I don't think there is a formula to work this out. It needs to be done by trial and error. Start with the fact that the square of 12 is 122 and work backwards through the possible combinations. It shouldn't take long to arrive at the solution. HINT: You shouldn't need to go below 5. ------------------------------------------------------------------------------------------------------------ EDIT: This is an algebra problem that is not simply solved by 'Trial and Error'. It is as easy as creating equations to solve it. LET: x=smaller odd number, y=larger odd number. y=x+2 (Equation1) We know that if 2 is added to an odd number, we get the next odd number. x^2 + y^2 = 130 (Equation2) [''^2'' is ''squared''] We need to have only one variable in the equation to solve the equation easier. Since we know from Equation1 that y=x+2, we replace all y's in Equation2 with x+2 . So, x^2 + (x+2)^2 = 130 2x^2 - 4x - 126 = 0 2(x^2 - 2x - 63) = 0 2(x+7)(x-9) = 0 x = -7; y = -7 + 2 = -5 or x = 9; y = 9 + 2 = 11
11 and 12
all numbers in the range -√130 to √130, or approximately -11.4 to 11.4 lie between the square root of 130. So two numbers that lie between the square root of 130 are 1 and 2. ---------------------------------------- If you meant: What two whole numbers does the square root of 130 lie between? Then √130 lies between 11 and 12.
11 and 12
The numbers that go into 130 are the numbers that divide 130 evenly without leaving a remainder. These numbers are 1, 2, 5, 10, 13, 26, 65, and 130. This is because these numbers can be multiplied by another number to equal 130.
36 49 64 81 100 121
11 and 12
The numbers are -134, -132 and -130.
7 and 9 72+92 = 130
To find a square number between 110 and 130, we need to calculate the square root of the numbers in that range. The square root of 110 is approximately 10.49, and the square root of 130 is approximately 11.40. Therefore, the square number between 110 and 130 is 121, which is the square of 11.
They are 128, 129 and 130.
all numbers in the range -√130 to √130, or approximately -11.4 to 11.4 lie between the square root of 130. So two numbers that lie between the square root of 130 are 1 and 2. ---------------------------------------- If you meant: What two whole numbers does the square root of 130 lie between? Then √130 lies between 11 and 12.
If you have two consecutive integers then one of them must be odd and the other must be even. The square of an odd integer must be odd, the square of an even integer must be even. The sum of an odd number and an even number must be odd. Thus, the sum of squares of any two consecutive numbers must be odd. Therefore, the question has no valid answer.
They are 31, 32, 33 and 34.The numbers are 31, 32, 33 and 34.
There are two consecutive even integers of negative 130: -66 and -64.
11 and 12
Since 13 is not an even number, a list of even numbers cannot start with 13. Perhaps you mean 3 consecutive even numbers greater than 13: 14, 16, 18. Or perhaps you mean 3 consecutive even numbers starting with the digits 1 and 3: 130, 132, 134.
Just calculate the square numbers: 1x1, 2x2, 3x3, 4x4, etc., until you get a square that is larger than 130.