Melody

@@ End of Day Wrap : Mon and Tues 15/7/14   Sydney, Australia   Time 7:50pm    ♬        

Hi everone,

Great answers have been provided the last couple of days by CPhill, Rose98, Alan, Rosala, DragonSlayer554, NinjaDevo, Quazars, Zegroes, Kitty<3, Needhelp, GoldenLeaf and Mari379.

Thank all.  

Ninja changed the name of our reference thread to "Reference Material" .  Thanks Ninja, that was smart thinking. 

Now for some interesting threads,

1) I like questions like this   (Thanks CPhill )

1/17=17*1/(17^2)

$$\begin{array}{rll}
0.05^{1/n}&=&0.8609\\\\
log(0.05^{1/n})&=&log0.8609\\\\
(1/n)log(0.05)&=&log0.8609\\\\
n&=&\frac{log0.05}{log0.8609}
\end{array}$$

$${\frac{{log}_{10}\left({\mathtt{0.05}}\right)}{{log}_{10}\left({\mathtt{0.860\: \!9}}\right)}} = {\mathtt{20.001\: \!293\: \!705\: \!215\: \!387\: \!9}}$$

This is a difference of 2 squares question.

(x+5)(x-5)=x²-5² = x²-25

Is this the question Nataszaa

cubed sqrt128x^7

$$(\sqrt{128}x^7)^3\\\\
=(8\sqrt{2}x^7)^3\\\\
=512\times2\sqrt{2}x^{21}\\\\
=512\times2\sqrt{2}x^{21}\\\\
=1024\sqrt{2}x^{21}\\\\$$

Well mari379, don't you have the date displayed on your computer or phone?

Also, every post has got a date stamp on it.

Also,  When you asked, it was the 14th for many people but the 15th for me.

So someone could have answered you corectly from their perspective but it would be incorrect from your perspective.

The trials and tribulations of being on an international forum!  

The value of 4 is 4

$$\frac {48+\frac{(2-1)}{7}+50+(2-2)}{2}-\sqrt{4}+50\\\\
=\frac {48+\frac{1}{7}+50+(0)}{2}-2+50\\\\
=\frac {48+\frac{1}{7}+50}{2}-2+50\\\\
=\frac {98+\frac{1}{7}}{2}-2+50\\\\$$

$${\frac{\left({\mathtt{98}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}}\right)}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{50}} = {\frac{{\mathtt{1\,359}}}{{\mathtt{14}}}} = {\mathtt{97.071\: \!428\: \!571\: \!428\: \!571\: \!4}}$$

Umm different answer from zegroes - lets try the site calc.

$$\left({\mathtt{48}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{50}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right)}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{50}}\right) = {\frac{{\mathtt{848}}}{{\mathtt{7}}}} = {\mathtt{121.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

ok lots of different answers happening here.  One of the reasons is that there is a bracket missing in the questions and it keeps getting interpreted differently.

Fix up the bracket and we will fix up our anwers.    

Thanks Ninja,

I would have just said

$$\dfrac{26\frac{1}{3}-8\frac{1}{4}}{2}$$

$${\frac{\left(\left({\mathtt{26}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}\left({\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)\right)}{{\mathtt{2}}}} = {\frac{{\mathtt{217}}}{{\mathtt{24}}}} = {\mathtt{9.041\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$${\mathtt{217}} \,{mod}\, {\mathtt{24}} = {\mathtt{1}}$$   

 (I just wanted to try out the mod function on the calcuator - I haven't used it before  

Answer = 9 and 1/24 inches

You cannot take sin inverse 5 because that means 

$$sin\theta=\frac{5}{1}=\frac{opposite}{hypotenuse}$$

but the hypotenuse has to be the longest side so this doesn't make sense!

Hence,

    $$\\So\;\; sin^{-1}x \;\;\mbox{Only exists for }-1\le x\le1\\
\mbox{This is also true for } cos^{-1}x\\\\$$

But

$$tan\theta = \frac{opp}{adj}$$

Either one of these can be the biggest one so of course you can take  $$tan^{-1}5$$   That is perfectly fine.