Which of the following points is in the solution set of y>-x^2+5?
1 answer:
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This question is incomplete, the complete question is;
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.
In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?
Answer:
the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Step-by-step explanation:
Given the data in the question;
Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1
this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.
So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'
Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;
= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]
= [( n + 4 )!] / [ 5!( n-1 )! ]
= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Answer:
n = , n =
Step-by-step explanation:
6n² - 5n - 7 = - 8 ( add 8 to both sides )
6n² - 5n + 1 = 0 ← in standard form
Consider the product of the factors of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term
product = 6 × 1 = 6 and sum = - 5
The factors are - 3 and - 2
Use these factors to split the n- term
6n² - 3n - 2n + 1 = 0 ( factor the first/second and third/fourth terms )
3n(2n - 1) - 1(2n - 1) = 0 ← factor out (2n - 1) from each term
(2n - 1)(3n - 1) = 0 ← in factored form
Equate each factor to zero and solve for n
3n - 1 = 0 ⇒ 3n = 1 ⇒ n =
2n - 1 = 0 ⇒ 2n = 1 ⇒ n =