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3 years ago

What is the sum of 7x/(x^2-4) and 2/(x+2)

Mathematics

2 answers:

pishuonlain [190]3 years ago

8 0

Answer: 9x - 4 / x^2 - 4

Step-by-step explanation:

ioda3 years ago

7 0

Answer:

The answer is 9x-4/(x+2)(x-2)

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3 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

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

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

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

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

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

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3 years ago

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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

Evan is making a table that will be created in the shape of the figure below. The table top is a triangle attached to a rectangl

emmainna [20.7K]

Area of a triangle:
A (triangle )= 4 · 6 / 2 = 12 ft²
150 - 12 = 138 ft² ( the maximum area of the rectangle )
L = ?
W = 6 ft
A ( rectangle ) = L · W
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L = 138 : 6 = 23 ft.
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3 years ago

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What is the sum of 7x/(x^2-4) and 2/(x+2)​

What is the sum of 7x/(x^2-4) and 2/(x+2)