36 is a perfect square number.
The formula for finding the nth square number is n^2. Therefore, the 3rd square number is 3^2 = 9, and the 6th square number is 6^2 = 36. To find the sum of the 3rd and 6th square numbers, you add them together: 9 + 36 = 45. So, the sum of the 3rd and 6th square numbers is 45.
They are: 36 and 64
How about: 36+61 = 97
x2 + 9 = 36
The number is 4 (4+3) * (4+3) = 49 (4+2) * (4+2) = 36 49 - 36 = 13
"the sum of 2" and "twice" "a number" "is" "36"2+2x=36 â–
sum of 14th square number and 10th square number
64 = 28 + 36
Oh, dude, the sum of the square of 6 and 8 is 36 + 64, which equals 100. So, like, if you square 6, you get 36, and if you square 8, you get 64. Add them together, and boom, you've got 100. Easy peasy lemon squeezy.
Part1: Finding probability of getting sum as a perfect square. Maximum sum of both the dice is (6+6) equal to 12. Up to 12, the perfect squares are: 1, 4 and 9. Getting a sum of 1 from two dice is not possible. So, we are left with 4 and 9. To get 4, the combination can be: (2,2) or (1,3) or (3,1). This means, to get the sum as 4, the probability is [3/36]. To get 9, the combination can be: (3,6) or (6,3) or (5,4) or (4,5). This means, to get the sum as 9, the probability is [4/36]. Therefore,the total probability of getting the sum as a perfect square is: [(3/36)+(4/36)]=[7/36]. Part2: Finding the probability of getting sum as an even number. The possible even numbers can be 2, 4, 6, 8, 10 and 12. But, as 4 is already considered in part1, it should be ignored in this case. The probability of getting sum as 2 is: [1/36] The probability of getting sum as 6 is: [5/36] The probability of getting sum as 8 is: [5/36] The probability of getting sum as 10 is: [3/36] By adding all the above, the probability of getting sum as an even number (ignoring 4) is: [(1/36)+(5/36)+(5/36)+(3/36)]=[14/36]. From part 1 and part 2, we get the total probability as [(7/36)+(14/36)]=[7/12]=0.583333.
10^2 = 6^2 + 8^2 ie 100 = 36 + 64
the sum of 8 and 4 times a number is 36