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3 years ago

Solve this problem???

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

The denominator of the first term is a difference of squares, such that

4a ² - b ² = (2a)² - b ² = (2a - b) (2a + b)

So you can write the fractions as

(4a ² + b ²)/((2a - b) (2a + b)) - (2a - b)/(2a + b)

Multiply through the second fraction by 2a - b to get a common denominator:

(4a ² + b ²)/((2a - b) (2a + b)) - (2a - b)²/((2a + b) (2a - b))

((4a ² + b ²) - (2a - b)²) / ((2a - b) (2a + b))

Expand the numerator:

(4a ² + b ²) - (2a - b)²

(4a ² + b ²) - (4a ² - 4ab + b ²)

4ab

So the original expression reduces to

4ab / ((2a - b) (2a + b))

4ab / (4a ² - b ²)

upon condensing the denominator again.

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In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

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Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

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Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

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