The new mean would be 7. The mean is the average of the data. (x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10=21 [(x1/3)+(x2/3)+(x3/3)+(x4/3)+(x5/3)+(x6/3)+(x7/3)+(x8/3)+(x9/3)+(x10/3)]/10=? [(1/3)(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)]/10=? [(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10]/3= 21/3=7
There are infinite varieties of dimensions as 4'x6' , 3'x8', 12'x2', ..., etc
(3x - 5x) + (x2 - 6x2) = -2x -5x2 = -x(2 + 5x)
x^2+x^2=2x^2
Cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is measured in radians
x8
(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
(x1+x2+...+x5)/5 = 34 (x1+x2+...+x5)= 5 * 34 = 170 [(x1+x2+...+x5) + x6]/6 = 35 170 + x6 = 6*35 = 210 x6 =210-170=40 Answer is 40
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
It is 36.
The greatest common factor of a variable, raised to different powers, is the LOWEST of the powers - in this case, x to the power 2.
The infinite series is 1 - x2/2! + x4/4! - x6/6! + ...
x6 - 27 = (x2 - 3)*(x4 + 3x2 + 9) Danny rocks
x12 = (x6)2 = (x2)6 = x2x2x2x2x2x2
The new mean would be 7. The mean is the average of the data. (x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10=21 [(x1/3)+(x2/3)+(x3/3)+(x4/3)+(x5/3)+(x6/3)+(x7/3)+(x8/3)+(x9/3)+(x10/3)]/10=? [(1/3)(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)]/10=? [(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10]/3= 21/3=7
There are infinite varieties of dimensions as 4'x6' , 3'x8', 12'x2', ..., etc
It is an expression that can be simplified to: 4x2-42