Find the cube roots of 8. Write the answer in a + bi form.

z = 2, -1 + sqrt(3)i, -1 + sqrt(3)i

Step-by-step explanation:

z³ = 8[cos(0) + isin(0)]

z = 8^⅓ = 2

z = {8[cos(2pi) + isin(2pi)]}^⅓

z = 2[cos(2pi/3) + isin(2pi/3)]

z = -1 + sqrt(3)i

z = {8[cos(4pi) + isin(4pi)]}^⅓

z = 2[cos(4pi/3) + isin(4pi/3)]

z = -1 - sqrt(3)i

56

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?