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people got to do it.... Unless you're playing video games... Then you're on your own

Behave yourself Dragon!

HOW DO I BECOME A POWER USER? PLZ HELP!!!!

If you are serious Kool-aid take away the points that you have given yourself and I will try to help you.

If you are not serious and you are just having fun then take away your own points anyway.  ♬      

Hi Melody,

Hi Heureka, 

So I am going for     $$(2,-2\sqrt3)$$

Convert (4, 300°) to rectangular form

$$\\x= 4 * \cos{(300\ensuremath{^\circ}} )}=4*\cos{(60\ensuremath{^\circ}} )}=4*\frac{1}{2}=2\\
y=4*\sin{(300\ensuremath{^\circ}} )}=4*( -\sin{(60\ensuremath{^\circ}} )}
)=-4*\frac{1}{2}*\sqrt{3}=-2\sqrt{3}\\
(2,-2\sqrt{3})$$

WHAT IM ASKING IS I TYPED IT IN BUT IT WONT ANSWER MY QUESTION.(MABY ITS DEFECTIVE)

HOW DO I ENTER THE EQUATION?

You are not talking about everyone in the whole world Dragon.  Not everyone comes from the US you know!  Enjoy your holidays though! ♬

Now Kool-aid,

I know that you fingers are not broken or you would not have been able to type in your questions.

Is the site calculator broken? (It is on the home page)  There is even a fraction input page, you will find it in the drop down menu in the top right corner of the calculator!  Check it out!

Could you please check it out for me and report back?  

You could test it with your questions.  

Thanks so much Kool-Aid. 

Is there anything else that I can help you with?

This site is only relevant in the school year and summer school periods. Today is my last day of summer school btw, not so sure about ornstein....

No, sorry.

._. i leave for 3 day- THIS IS WHY WE CAN"T HAVE NICE THINGS!

No one claimed to k**l me?

Absolutely nothing.  

$$\\
\dfrac{1}{2}*
x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}}
}
=\lambda p_1\\\\
\dfrac{1}{3}*
x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}}
}
=\lambda p_2$$

$$\dfrac{
\dfrac{1}{2}*
x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}}
}
{
\dfrac{1}{3}*
x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}}
}
=\dfrac{\lambda p_1}{\lambda p_2}$$

$$\Rightarrow (1) \quad \dfrac {p_1}{p_2}=\dfrac{3}{2}\dfrac{x_2}{x_1}$$

$$\begin{array}{rcccc}
&I.&&II.&\\\\
(2)\qquad p_1x_1+p_2x_2=m \quad | \quad &\times&\dfrac{1}{p_2} \quad | \quad &\times&\dfrac{1}{p_1}
\end{array}$$

$$I.\quad \dfrac{p_1}{p_2}x_1+x_2 = \dfrac{m}{p_2} \quad
\Rightarrow \quad \dfrac{3}{2}\dfrac{x_2}{x_1}x_1+x_2=\dfrac{m}{p_2}\quad \Rightarrow \quad \dfrac{5}{2}x_2=\dfrac{m}{p_2}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_2=\dfrac{2}{5}\dfrac{m}{p_2}}}$$

$$II.\quad x_1+\dfrac{p_2}{p_1}x_2 = \dfrac{m}{p_1}\quad\Rightarrow \quad x_1+ \dfrac{2}{3}\dfrac{x_1}{x_2}x_2=\dfrac{m}{p_1} \quad \Rightarrow \quad \dfrac{5}{3}x_1=\dfrac{m}{p_1}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_1=\dfrac{3}{5}\dfrac{m}{p_1}} }$$

Yes Heureka, your method was perfecty valid 

That's neat Chris, thank you and thumbs up from me.

We can write

log594750⁹⁷²³        and by a property of logs, we have

9723log594750 =

9723*(5.774334450709304) = 56143.853864246562792 ...... And in scientific notation......

594750⁹⁷²³  =

10^{.853864246562792} x  10⁵⁶¹⁴³   ≈

7.142 x  10⁵⁶¹⁴³    

And, as Melody has said........that's a REALLY big number containing 56144 digits !!

5 cm : 1.3 m  ?

$$\\\dfrac{5{cm} }{1.3m}=\dfrac{5\not{c}\not{m}*\dfrac{1m}{100\not{c}\not{m}}} {1.3m}=\dfrac{5\not{m}}{100*1.3\not{m}}=\dfrac{5}{130}\\\\
=\dfrac{1}{26}= 0.03846153846$$

Here's my take on this:

one over eight of the books in a library are tamil books and five out of twenty-four of them are chinese books. seven eight of the remaining books are English books.There are 840 english books.Find the total number of books in the library.

Good question.  Let's look at this.

$${\frac{{\mathtt{1}}}{{\mathtt{8}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{24}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

So one third are either Tamil or chinese.  So 2/3 are not.

7/8 of 2/3 are English books

$${\frac{{\mathtt{7}}}{{\mathtt{8}}}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right) = {\frac{{\mathtt{7}}}{{\mathtt{12}}}} = {\mathtt{0.583\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

So 7/12 of total is 840 English books

$$\frac{7}{12}T=840\\\\
T=840\times \frac{12}{7}\\\\
T=1440$$

There are 1440 books in the library