All Answers 358091 Answers

The radius of a circle is a half of the diameter.

The forumula for circumference is

  $$\\C=2\pi r\\\\
\mbox{This is }\\\\\quad C=2\times \pi \times r$$

You should be able to do this yourself.  Let us know if you have more problems.  

Thurs 21/8/14

* Woking with speed (This is a rates question Ninja)  

CPhill recommended this question.  It is an interesting one.  

@@ End of Day Wrap :     Wed 20/8/14          Sydney, Australia       Time 9:00pm      ♬  

Hello all,

Lots of great answers have been posted by Alan, AzizHusain, Kitty<3, CPhill, NinjaDevo, Rosala, Latino_King, Sasini, Dragonslayer554.  Thanks all.  

If you would like to reference a particular post within a thread the address is linked to the "post date" on the top right hand side of the post.  If you click on this, the address in the address bar will change to the address of this particular post.  You can then copy it and paste it as your reference address.

Moderators now have the ability to categorise questions.

At present the categories are Fun, Mathematics, Off Topic and Physics.

Mathematics is then split into many sub categories.  

The idea is great but the topic headings do not work well.  I think if we can put together a categories list that will work for us, Mr Massow would be happy to consider it.

I am requesting that answerers give this some thought.  I think I will start a new Sticky Topic thread where you can post your ideas.  You can also private message me if you would prefer.

I have only  collected 2 interest posts.

1) Lets end the day with a laugh - Thanks Rosala

$${\frac{{\left(-{\mathtt{6}}\right)}^{{\mathtt{3}}}}{{\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{2}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{\frac{{\left(-{\mathtt{7}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{2}}\right)}{{\left({\mathtt{3}}\right)}^{{\mathtt{2}}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{674}}}{{\mathtt{9}}}} = -{\mathtt{74.888\: \!888\: \!888\: \!888\: \!888\: \!9}}$$

$$\\(-6)^3/(-3)^2+(-2)^3(5)-(-7)^2(2)/(3)^2\\\\
=-216/9+-8*(5)-49*(2)/9\\\\
=-216/9-40-98/9\\\\
=-24-40-10\frac{8}{9}\\\\
=-64-10\frac{8}{9}\\\\
=-74\frac{8}{9}$$

Yes you do not have the right number of brackets.  

I have referenced this thread under Puzzles.  

Please explain this information for me--> 

Question 7 The figure shows four pairs of charged particles. For each pair, let  V = 0 at infinity and consider  V net at points on the x axis.

Your answer is incorrect.  Try again.     For which pairs is there a point at which Vnet = 0 between the particles? (1) (2) (3) (4) None       SHOW HINT

Your answer is incorrect.  Try again.     For which pairs is there a point at which Vnet = 0 to the right of the particles? (1) (2) (3) (4) None       SHOW HINT

Your answer is correct.     At such a point is  due to the particles equal to zero? Yes No

Your answer is correct.     For each pair, are there off-axis points (other than at infinity) where Vnet = 0? Mark the pairs where this is true. (1) (2) (3) (4) None of the pairs exhibit this quality.

DragonSlayer's riddles.  (16/8/14)

Unsure about this. It is similar to last weeks question. I don't know what is being asked when I read it. 

A spherical conducting shell has a charge of -37 μC on its outer surface and a charged particle in its hollow. If the net charge on the shell is -31 μC, what is the charge (a) on the inner surface of the shell and (b) of the particle?

(a)

Number

Units

° (degrees) This answer has no units m kg s m/s m/s^2 N J W N/m A V μC N/C C/m C/m^2 C/m^3 N·m^2/C

(b)

Number

Units

° (degrees) This answer has no units m kg s m/s m/s^2 N J W N/m A V μC N/C C/m C/m^2 C/m^3 N·m^2/C

I liked that one too :)

Thanks Rosala. 

Aziz has foundand approximation.  Is that what you wanted?  Perhaps you just wanted it simplified.

$$8\pm \frac{4\sqrt3}{2}\\\\
=8\pm 2\sqrt3\\\\\ or \\\\
=4(2\pm \sqrt3)$$

thanks, and yes t is time. sorry i forgot to add that in. thanks again.

NOTE: This is an approximation. $${\sqrt{{\mathtt{7}}}}$$  is an irrational number and it cannnot be expressed exactly in digits.

Where:

x=sin(t)

y=t+1

z=cos(t)

$$\\\frac{dx}{dt}=cos(t)\\\\
\frac{dy}{dt}=1\\\\
\frac{dz}{dt}=-sin(t)\\\\$$

I think this is right.  t is time.  

$$\begin{array}{rlll}
4(3x-5)&=&-2(-x+8)-6x \qquad \mbox{First expand and simplify both sides seperately }\\
12x-20&=&2x-16-6x \qquad \mbox{}\\
12x-20&=&-4x-16 \qquad \mbox{}\\
12x-20+4x&=&-4x-16+4x \qquad \mbox{}\\
16x-20&=&-16 \qquad \mbox{}\\
16x-20+20&=&-16+20 \qquad \mbox{}\\
16x&=&4 \qquad \mbox{}\\
16x/16&=&4/16 \qquad \mbox{}\\
x&=&\frac{1}{4} \qquad \mbox{}\\


\end{array}$$

This is a huge number.  It is not infinity 

$$The imaginary number system is based on $i=\sqrt{-1}$\\
So \\
3i = \sqrt9\times \sqrt{-1} = \sqrt{9*-1} = \sqrt{-9}$$

You can enter it just as you have written it.

$$$\frac{x}{8}=2$\\\\
Just multiply both sides by 8\\\\
$\frac{x}{\not{8}}\times \frac{\not{8}}{1}=2\times 8$\\\\
$x=16$$

Note that the area should have units of in² not just in

Multiply both sides by 8.

$$\frac{x}{8}\times 8=2\times 8\\\\
x = 16$$

Do you  mean how do you enter 34,547 into the calculator?  Just leave out the comma.  It is not needed.

Also, this is an international forum and the comma notation is very comfusing for a lot of people because some countries use a common instead of a decimal point when they write a decimal.

So a lot of people would think 34,547 means 34.547

I remember the plastic sheet overhead projectors as well and yes I have used those extensively but I was referring to a proper modern projector for computers.  They were mounted in the ceiling of the classrooms but I was one of the few who actually used them. 

I remember overhead projections in school...they were neat and reusable so that was great! I remember only absent students could see the overheads at my school or else had to get notes from a friend.

Yes, an iPad and Windows 8 w/ touch screen would work. There is also a laser keyboard, so it's cool! With nothing, you get everything!

year = 4 = Months = 3 Days = 1 Deaths = 300 000

What you want is not clear to us but this might be it.

$${\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{365}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{30}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{1\,551}}$$ days (approximately)

$${\frac{{\mathtt{300\,000}}}{{\mathtt{1\,551}}}} = {\mathtt{193.423\: \!597\: \!678\: \!916\: \!827\: \!9}}$$

Approximately 193 deaths per day.