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EastWind [94]
3 years ago
8

I Need Help Like Now!!

Mathematics
1 answer:
kogti [31]3 years ago
5 0

Answer:

\frac{136}{999}

Step-by-step explanation:

We require 2 equations with the repeating decimal after the decimal point in both equations.

The bar above 136 indicates that 136 is being repeated

let x = 0.136136...... → (1) Multiply both sides by 1000

1000x = 136.136.... → (2)

Subtract (1) from (2) to eliminate the decimal part

999x = 136 ( divide both sides by 999 )

x = \frac{136}{999}

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Subtract − 9 x − 7 −9x−7 from 8 x 2 − 5 x + 9 8x 2 −5x+9
stiks02 [169]

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1x1+9x1/10+8x1/100+1x1/1000
xxMikexx [17]

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3 years ago
2b=6a-14,3a-b=7 how to work it out
pshichka [43]
So, since there are two equations, we can substitute one of them into the other. In this case, it would be easiest to substitute 2b=6a-14 into the other equations. But first, simplify by dividing both sides by 2. You bet b=3a-1

We can now plug this into the other equation
Sine the equation b=3a-1 results in the value of b, we have to plug in for the value of b in the other equation

So this is what we get after plugging in:
3a-(3a-1)=7
Now, simplify. 3a-3a+1=7
Since 3a-3a = 0, this equation results in a no solution

Answer:
The result is a no-solution, or Ф

Hope this helped!! :D
5 0
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