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miv72 [106K]
4 years ago
9

Jonah makes a snack mix to take with him on vacation. His favorite snack mix is 5 parts raisins and 4 parts peanuts. He wants to

take 27 cups of snack mix. How many cups of raisins and peanuts does he need
Mathematics
1 answer:
Arisa [49]4 years ago
3 0
(3) 5 parts raisins
(3) 4 parts peanuts
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The sum of a number and 17 is no less than 26
Tema [17]
We put this into an algebraic equation, where <span>x = the number .
x + 17 > 26
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This </span>> is the greater than sign, we use this because it's no less than 26. 
Ideally we would use this sign:
<span>http://etc.usf.edu/clipart/41700/41712/fc_greateq_41712_md.gif 
because it could be equal to 26.

</span><span>The number is <u>greater or equal to 9</u>

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Seven times the first number plus 6 times the second number equals 36. Three times the first number minus ten times the second n
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First number is 6, second number is -1. (6,-1)
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PLEASE HELP Two prisms are composed to form a V shape. The thickness of each prism is 1 unit, and width of each prism is 2 units
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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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3 years ago
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