Angelique says that finding the absolute value of a number is the same as finding the opposite
1 answer:
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Ok, so it seems to be the square root of the cube root of 2
we just convert to exponential
remember
and
![\sqrt[n]{x^m} =x^ \frac{m}{n}](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%3Dx%5E%20%5Cfrac%7Bm%7D%7Bn%7D%20)
therfor
![\sqrt{ \sqrt[3]{2} }= \sqrt{2^ \frac{1}{3} } =( 2^ \frac{1}{3})^ \frac{1}{2} =2^ \frac{1}{6}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Csqrt%5B3%5D%7B2%7D%20%7D%3D%20%5Csqrt%7B2%5E%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%28%202%5E%20%5Cfrac%7B1%7D%7B3%7D%29%5E%20%5Cfrac%7B1%7D%7B2%7D%20%3D2%5E%20%5Cfrac%7B1%7D%7B6%7D%20%20)
last choice is correct
we just convert to exponential
remember
therfor
last choice is correct
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