There are infinitely many polynomials of order 4 that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
One possible answer, based on a polynomial of order 3 is:U(n) = (4963*n^3 - 29652*n^2 + 54236*n - 39541)/3 for n = 1, 2, 3, ...
% rate= 9/450 * 100%= 0.02 * 100%= 2%9 as percentage of 450 is 2%
2*-2, 3*-3,4*-4,etc. * is a times sign.
1^2 = 13^2 = 95^2 = 25so 7^2 = 49 (missing number)9^2 = 81Following the same pattern of odd numbers squared,the next number should be 11^2 = 121, not 10^2 = 100.The (100) seems to be out of sequence according to thepattern.
+2+4+5+6+2+4+2+4+5+6+8+9+3+1+2+6+7
It is: 2/9 times 100/1 = 22.22%
% rate= 9/450 * 100%= 0.02 * 100%= 2%9 as percentage of 450 is 2%
(9/2) + 8 - 100 = 4.5 + 8 - 100 = 87.5
9^2+9+9 +9/9 81+9+9+1=100
2*-2, 3*-3,4*-4,etc. * is a times sign.
1^2 = 13^2 = 95^2 = 25so 7^2 = 49 (missing number)9^2 = 81Following the same pattern of odd numbers squared,the next number should be 11^2 = 121, not 10^2 = 100.The (100) seems to be out of sequence according to thepattern.
2 / 9 x 100% = 22.222%
(9 + 1)^(4/2) = 10^2 = 100.
percentage = 22.22%% rate:= 2/9 * 100%= 0.2222 * 100%= 22.22%
9/100 is out of 100 right? You need it out of 200, so multiply top and bottom by 2. 9/100 x 2/2 = 18/200 So the answer is 18.
+2, +3, +2
+2+4+5+6+2+4+2+4+5+6+8+9+3+1+2+6+7
The lowest common multiple of 4, 2, 9, 5 and 100 is 900. Since 4, 2 and 5 are factors of 100, you can ignore them. Then all you have to do is multiply 9 times 100.