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kkurt [141]
3 years ago
6

C/- 1 - 5 help me please

Mathematics
1 answer:
tia_tia [17]3 years ago
7 0

Answer:

-6

Step-by-step explanation:

-1 - 5 = -6

Hopefully that's what you meant...

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The difference of any two negative integers is a negative integer
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Answer: look at the bottom

Step-by-step explanation:

Difference of two negative numbers may or may not be negative. It depends on the numbers taken as well as on the sequence in which they are taken. Then, a-b=b-a=0, (which is neither negative nor positive..!!!) So difference of two negative numbers may be positive, negative or even zero.

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84 out of 240 what is the percentage
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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
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

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Answer:

Hope this solution helps you

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Brooklyn is a waitress at a restaurant. Each day she works, Brooklyn will make a guaranteed wage of $30, however the additional
Elodia [21]

Answer:

153, 9

Step-by-step explanation:

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