(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
The infinite series is 1 - x2/2! + x4/4! - x6/6! + ...
I will use the quotient rule here. d/dx(f(x)/g(x) = g(x)*f'(x) - f(x)*g'(x)/[g(x)]2 x3*1/x - ln(x)*3x2/(x3)2 x3/x - 3ln(x)x2/x6 x2 - 3ln(x)x2/x6 = - 3ln(x)/x4 =========
The GCF is x2
x8
The GCF is x2.
The GCF is 1.
(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
Guessing they are powers and no operational symbol is between the two (ie a multiplication is assumed), then: x6 x2 = x6 + 2 = x8 Otherwise, re-ask with words for missing symbols, eg "what is x to the power 6 plus x to the power 2 simplified"
(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 GCF is x2
The GCF of anything compared to itself is itself.
The infinite series is 1 - x2/2! + x4/4! - x6/6! + ...
x6 - 27 = (x2 - 3)*(x4 + 3x2 + 9) Danny rocks