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)
Answer:
12.35
Step-by-step explanation:
95
*13
____
285
+950
____
1235
then you need to move the decimal number 2 spaces, because you removed a total of 2 space from the original numbers.
12.35
1: m=-4
2: b=-4
3: -9=k
4: C=-11.4
5:X<10
I’m in middle school so if u show me I’ll be able to help I’m one of those people
