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:
Step-by-step explanation: I think you skip count by 50s or something like that. So, it has to be 64.
Answer:
a. 400
b. 4$
Step-by-step explanation:
2$ equals 0 Quantity
3$ equals 100 Quantity
4$ equals 200 Quantity
We can see that starting from 2$, a dollar will increase the Quantity by 100
Number of combinations = (6x5) ÷ 2 = 15
Answer:There are 15 combinations
The answer is an intersect between both lines.
