23*5 = 115 So pick any four numbers: integers, fractions, irrational numbers, whatever. Add them together to give a sum S. Let the 5th number be 115 - S Then: sum of all five numbers = (sum of the first four numbers) + (the fifth number) = S + (115 - S) = 115 And the count of numbers = 5 So mean = Sum/Count = 115/5 = 23
There are 115 of them.
The perimeter of a square with side 115 = 115 X 4 = 460.
the sum of eight is 4+4
(4 - 3) x 10 x 12 - 5 = 115.
115 over 115 plus 2 equals two.
1385. If something plus 115 equals 1500 then 1500 minus 115 equals the something: 1500 - 115 = 1385
23*5 = 115 So pick any four numbers: integers, fractions, irrational numbers, whatever. Add them together to give a sum S. Let the 5th number be 115 - S Then: sum of all five numbers = (sum of the first four numbers) + (the fifth number) = S + (115 - S) = 115 And the count of numbers = 5 So mean = Sum/Count = 115/5 = 23
182. Just curious why you ask so simple sum by internet?
There are 115 of them.
115 ÷ 4 = 28 with remainder 3
The perimeter of a square with side 115 = 115 X 4 = 460.
28.75
The least common multiple of the numbers 115 and 4 is 460.
24/5 - 4/23. You must find a common denominator. Both 5 and 23 are prime, so the LCD is 5 x 23 = 115. So 24/5 = 552/115 & 4/23 = 20/115. 552/115 - 20/115 = 532/115 = 4 72/115
the sum is example 2+2=4, 4 is the sum
Let 'n' be the first odd number,. Then n+ 2 & n + 4 are the second and third odd number. Hence n + (n+2) + ( n + 4) = 339 Add 3n + 6 = 339 3n = 333 n = 111 n + 2 = 113 n + 4 = 115 Hence the three odd consecutive numbers are 111,113 & 115.