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tangare [24]
4 years ago
7

What is the solution to the equation below?

Mathematics
1 answer:
Nat2105 [25]4 years ago
5 0

Answer: OPTION C

Step-by-step explanation:

 Given the equation \frac{\sqrt{3-2x} }{\sqrt{4x} } =2, you need to solve for  the variable "x".

First, you need to multiply both sides of the equation by \sqrt{4x}:

(\frac{\sqrt{3-2x}}{\sqrt{4x}})(\sqrt{4x} })=2(\sqrt{4x} })\\\\\sqrt{3-2x}=2\sqrt{4x}

Now you need to square both sides of the equation:

(\sqrt{3-2x})^2=(2\sqrt{4x})^2\\\\3-2x=4(4x)\\\\3-2x=16x

Subtrac 3 and 16x from both sides:

3-2x-(3)-(16x)=16x-(3)-(16x)\\\\-18x=-3\\

Divide both sides by -18:

\frac{-18x}{-18}=\frac{-3}{-18}\\\\x=\frac{1}{6}

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To prove that a square root number lies between which two consecutive integers do that

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Learn more:

You can learn more about the numbers in brainly.com/question/9621364

#LearnwithBrainly

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