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timurjin [86]
3 years ago
14

Solve x+4/4 = x/x-2

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

Answer:

After solving the equation \frac{x+4}{4}=\frac{x}{x-2} we get x=-2 and x= 4 i.e {-2,4}.

Option B is correct.

Step-by-step explanation:

We need to solve \frac{x+4}{4}=\frac{x}{x-2}

Solving:

\frac{x+4}{4}=\frac{x}{x-2}

Cross multiply

(x-2)(x+4)=x*4

Multiplying:

x(x+4)-2(x+4)=4x\\x^2+4x-2x-8=4x\\x^2+2x-8-4x=0\\x^2-2x-8=0

Now, we have got quadratic equation x^2-2x-8=0

Solving this quadratic equation using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\

We have a=1, b=-2 and c=-8 Putting values

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-8)}}{2(1)}\\x=\frac{2\pm\sqrt{4+32}}{2}\\x=\frac{2\pm\sqrt{36}}{2}\\x=\frac{2\pm6}{2}\\x=\frac{2+6}{2} \ or \ x=\frac{2-6}{2}\\x=\frac{8}{2} \ or \ x=\frac{-4}{2}\\x=4 \ or \ x=-2

So,after solving the equation \frac{x+4}{4}=\frac{x}{x-2} we get x=-2 and x= 4 i.e {-2,4}.

Option B is correct.

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Answer:

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Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation, also called standard error s = \frac{\sigma}{\sqrt{n}}.

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