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3 years ago

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Item 15 A cube has an edge length of 5 inches. You triple the side lengths. How many times greater is the volume of the new cube

?

1 answer:

Inga [223]3 years ago

6 0

The answer to this is 3,250 inches cubed

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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 that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

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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You estimate the distance from your house to your school to be 1.6 miles. The actual distance is 1.9 miles. Find the percent err

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To find percent error you use the following formula:
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% error = (1.6 - 1.9)/1.9 * 100% = 15.79%

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Goryan [66]

Answer:

Step-by-step explanation:

BDC=95

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Answer:

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Step-by-step explanation:

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