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Neko [114]
3 years ago
10

Make r the subject of the formula t=

r-3}" alt="\frac{r}{r-3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

r=\frac{3t}{t-1}

Step-by-step explanation:

So we have the formula:

t=\frac{r}{r-3}

And we want to make r the subject of the formula.

So, first, let's multiply both sides by (r-3). The right will cancel. So:

t(r-3)=r

Distribute the t:

rt-3t=r

Subtract rt from both sides. The left cancels:

-3t=r-rt

Factor out an r from the right:

-3t=r(1-t)

Divide both sides by (1-t). The right cancels:

\frac{-3t}{1-t}=r

Simplify the negatives:

\frac{3t}{t-1}=r

Flip:

r=\frac{3t}{t-1}

And we're done!

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The product below is equivalent when x > 0 is <u>1/9</u>

       

<h3>Resolution - Explanation</h3>

Square root is a real number x multiplied by itself - which results in a perfect value, where it is possible to calculate the real proof (which is the square root).

             

Given the expression, \large \sf \sqrt{\dfrac{1}{x^{2} } }  \cdot \sqrt{\dfrac{x^{2} }{81} }, first step: we will calculate the root of the numerator and denominator of this fraction:

<u />

<u />\\\large \sf \sqrt{\dfrac{1}{x^{2} } }  \cdot \sqrt{\dfrac{x^{2} }{81} }

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Step two: rearranging the expression and canceling the common factors x, we will have,:

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