Can I please have some help.?
1 answer:
check the picture below.
we know the perimeter is 28, the values we know are 4, 2 and 3.
the right-side of the figure, is just 4 + 2, therefore 6.
the bottom-right side of the figure, is just a mirror length of above, therefore 3.
the other two lengths, are twins, and since if we pluck out the values from the perimeter, we have
28 - 4 - 2 - 3 - 6 - 3 = 20.
since the last two lengths are twins, then each one is 20/10 or 10, as you see there in the picture.
now, we can get the full are of the figure, because is just two rectangles, as you see there, so the area is just
(4*10) + (3*6), namely 58.
notice, the figure there, has two triangles.
one on the left-hand-side, has a base of (10+3) and a height of 4.
the other triangle on the right-hand-side has a base of 3, and a height of 6.
we can just get the area of both triangles, and then subtract them from the total area of the figure, what's leftover, is the shaded section.
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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