2(5)+1+3 =14
-2, -3, -4, -5. When these numbers are added together you will get -14.
There are 14C5 = 14*13*12*11*10/(5*4*3*2*1) = 2002
9and5
-2 and -3Check:(-2) + (-3) = -5(-2)(-3) = 6Thus -2 and -3 are not the required numbers. let's find them: x + y = -6xy = -5 y = -x -6x(-x - 6) = -5-x^2 - 6x = -5x^2 + 6x = 5x^2 + 6x + 9 = 5 + 9(x + 3)^2 = 14x + 3 = (+ & -)square root of 14x = -3 (+ & -)square root of 14x = -3 + square root of 14 or x = - 3 - square root of 14y = -x - 6y = 3 - square root of 14 - 6 or y = 3 + square root of 14 - 6y = -3 - square root of 14 or y = -3 + square root of 14Check:(-3 + square root of 14) + (-3 - square root of 14) = -6(-3 + square root of 14)(-3 - square root of 14) = -5 ?(-3)^2 - (square root of 14)^2 = -5 ?9 - 14 = -5Check also tow other numbers.
perimeter= add all numbers So, its 3+2+4+5=14
2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 3 + 7 = 10 5 + 7 = 12 7 + 7 = 14
The 2 numbers in a row are multiplied, then you subtract one and that's the next number in the sequence. EG, 2x2 = 4 4-1=3 2x3=6 6-1=5 3x5=15 15-1=14 number sequence is 2, 2, 3, 5, 14
The set of numbers is bimodal with modes 3 and 5.
n*3; n-1 {Where "n" is the previous answer starting at 1. 1, 1*3=3, 3-1=2, 2*3=6, 6-1=5, 5*3=15, 15-1=14, 14*3=42, 42-1=41.
The prime numbers in the range 1-14 are 2, 3, 5, 7, 11 and 13.
Eleven plus five minus two equals foureen. 11+5-2=14.