Let r(cos O + i sin O) be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3 = 125(cos 288 + i sin 28)
so r^3 = 125 and cos 3O + i sin 3O = cos 288 + i sin 288
so r = 5 and 3O = 288 + 360p and O = 96 + 120p
so one cube root is 5 (cos 96 + i sin 96)
Im a little rusty at this stuff Its been a long time.
Im not sure of the other 2 roots
sorry cant help you any more
125-45=M would be a simple way of writing the amount taken from the original
sin(4π21). Explanation: Notice that this fits the form of the sine subtraction formula: sin(A−B)=sin(A)cos(B)−cos(A)sin(B).
Answer:
8.356 rounded to the nearest hundredth=8.360
X less 16.5 as it implies that 16.5 is being reduced from x
