All Answers 358131 Answers

2*c=8,8*15

c=8,8*15/2

c=66

Nevermind.....I see  it.....(1/2) of the expected probability of a "normal" coin = 2 x the probability of our "biased" coin....

 BTW....nice answer....!!!

1 / [1 + tan^2x]  + 1 / [ 1 + cot^2x]  =

1/sec^2x + 1/csc^2x =

cos^2x + sin^2x = 

1

Where did you find that grapher, Melody ????   ....pretty neat...

First term  =  -5*3^(1-1) = -5*3^0   = -5*(1)  = -5

Fourth term =  -5*3^(4-1)  = -5*3^(3)  = -5*27  = -135

I'll leave it to you to find the eighth term.......it shouldn't be difficult.....!!!

Call x the number of pages....so

(3/7)x = 12    multiply both sides by 7/3

x = 12(7/3)  =   4 x 7  = 28 pages total

cos^2x-sin^2x/ctg^2x-tg^2x=sin^2x*cos^2x

First, notice that :

ctg^2x - tan^2x =    cos^2x/sin^2x - sin^2x/cos^2x  = [cos^4x - sin^4x]/[sin^2x cos^2x)..so we have..

[cos^2x - sin^2x] [sin^2xcos^2x]  / [cos^4x - sin^4x]  = 

[cos^2x - sin^2x] [sin^2xcos^2x] / [ (cos^2x - sin^2x) (cos^2x + sin^2x) ] =

[sin^2xcos^2x] / [sin^2x + cos^2x]              and [sin^2x + cos^2x ]  = 1  .....so....

sin^2xcos^2x  =  sin^2xcos^2x

Nah......that seems pretty logical to me, Melody.

7 different groups each seated 5! ways......

I agree with your answer......and we're NEVER incorrect on these counting probs, are we ???  LOL!!!

Hey, Melody....a brief question......what was the motivation for multiplying by "2" on the left hand side in the top equation ???

Go slow with that explanation......you know me and "counting" problems....LOL!!!!

Notice that, if A comes first, there are 24 ways to form a two-letter combo. {We can't use O}

And if B comes first we have 23 possibilities.

Up to the letter O, we have :

24 + 23 + 22+ .....+ 12 + 11 possibiilities

And after O, we have  10 + 9 + 8 +....+ 2 + 1 possibilities

So, the total possibilities is just (24)(25)/ 2 = 300 possibilities

And for the numbers, since 0 can't be used, we have :

8 + 7 + 6 + ......2 + 1  =  (8)(9)/ 2 = 36 possibilities

So, the total ways of forming a license plate is given by ;

300 x 36 = 10,800 combos

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@@ End of Day Wrap   Thurs 2/4/15   Sydney, Australia   Time 3:55 am [Definitely Fri :)] ♪ ♫

Hi all,

Great answers today from MathsGod1, CPhill, Alan, Eannehail, Bertie, Mathcad, Willie, Garmoon, Dgem250, Geno3141 and Civonamzuk.

Interest posts:

3/4/15

FTJ means "For the Juniors"

That looks impressive Bertie but where is my picture 

I did graph it and I got a very interesting landscape but I didn't know what to do with A.

I didn't find a 3D grapher that would plot both the graph and the point as well.

I'd really like to see them together.   Maybe I could get a 'feel' for the question that way.

That is hopeful thinking that is :)

Actually, this graph is not too bad:)

Here is a graph of the terrain around landing spot.  :)

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I want you to do them WITHOUT the calculator please MG    

Here is some more - only do them if you want to.  :)

Hint:  Don't make a big deal out of these - they are quite simple.

$$\\1)\;\; 6-\frac{2}{3}\\\\
2) \;\;12-3\frac{5}{9}\\\\$$

Hi anon, thanks for getting involved - we like that.

However, the minimum number is 9.     5 across and 4 down  

I  don't know where you got 8 from?      :)

Hi Chris,

Yes i thought that might be Nauseated's formula,  It is nifty, 

There are so many formuas for so many things i cannot remember them all.

I suppose i should try and organise a file for some of these new ones. 

Sin^3x(1+Ctgx)+cos^3x(1+tgx)

$$\\sin^3x(1+\frac{cosx}{sinx})+cos^3x(1+\frac{sinx}{cosx})\\\\
=sin^3x(\frac{sinx+cosx}{sinx})+cos^3x(\frac{cosx+sinx}{cosx})\\\\
=sin^2x(\frac{sinx+cosx}{1})+cos^2x(\frac{cosx+sinx}{1})\\\\
=(sinx+cosx)(sin^2x+cos^2x)\\\\
=(sinx+cosx)\times 1\\\\
=sinx+cosx\\\\$$

You just have to put brackets around it and divide the numerator by the denominator. 

Ex. 3 quarters would be entered as (3/4), where / is the symbol for divison.

Here is another question for you MG    

Hint: be careful, it is a bit tricky :)

$$5\frac{1}{4}-2\frac{5}{6}$$

ok I am just trying to trick you and to make you think

now i will tell you properly.

$$(+3)^2=3*3=+9$$

$$(-3)^2=-3*-3=+9$$

$$-3^2 = -3*3=-9$$      you can think of it like   $$-3^2=-1*3^2 = -1*9 = -9$$

$$\sqrt{9}=+3$$

$$\\\sqrt{9}\ne -3\\\\
$By convention the square root answer has to be positive$$$.

$$\\\frac{500 N}{m^3} \times \frac{5m}{1}\\\\
=\frac{500 N}{m^2} \times \frac{5}{1}\\\\
=\frac{2500 N}{m^2} \\\\
=2500\;N/m^2$$

well there is 7 ways to leave someone out so that makes 7 ways to choose 6 people.

there are 5! ways to seat those 6 people.

so maybe    7*5!   way.

It is probably wrong.    :)

$$\frac{12*8}{20C2}$$

$${\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{8}}\right)}{{\left({\frac{{\mathtt{20}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}} = {\frac{{\mathtt{48}}}{{\mathtt{95}}}} = {\mathtt{0.505\: \!263\: \!157\: \!894\: \!736\: \!8}}$$

0.505  maybe