They are 14, 16 and 18.
9 and 10 9 + 2(10) = 29
9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70
18, 20 and 22
55 and 57
They are 14, 16 and 18.
The numbers are 14, 16 and 18.
9 and 10 9 + 2(10) = 29
9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70
The answer would be 10 12 and 14... 14 x 3 = 42 and 2(10 + 12) = 44. So the product of the largest integer and three is two less than twice the sum of the lower integers.
Let's represent the three consecutive odd integers as ( 2n-1 ), ( 2n+1 ), and ( 2n+3 ), where ( n ) is an integer. According to the given information, twice the smallest integer ( 2(2n-1) ) is equal to seven more than the largest integer ( 2n+3 ). Setting up the equation, we have ( 4n-2 = 2n+3+7 ). Solving this equation gives us ( n = 6 ). Therefore, the three consecutive odd integers are 11, 13, and 15.
18, 20 and 22
Suppose the smallest integer is A. The next two even numbers are A+2 and A+4. Using the information supplied we can form an equation: 2A - 14 = A + A+2 + A+4 Rearranging: 2A - 14 = 3A + 6 -20 = A So the three integers are -20, -18 and -16.
The integers are 14 and 7.
no one wants to know the answer. its freaking math
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If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.