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mestny [16]
3 years ago
9

A rectangular prism has a volume of 16y^4+16^y^3+48y^2 cubic units. Which could be the base area and height of the prism?

Mathematics
1 answer:
Liula [17]3 years ago
5 0
\bf \textit{volume of a rectangular prism}\\\\
V=Bh~~
\begin{cases}
B=area~of\\
\qquad its~base\\
h=height\\
-----------\\
V=16y^4+16y^3+48y^2
\end{cases}
\\\\\\
16y^4+16y^3+48y^2=Bh\implies 16y^2(y^2+y+3)=Bh

so since its just two factors, you can pick either for the "B" or "h", or one can say that the trinomial is the "B", because it can be factored into a couple more binomials, either way will work.
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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 ar
sleet_krkn [62]

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;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( 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

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