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lidiya [134]
3 years ago
11

1) If B is between A and C, AB=10 and AC=28, what is the value of BC?

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0

One

AB + BC = 28

AB = 10

10 + BC = 28

BC = 28 - 10

BC = 18

Two

AB + BC = AC

AB = 6x

BC = 8x

6x + 8x = 42

14x = 42

x = 42/14 = 3

Three

AB = 2x - 1

BC = 3x + 5

AB + BC = 24

2x - 1 + 3x + 5 = 24

5x + 4 = 24

5x = 24 - 4

5x = 20

x = 20/5

x = 4

AB = 2x - 1

AB = 2*4 - 1

AB = 7

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Answer:

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Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

\frac{x}{x^2-16}-\frac{3}{x-4}

To solve the given expression

\frac{x}{x^2-16}-\frac{3}{x-4}

=\frac{x}{x^2-4^2}-\frac{3}{x-4}

=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}  ( using the property a^2-b^2=(a+b)(a-b) )

=\frac{x-3(x+4)}{(x+4)(x-4)}

=\frac{x-3x-12}{(x+4)(x-4)} ( by using distributive property )

=\frac{-2x-12}{(x+4)(x-4)}

=\frac{-2(x+6)}{(x+4)(x-4)}

\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is \frac{-2(x+6)}{(x+4)(x-4)}

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Answer:

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