Number System and Simplification

$\displaystyle (64)^{-2/3} = 4^3 \times 1^{-2/3} = 4^{-2}$
                 $\displaystyle = \frac{1}{4^2} = \frac{1}{16}$   

$\displaystyle \frac{(27)^{-2/3}}{ ( 64)^{-2/3}} = \frac{\displaystyle \frac{1}{9}}{ \displaystyle \frac{14}{16}} = \frac{16}{9}$

$\displaystyle 6\cdot 8793\times 10^{4}=68,793$

This expression uses the Sophie Germain identity: a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab). Here, 324 = 4 * 81 = 4 * 3^4, so b=3. Applying this to each term allows for cancellation of factors between the numerator and denominator, leaving only the ratio of the remaining terms.

Consider the given expression,

  $ \Rightarrow {{9}^{\dfrac{3}{2}\div {{\left( 243 \right)}^{-\,\dfrac{2}{3}}}}}={{9}^{\dfrac{3}{2}\times {{\left( 243 \right)}^{\dfrac{2}{3}}}}} $

 $ ={{9}^{\dfrac{3}{2}\times {{\left( {{3}^{5}} \right)}^{\dfrac{2}{3}}}}}={{9}^{\dfrac{3}{2}\times {{3}^{\dfrac{10}{3}}}}} $

 $ ={{\left( {{3}^{2}} \right)}^{\dfrac{3}{2}\times {{3}^{\dfrac{10}{3}}}}}={{\left( 3 \right)}^{3\times {{3}^{\dfrac{10}{3}}}}} $

 $ ={{3}^{{{3}^{1+\dfrac{10}{3}}}}}={{3}^{{{3}^{\dfrac{13}{3}}}}} $

Hence, this is the answer. 

$\displaystyle 5\times 10^{-8}=0\cdot 00000005$

$\displaystyle 4\cdot 56\times 10^{-5}=0\cdot 00000456$

$\displaystyle 0\cdot 00044= \frac{44}{100000}= \frac{4\cdot 4\times10^{1}}{10^{5}}$
$=\displaystyle 4\cdot 4\times 10^{-4}$
Now
$\displaystyle 4\cdot 4\times 10^{-4}=4\cdot 4\times 10^{n}$
$\displaystyle \therefore n=-4$

$\displaystyle \frac{1}{10000000}= \frac{1}{10^{7}}= 1\times 10^{-7}$

$2.73 \times 10^{12} $$=2730000000000$

$\dfrac{1}{m}$ $=$ $\dfrac{-1}{3}$

$4\times12$ $=$ $48$

$30\times25$ $=$ $750$

Using the formulas r1 = s*tan(A/2), r2 = s*tan(B/2), r3 = s*tan(C/2), and r = s*tan(A/2)*tan(B/2)*tan(C/2), the expression simplifies to 0 based on trigonometric identities in a triangle.