Let one of the number be y and other be y+2
According to question:
y(y+2) = 360
y2 + 2y = 360
y2 + 2y - 360 = 0
which is of the form ay2 + by + c = 0
Value of y can be calculated using the formula: y = {-b ± (b2 - 4ac)1/2}/2a
Here a = 1, b = 2 and c = -360
Putting these values in the formula:
y = {-2 ± (22 - 4x1x-360)1/2}/2x1 = {-2 ± (4 + 1440)1/2}/2 = {-2 ± 38}/2
y = (-2 + 38)/2 = 18 or y = (-2 - 38)/2 = -20
If y = 18 then the other number = y + 2 = 20
If y = -20 then the other number = y + 2 = -18
So, we have two solutions (18, 20) and (-20, -18).
9 and -40 9 + -40 = -31 9 x -40 = -360
If consecutive angles of a 4-sided figure were complementary, then all four of themwould add up to 180°. But that's a problem, because we all know that the sum of allfour angles in any 4-sided figure is always 360°. So the answer must be: 'No'.
360 degrees = Total number, T. So a sector of x degrees is equivalent to x*T/360 units.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
The same as the greatest common divisor between 360 and 72 (72 is the remainder of the division of 432 and 360).Apply this reasoning repeatedly (that's the Euclidian algorithm), until one of the two numbers is zero.
The numbers are 18 and 20.
(6)(5)(4)(3) = 360
The numbers are 352, 354, 356, 358 and 360.
The value of the greatest integer is 76. This occurs when the set of five consecutive integers is all even numbers.
No four consecutive whole numbers have a sum of 360. 88 + 89 + 90 + 91 = 358 . . . too small 89 + 90 + 91 + 92 = 362 . . . too big
The numbers are: 101.451506616 and 3.548493384
510
They are are: 6 and 60
The two numbers are [-A ± sqrt(A2 - 1440)]/2
The average must be 360/3 ie 120 so the integers are 118, 120 and 122.
36x10 would be the easiest sum to get 360. Or even 360x1!!
They would be 30 and 12(30+12=42;30*12=360)*=This means times,or multiply by.