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cos(17pi/6)

We know that pi/6 = 30 degrees, so 17pi/6 must be 30*17 = 510 degrees.

Also, note that pi = 180 degrees which means 2pi = 360 degrees.

Now, 510 - 360 degrees = 150 degrees, which means we have made a full 2pi revolution and are left with 150 degrees --> which is pi/6*5 [30 degrees*5 = 150 degrees] --> 5pi/6.

Now, cos(5pi/6) can be evaluated as 150 degrees or the 2nd quadrant with a negative x value since pi/6 is positive sqrt3/2 and lies in the 1st quadrant.

Therefore, we have cos(5pi/6) = -sqrt(3)/2.

46.2*10^-2 --> You have to move the decimal place to the left twice (since it is ^-2) instead of right twice, so the answer will be 0.462.

2x^2+3cy-4y^2; x = 2 & y = -4

2(2^2) + 3c(-4) - 4(-4^2) -->

8 -12c - 64 --> 

We need to know: c = ?

It depends upon what "n" might be.......as is.....all we can say is that it equals 2n

Wow! That actually helped alot!

Thanks Melody. :)

This message was sent to me by NinjaDevo via private message.

I am copying it here because it will probably be of interest to other people as well.

NINJA'S QUESTION:

I was looking through those LaTeX videos you posted in the LaTeX thread. 
I was wondering though, what's the difference between beginning a document and an array? For example, I see in some of your posts you did "\begin{array}{rll}" What does this mean and what does the rll mean in the brackets? 

Thanks, 
~Ninja 

MELODY'S ANSWER:

There a many different versions of LaTex which is really annoying. I am still glad I started with the videos, although i have actually not studied all of them. I should, it would probably help a lot.

When you write in Texmaker you have to enter the whole program, there are shortcuts to help you but you can still still see the whole thing. On any maths forum it is different. Some of it is pre-written so that you do not need to enter lines such as \begin{document} They have already been entered for you.

array is the term used when you want to line up objects such as equal signs.  There are a number of different types of arrays but this forum only seems to have the one.  It does everything the others do it is just a little harder to use and to be honest i haven't totally worked it out yet.

If i was going to wrtite a standard equation with the equal signs lined up then I woudl use 

I have added a litle equation solution to show you how it works

\begin{array}{rll}   note this must be ended with     

x-6&=&12\\

x&=&18

\end{array}    or it will not work!

Now array means you are lining things up.  r stands for right, l for left and c for centre.

The equal signs are going to line up.

The first input is going to be right aligned against the first&

The equal sign is pushed left so it butts up against the invisable first&

Then the other side of the equation is pushed left as well to butt up against the second &

There is always one more alignment command (l,r, or c) than the number of ampersands & used.

Now let's see if my code works.

$$\begin{array}{rll}
x-6&=&12\\
x&=&18
\end{array}$$

Good it worked well.     

Does that help Ninja?

Thanks anonymous.

If we're not carefull, be can just solve equations like these in a breeze.

Thanks for the correction, I didn't even think about that.

Thurs 31/7/14

The diameter is twice the radius.....however.....if you're asking how to construct a diameter in a circle, here's one way using Euclid's simple method for finding the center of a circle.....the line CE that bisects the chord AB at right angles is a diameter........

@@ End of Day Wrap :   Wed 30/7/14     Sydney, Australia     Time 12:45am   (Really Thursday morning)   ♬

Hi all,        ♫♪

Today our great answers were delivered by CPhill, Will85237, Heureka, AzizHusain, DragonSlayer554, Alan and NinjaDevo.  Thank you all - Your blood's worth bottling.     If you do not know what this means then look here.                    

You can do both of these with the on-site calculator......

LaTex is a program for writing mathematical things.  It is what we use for all the presentations shown above.

If you would like to start learning you can start a new thread in the main part of the forum asking for some beginner lessons.  Please don't post in here any more because this thread is just for Latex reference not for learning with member help if that makes sense.

If anyone is really serious about learning, here is a good place to start

I also really liked the fact that this was a serious thread and that no one posted any irrelevant, distracting posts in the middle of it.  Thank you everyone.

Yes, that's fine. Pleased to be able to help.

Anonymous is correct!

I did it the right way but I forgot that the answers to these types of questions always have to be checked.

and x=-4 simply does not work!

Taking my own little equation

|2x-8| = 9X+20

LHS = |2*-4-8|=16

RHS = 9*-4+20 = -16  not equal LHS

So as Anonmous has so adamantly stated, the ONLY answer is x=-12/11

THANK YOU ANONYMOUS!

Note: I should have checked x=-12/11 too, but it will be correct.

Does this answer satisfy you Anonymous.  If not please say so - we are all here to learn.

Suppose that the number is -4.

What is twice that number ? Answer -8.

What is the difference between -8 and 8 ? Answer 16.

This is 20 more than 9 times the number, so what is 9 times the number ? Answer -4.

So 9 times the number is -4 ?

Wrong, 9 times the number is -36.

The difference between twice the number and 8 can be written (should be written) as    |2x -  8| and the equation will be |2x - 8| = 9x + 20, (as you say).

If the expression (2x - 8) is positive, the equation can be written as 2x - 8 = 9x + 20. However, this leads to a contradiction, since x = -4 makes the expression negative.

Look at it this way. Substitute your x = -4 into the equation |2x - 8| = 9x + 20 and see what you get. |2(-4) - 8| = 9(-4) + 20, |-16| = -36 + 20, 16 = -16. It doesn't work,      x = -4 is not a solution of the equation.

$$csc^{-1}(2)\\\\
=sin^{-1}\left(\frac{1}{2}\right)\\\\
= 30^0 \quad or \quad \frac{\pi}{6}\;radians\\\\\\

cot^{-1}(-\sqrt3)\\\\
=tan^{-1}\left(\frac{-1}{\sqrt3}\right)\\\\
=tan^{-1}\left(\frac{-1}{\sqrt3}\right)\\\\
= -30^0 \quad or \quad \frac{-\pi}{6}\;radians\\\\\\$$

You do not need to take points off yourself Chris, I thought your answer was worth a tick!      

It was a good point to reiterate - and yours was perhaps a little clearer.  

Logs are expontents (or powers or indices- these are all the same thing)

$$10^2=100\\
log_{10}100=2$$

See the power is 2, the log is 2  THE LOG IS THE POWER

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

You use logs when you need to find an exponent (index).

eg      $$17=10^x$$

We need to get rid of the 10. We do that by taking the log (base 10) of the RHS but we have to keep the equation balanced so we do it to the left side as well

$$log_{10}17 = log_{10}10^x\\

log_{10}17 = x\\

x= 1.2304489 \qquad\mbox{(correct to 7dp)}$$

I saw that after I had posted...let me give Aziz full credit too!!   (I won't take any points for my repeated answer!!.......)

Thanks Chris,

Aziz had already done that but perhaps your working is a bit clearer.

I agree with Melody's answer......however......I'll bet that 99% of us would have worked it just like ND did.......(I know I would have !!!)

Let me add one thing to what Melody is saying......we can always "test" whether a function is even by replacing x with -x.....if the reults are the same, then the function is even.

Then, is 2^(x) = 2^(-x)  ???........NO !!!

Note that a function like x² is even, because...

x² = (-x)²  

And this is exactly what Melody said......

Thank you Anonymous for pointing that out  

I think that you each have half the solution.

The difference between 5 and 3 =2

The difference between 3 and 5 is also 2

So the difference between 2 numbers n1 and n2 is  |n1-n2|

So the problems becomes

|2x-8| = 9X+20

This has 2 solutions

$$\begin{array}{rllrrll}
2x-8&=&9x+20 &or& \qquad 2x-8&=&-(9x+20)\\\\
-7x&=&28 &or& 2x-8&=&-9x-20\\\\
x&=&-4 &or& 11x&=&-12\\\\
x&=&-4 &or& x&=&\frac{-12}{11}\\\\


\end{array}$$

I'm assuming that you might be trying to find the missing side of a right triangle with a leg of 4cm and a hypoteneuse of 64cm.....so, squaring both sides, we have......

4096 = x^2 + 16   subract 16 from both sides

4080 = x^2      take the positive root of both sides

√4080 = x = √(16*255) = 4√255 cm..... or about 63.87 cm

Thanks, CPhil!