For the numbers in
form, convert to polar form:

By DeMoivre's theorem,





For the numbers already in polar form, DeMoivre's theorem can be applied directly:


At second glance, I think the 2s in the last two numbers should also be getting raised to the 3rd and 4th powers:


Answer:36 times for 8 dollars
Step-by-step explanation:
k so first you divide 54 and 12 =4.5 per dollar so you then multiply 4.5 and 8 to get the answer that =36 times for 8 dollars
Answer:
The correct answer is 9:50.
Step-by-step explanation:
9:05+0:45=9:50
Answer:
that answe is b
Step-by-step explanation:
b because i took this test