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3 years ago

I don’t know the answer

Mathematics

1 answer:

Andreas93 [3]3 years ago

5 0

I believe the answer would be yes. Since your dealing with Ratios it would be simple to understand that they are proportional. I'll explain a little better.

This means that if you double this ratio, it will still be equivalent to 1:3

1:3

This is also proportional because this number will always be equivalent to itself.

7:21

-----------------------------------------------------------------------------------------------------------------Hope this helps

-Miri

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3 years ago

Read 2 more answers

P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }

balandron [24]

The solution to the division of the given surd is: $\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x} }$

<h3>Division of Surds.</h3>

The division of surds follows a systemic approach whereby we divide the whole numbers separately and the root(s) are being divided by each other.

Given that:

$\mathbf{P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }}$

i.e.

$\mathbf{=\dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )}{ \frac{4x}{(x-1)^{2} }} }$

Using the fraction rule:

$\mathbf{\dfrac{a}{\dfrac{b}{c}}= \dfrac{a\times c}{b}}$

$\mathbf{\implies \dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )(x-1)^{2}}{4x}} }$

By simplification, we have:

$\mathbf{ =\dfrac{\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{2} }{4x} }$

$\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x} }$

Learn more about evaluating the division of surds here:

https://brainly.in/question/27942899

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