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n^^ = n! - n^2, where n is a positive integer. For how many
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01 Oct 2017, 04:43
01 Oct 2017, 04:43
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Question Stats:
64% (01:01) correct
35% (01:53) wrong based on 17 sessions
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n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?
A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^
Kudos for correct solution.
A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^
Kudos for correct solution.
Re: n^^ = n! - n^2, where n is a positive integer. For how many
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01 Oct 2017, 07:20
01 Oct 2017, 07:20
Bunuel wrote:
n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?
A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^
Kudos for correct solution.
A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^
Kudos for correct solution.
Not clear with the ques?
It has asked for how many values of n, n^^<0, or should it be; for what value of n, n^^<0. PLz clarify??
Re: n^^ = n! - n^2, where n is a positive integer. For how many
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01 Oct 2017, 07:59
01 Oct 2017, 07:59
2
I interpreted the first way.
Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18
So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.
Thus, the answer is B!
Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18
So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.
Thus, the answer is B!
