were to make these odd integers (add 1 to each), we would get 153, not 151. Sum of 3 consecutive ODD integers = 151 49 + 51 + 53 = 153 47 + 49 + 51 = 147 We can see that, whenever we slide our three consecutive integers up one, we raise the value of the sum up 3, and if we jump from odd to odd, or even to even, we raise each integer by 2, so we raise the sum by 6.
The three consecutive even numbers with a sum of 138 are 44, 46 and 48.
There are no two consecutive integers which sum to make 48. The closest you can get is if you drop the integer part of the problem and state simply that they have to have a difference of 1 between them. In which case it would be 23.5 and 24.5.
Three consecutive integers will always total a multiple of three. Specifically, they will always add up to three times the middle number. Divide 48 by 3. Choose the the integers on either side of that. 15 + 16 + 17 = 48 The three numbers could be described as x, x + 1 and x - 1. That adds up to 3x. 3x = 48 x = 16 x + 1 = 17 x - 1 = 15
The numbers are 46, 48 and 50.