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@@ End of Day Wrap - Wednesday 14/5/14   Sydney, Australia Time 21:30

hi all,

Another busy day! Great answers from Alan, CPhill, Heureka, problem, zegroes, Goldenaleaf, gamrgrrl156, and Kitty<3.  Thank you everyone.  You all do a great job.

Here are some interesting posts for the day.

Polar co-ordinates

Thanks Alan.

Some calculators, such as the TI-Nspire with CAS, will simplify this.

Maybe look here

http://www.mathsisfun.com/data/binomial-distribution.html

I will use the same symbols as the last answerer.

B=P(1+r)ⁿ 

P=70,000

r = interest rate per compounding period, written as a decimal

n=number of compounding periods in total. 

eg 5%per annum compounded yearly r=0.05

6% per annum compounded monthly r=0.06/12

-----------------------------------------------------------

If the money is compounded yearly then the first answer is correct

B=5000(1+0.09)⁹

$${\mathtt{5\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1.09}}\right)}^{{\mathtt{9}}} = {\mathtt{10\,859.466\: \!397\: \!211\: \!546\: \!945}}$$

Thats $10859.47

If the money is compounding monthly then this would be the calculation

B=5000(1+(0.09/12))^9*12

$${\mathtt{5\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{0.09}}}{{\mathtt{12}}}}\right)\right)}^{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)} = {\mathtt{11\,205.620\: \!861\: \!161\: \!185\: \!437\: \!5}}$$

That's $11205.62

etc

You can't solve that - for starters there is no equal signs.

you can simplify it by hand but I don't think that you can on any remotely normal calculator.

(i could be wrong)

You are older  than your teeth and younger than your tongue!     

Yes and it has got depth as well. (Considered minimal for this exercise)

I thought that the poster could double the number all by him/herself!

Oh yea,

I had forgotten about that.

(I think they are just words but I was a bit put out when I first saw them too)

     line up the decimal points and then subtract like normal.

             $$1.000\_

{0.165}$$

This is an excellent answer.

Why don't you sign up and get credit for your great answers.  It is free and easy!

Me, Melody,admin,Cphill,Alan,Rom,kitty<3,reinout-g,scrutinizer,radix,GoldenLeaf !

Just use the calculator here to find x.  x = sin^-1(3/4)   This may also be written as asin(3/4).

You should find that x = 48.59° (In the first quadrant.  It could also be 180° - 48.59° = 131.41° in the second quadrant).

$${\mathtt{x}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{48.590\: \!377\: \!890\: \!729^{\circ}}}$$

I think "hobo" is not just an Australian term.  Here's an extract from Wikipedia:

Thanks Alan,

In some school in Australia they swap the hags for hobos because hags is derogatory.

But, I thought a lot of people in here would not know what a hobo was.  I think it is just Australian slang.  

Thank you zegroes that is really nice of you.   

We do try hard to help every one!

and welcome to Web2.0calc forum. We hope you continue to like it here.

There is no friends request system at the moment but there is a message system that works well.

If you want to send a note to someone and you cant find their name in the contact list, you can just click on their name anywhere in the forum. (like on one of their posts) and it will take you to their profile and ask you if you would like to send them a message.

After that it is easy because their name will be on an old message.

What about if a pile of mathematics text books fall and squash you!

Nice one Melody!!

Might be simpler to factor the 20 first:

37 + 20*(log6000 + log30)

37 + 20*log(6000*30)

37 + 20*log180000

$${\mathtt{37}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{180\,000}}\right) = {\mathtt{142.105\: \!450\: \!102\: \!066\: \!13}}$$

I've got another one - use the first letters!

----------------------

Some old hags 

Can't always hide

Their old age

------------------

    $$Sin\theta=\frac{opp}{hyp}\\

Cos\theta=\frac{adj}{hyp}\\

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

Doesn't a collar have two sides?  

I know, you very clearly specified the answer for one side!

I take it this means the normal cumulative distribution function.  Here is a graph of the normal cdf for a mean of 0 and standard deviation of 1:

If you have a data set x1, x2, etc.  for which you know the mean, xav, and standard deviation, s, and want to find the cumulative probability for a particular value of x, then calculate  z = (x-xav)/s and look up the corresponding value of cumulative probability (called cnorm(z) on the graph above).  You should be able to find better graphs if you search the internet.  Alternatively, use a piece of software which will calculate it for you, given the values of xav and s (again you can almost certainly find something on the internet to do this).

Or, if you are old-fashioned, use tables of values that are printed in books!

If this is meant to be log to the base 5 of (x-2) is 3 then:

log₅(x-2)=3 means  x-2 = 5³, so x = 5³+2, or x=125+2, or x = 127.

If it is meant to be log to the base 10 of 5*(x-2) =3 then:

log₁₀(5*(x-2)) = 3 means  5*(x-2) = 10³, so x-2 = 200, so x = 202

If it is meant to be log to the base 10 of 5, all multiplied by (x-2) = 3 then:

(x-2)*log₁₀(5)=3, so x-2 = 3/log₁₀(5) so x = 3/log₁₀(5)+2 or $${\mathtt{x}} = {\frac{{\mathtt{3}}}{{log}_{10}\left({\mathtt{5}}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} \Rightarrow {\mathtt{x}} = {\mathtt{6.292\: \!029\: \!674\: \!220\: \!178\: \!6}}$$

CPhill is just kidding around.

Very funny Chris. 

I am sure that is not what was meant Kitty but it doesn't matter - You get 3 more points.