All Answers 358116 Answers

Alright so first let's look at what we have.

First work away the brackets, for example;

$$3(x-4) = 6(2x+1) \Rightarrow 3x-12 = 12x+6$$

Now substract $$12x$$ from both sides

$$-9x-12 = 6$$

Add $$12$$ to both sides

$$-9x = 18$$

Divide both sides by -9

$$x = -2$$

voila! 

$$(1+secx)(1-cosx)=tan^2xcosx\\\\
\begin{array}{rll}
LHS&=&(1+\frac{1}{cosx})(1-cosx)\\\\
&=&1-1-cosx+\frac{1}{cosx}\\\\
&=&\frac{1}{cosx}-cosx\\\\
&=&cosx(\frac{1}{cos^2x}-1)\\\\
&=&cosx(\frac{1-cos^2x}{cos^2x})\\\\
&=&cosx\times\frac{sin^2x}{cos^2x}\\\\
&=&cosx\:tan^2x\\\\
&=&RHS
\end{array}$$

This is the best I can do for you

Wolfram is now building a timemachine so they can ask einstein about question c.

Reinout

Okay - I just ran it through Wolfram|Alpha again and now I'm even more embarrased.

Thanks reinout and Bertie,

Okay. I'm embarrased.  I didn't actually try doing it.  I ran it through Wolfram|Alpha and obviously misinterpreted the output that I was given.  I will take another look.  

Pity. I like doing these questions - I was just low on time!

 (Believe it or not, suit yourself, I'm sticking with my story - does pinoccio have a twin?  lol)

Do you mean on the web2 calculator?  Would you like to elaborate?

lol... Good at maths....

I knew it could be done in that way.

I just preferred the longer route.

i guess its $4.1 .......... maybe.......im not that sure.......because im good at math

Okay Bertie, it is a permutation lock.

 Except, the letters are in a set postion and the numbers are in a set postition.

So why would you want to divide by 8!  I don't get it.  

(Okay, I just got it,you are just being padentic about the lay persons word combination )

In your example

Afterall, if I told you that  the three numbers on a simple lock were 2, 3 and 7 you wouldn't necessarily expect them to be in that order. You would have 3! = 6 different orders to try.  Yes that is true

It is a 3 digit tumbler - order obviously matters - the number of possiblilities is 10*10*10=1000

The right combination (lay speak)  is obviously a permution - order matters so the probablilty of getting the right combination is 1/1000

The probablity of getting 2,3 and 7 in any oder is 6/1000

Don't worry - I am really just talking to myself here - I do that.   

If order does not matter than the probability would be greater.  I'd multiply by 8! (not divide) but this would not be right I do not think.  It would be ok if every number/letter were different but what if all the numbers were 7 and all the letters were A then it would not work!  Plus, in Alan's implied scenario the letters occupy 4 specific spots and he numbers occupy a different 4 specific spots.

Ummm  

Oh Divine Zeus,

Supreme Ruler of the Gods,

Sir CPhill shall be beyond excitement!!!

Thy peoples of the realm shall be greatly pleased by such a mighty gift. 

To remove the curse of lightning.  How benevolent art thou!!

Thou glowing composition doth seem a little small.  Perhaps thou have more to giveth?

This feast should help.  Tis high in fibre I have been told.  Just what thou needs I doth think!

Sir CPhill's offering shall so grow.  Tis such an magnificent gift you bestow upon him!  

I am but your humble servant, Great Zeus

Lady Guinevere.

I giveth thee this glowing composition to be guideth to CPhill by thy most excelling rider.

Jupiter be curseth for not granting me the passage to him myself.

If thou triumphstedth I reward thee by not giving thy lands the curse of lightning.

Combination locks should really be called Permutation locks shouldn't they ?

Afterall, if I told you that  the three numbers on a simple lock were 2, 3 and 7 you wouldn't necessarily expect them to be in that order. You would have 3! = 6 different orders to try.

On that basis I think that the previous answer needs to be divided by 8! = 40320.

I bet that went over a head Reinout!  lol

$$\\\$4.50+ \$3.50+ \$5.50 +\$2.90 = \$16.40\\
\\
\dfrac{\$16.40}{4}=\$4.10\\
\\
\text{The average of }\; \$4.50, \$3.50, \$5.50 \text{ and } \$2.90 \text{ is } \$4.10$$

$$1+\tan^{2}\psi=\sec^{2}\psi

(Divide \quad\cos^{2}\psi+\sin^{2}\psi=1\quad\text{ by }\quad\cos^{2}\psi.)$$

$$\text{So,}\quad\frac{\tan^{2}\psi+2}{1+\tan^{2}\psi}=\frac{\sec^{2}\psi+1}{\sec^{2}\psi}=1+\cos^{2}\psi.$$

You can find the answer by adding all the values and dividing by the number of values (in this case 4)

So the answer is

$${\frac{\left({\mathtt{4.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.9}}\right)}{{\mathtt{4}}}} = {\frac{{\mathtt{41}}}{{\mathtt{10}}}} = {\mathtt{4.1}}$$

So your answer is $4.10

Reinout 

You don't want your answer now?

then I'll give it later...

Oh Divine Zeus,

Supreme Ruler of the Gods,

Thou hath humbled thine self to check upon my unworthy self!  Tis a great honour you bestow!

Thou hath visited but briefly but perhaps Artemis was about for the wind hath been a little wild.  Me thinks she found little to hunt and it was the day, the moon was but a memory, thus her proximity was relatively afar. Still, I fancy she was somewhere round about.

Does this mean that Sir CPhill must search about thine wholly excrement?  What a great honour you hath bestowed upon him.  Sisyphus’s  boulder should no longer be impenetrable.  Perhaps his quest for the roman zero will find its finale! How great shall be his joy!

I am but your humble servant, Great Zeus

Lady Guinevere.

$$tan\theta =\frac{10}{190}$$

$$\theta=tan^{-1}\left(\frac{10}{190}\right)$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{10}}}{{\mathtt{190}}}}\right)} = {\mathtt{3.012\: \!787\: \!504\: \!183^{\circ}}}$$

approximately 3 degrees.

Compound interest ot simple interest.

Simple interest

I=Prt

$${\mathtt{I}} = {\mathtt{250}}{\mathtt{\,\times\,}}{\mathtt{0.072}}{\mathtt{\,\times\,}}{\mathtt{5}} \Rightarrow {\mathtt{I}} = {\mathtt{90}}$$

Interest = $90

I think you meant 40% and 60%.

That would make sense.

Then the first answer is correct.

(Except in older British) a billion is a 10 digit number. Therefore 1.671 billion can be written as 1671000000.

This number has six zeroes.

Reinout