Asinging contest eliminates contestants after each round. the number of contestants that go into round n is equal to the number of contestants in the previous round raised to the n-1/n -th power, rounded to the nearest integer. the number of contestants in round 2 is 243. how many contestants will be in round 5?

9

Step-by-step explanation:

Since we know the number of contestants in Round 2, we could go step by step and calculate the number of remaining contestants in Round 3, then Round 4 and finally Round 5.

However, we can do it without calculating. So in Round 3, there will be:

R3 = R2^((n-1)/n), where n is number of the round

R3 = R2^(2/3)

For Round 4, the number will be:

R4 = (R2^(2/3))^(3/4)

And Round 5 will be:

R5 = ((R2^(2/3))^(3/4))^(4/5)

Threes and fives in exponents will cancel out and we'll remain only with:

R5 = R2^(2/5)

R5 = 249^(2/5)

R5 = 9.08

Rounded to the nearest integer, the number of contestants in Round 5 is 9.

correct answer:

i^31 = -i

im pretty sure its

(8,15)

step-by-step explanation: