Answer:
<h2>3(cos 336 + i sin 336)</h2>
Step-by-step explanation:
Fifth root of 243 = 3,
Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),
then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).
Equating equal parts and using de Moivre's theorem:
r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240
r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p
So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360
So there are 5 distinct solutions given by:
3(cos 48 + i sin 48),
3(cos 120 + i sin 120),
3(cos 192 + i sin 192),
3(cos 264 + i sin 264),
3(cos 336 + i sin 336)
The answer is B. 8x+6. You multiple 3 and x+4 then combine like terms
140 / 100 * X = 56
X = 56 * 100 / 140 = 5600 / 140 = 40
Answer:
x^2+8x+16
(x+4)^2
Step-by-step explanation:
x^2 +8x
Take the coefficient of the x term
8
Divide by 2
8/2 =4
Square it
4^2 =16
x^2+8x+16
We can factor it into
(x+8/2) ^2
(x+4)^2
