what is the missing number: 4, 5, 8, 17, 44
41/44 and 1 over 4
1
5-4 = 1 8-5 = 3 = 1*3 17-8 = 9 = 3*3 44-17= 27 = 9*3 let the number be x difference between 44 and x = 27*3 = 81 x = 44 + 81 = 125
Get the last two digits of the number. If that number is divisible by 4 then that number is divisible by 4. Example : 2393844 The last two digits are 44 and since 44/4 = 11 or 44 is divisible by 4 then the number 2,393,844 is divisible by 4.
4 and 4/10
44 = 256
125 (each number is increased by the exponential 3n from the previous: +1, +3, +9, +27, +81)
The sequence continues: 125, 368, 1097, 3284, 9845, 29528, 88577, 265724, 797165, ...
57
41/44 and 1 over 4
To find the missing number in the sequence 16, 4, 12, 36, 9, 27, 44, 11, we can look for a pattern. The first set of numbers appears to alternate between two sequences: the first sequence (16, 12, 9, 44) and the second sequence (4, 36, 27, 11). Following this pattern, the missing number, which follows the last number in the second sequence (11), should be 33. Thus, the missing number is 33.
Let the numbers be x and y:- If: x+3y = -4 and 3x+y = 44 Then: 3x+9y = -12 and 3x+y = 44 Subtracting 3x+9y = -12 from 3x+y = 44: -8y = 56 => y = -7 By substitution: x = 17 and y = -7 Check: 17+3(-7) = -4 Check: 3(17) -7 = 44 Therefore the two numbers are: 17 and -7
1
Let the numbers be x and y:- If: x+3y = -4 and 3x+y = 44 Solving the above simultaneous equation by elimination method: x = 17 and y = -7 Check: 17+(-21) = -4 Check: 51+(-7) = 44
5 - 4 = 1 = 3^0 8 - 5 = 3 = 3^1 17 - 8 = 9 = 3^2 44-17 = 27 = 3^3 X - 44 = 3^4 = 81 X = 125
5-4 = 1 8-5 = 3 = 1*3 17-8 = 9 = 3*3 44-17= 27 = 9*3 let the number be x difference between 44 and x = 27*3 = 81 x = 44 + 81 = 125
4 < 44 so 4/44 is a proper fraction, not a mixed number.