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Archy [21]
2 years ago
11

The volume of 10 drops of a liquid is 0.1 fluid ounces. What is the volume of 1000 drops?

Mathematics
1 answer:
harina [27]2 years ago
6 0
We know that 10 drops of liquid is .1 fluid ounces. What we could do is divide 0.1 by 10 to find what 1 drop of liquid is, 0.1 divided by 10 is equal to 0.01 fluid ounces.
What we can then do knowing what 1 fluid ounce is, is multiply it by 1000.

0.01 * 1000 = 10

The volume of 1000 drops is 10
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Answer:

see explanation

Step-by-step explanation:

A = πr² ← r is the radius and r = \frac{1}{2} d ← d is the diameter

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π (\frac{1}{2} d)² = A, that is

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Answer:

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Step-by-step explanation:

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2 years ago
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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