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
Answer:
( -
,
)
Step-by-step explanation:
Given the 2 equations
9x + 8y = 3 → (1)
6x - 12y = - 11 → (2)
To eliminate the y- term multiply (1) by 1.5
13.5x + 12y = 4.5 → (3)
Add (2) and (3) term by term
(6x + 13.5x) + (- 12y + 12y) = (- 11 + 4.5)
19.5x = - 6.5 ( divide both sides by 19.5 )
x =
= - 
Substitute this value into either of the 2 equations and solve for y
Using (1), then
- 3 + 8y = 3 ( add 3 to both sides )
8y = 6 ( divide both sides by 8 )
y =
= 
Solution is (-
,
)
Answer:
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Answer:
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Step-by-step explanation: