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OK......let's see if we can find a pattern

Suppose we had three people.....the third person shakes hands with persons one and two....the second person then shakes hands with person one. Note that person  one doesn't have anyone left to shake hands with. So, that's three total handskakes.

What about four people?? The fourth person shakes hands with the other three, the third person shakes hands with persons one and two, the second person shakes hands with person one, and again, person one has no one left. So, that's a total of 3+2+1 = 6 

Look at the patttern so far.......

2 people = 1 hand shake

3 people = 2 + 1 handshakes = 3

4 people = 3 +  2 + 1 handshakes = 6

So...it appears that with "n" people, the number of handshakes is just the sum of the first "n-1" counting numbers. There is a "formula" for this given by:

n* (n-1) /2 

where n is the number of people...To see that it's true, note that the number of handshakes between 4 people are:

(4) * (3) / 2 = 12/2 = 6...just as we found.....

So....with 20 people, we'd need...     (20)*(19)/2 = 190 handshakes

So 78 handshakes won't be enough...

Hope this helps........

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someone help please?

$$\begin{array}{rlll} sin(x)&=&cos(2x)\\ sin(x)&=&cos^2(x)-sin^2(x)\\ sin(x)&=&1-sin^2(x)-sin^2(x)\\ sin(x)&=&1-2sin^2(x)\\ 2sin^2(x)+sin(x)-1&=&0\qquad &\mbox{let y=sin(x)}\\ 2y^2\:\:+y\:\:-1&=&0\\ 2y^2\:\:+2y-1y\:\:-1&=&0\\ 2y(y+1)-1(y+1)&=&0\\ (2y-1)(y+1)&=&0\\ \end{array}$$

$$\begin{array}{rllrl} 
 2y-1&=&0\qquad y+1&=&0\\ y&=&\frac{1}{2}\qquad y&=&-1\\ sin(x)&=&\frac{1}{2}\qquad sin(x)&=&-1 \end{array}$$


$$x=\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{3\pi}{2}\qquad \mbox{for}\:\: 0\le x\le 2\pi}}$$

NOTE:You need the y= out the front or the question is meaningless!

CPhill has used the first bit of the quadratic formula to find the x value.

Another way is to factorise.

$$y=x^2+5x+6\\
y=(x+3)(x+2)$$

The roots are where y=0 and this is where x+3=0  or  x+2=0

Solving these gives you x=-3 and x=-2

The axis of symmetry will give you the x value of the vertex.

$$x=\frac{-3+-2}{2}=\frac{-5}{2}$$

Now the rest is exactly the same as CPhill's method.

Actually....if you ask a "Physiques" question........we'll answer you in French.....

Melody is right. There are few here who understand physics, but most here understand physiques.

if you ask a physiques question you will receive an answer quickly.

by Physiques Specialist

If you mean you want to find the inverse cosine, go to the calculator and hit the "2nd" button.....you will see the inverse cosine button labeled "acos." Make sure that you specify the correct mode that you want at the bottom left (either degrees or rads).......

So it does GoldenLeaf.  Congratulations!

Yes it is just like an x,  the question is telling you to replace all the ts with 4

That is a great signature you have there NBlueBird.

Look very carefully at your denominator Stu.  here is a copy of your last posted attempt:

The sin term should be multiplied by 4.005. You have added 4.005 to it.

Do you mean the probability of exactly 1 success in 4 trials given a probability of 0.1 of an individual success? If so, it is normal to put the 1 before the 4.

prob(k;n,p) = nCr(n,k)*p^k(1-p)^n-k    n = number of trials (4); k = number of successes (1); p = probability of individual success (0.1).

$${\mathtt{prob}} = {\left({\frac{{\mathtt{4}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.1}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.1}}\right)}^{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} \Rightarrow {\mathtt{prob}} = {\mathtt{0.291\: \!6}}$$

I really do not think the 3 line symbol is used much.

An equal sign with a wavy line on top means "Is congruent to". It is used in geometry to say that one shape or angle is the same as another one.

With numbers you say 'equal to'

With geometrical things you say 'congruent to'

http://rapidtables.com/math/symbols/Basic_Math_Symbols.htm

Yes, paper clips come in different lengths!

You were almost right Stu; just one small (but vital) mistake in the last line:  cos(-3) is +cos(3) not -cos(3).

Here is more on the same theme.

Thanks Chris that is a really good answer to a really good question.  Actually it is an excellent answer.

I don't know why I wrote more - sometimes I just can't help myself.

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I am going to have another take on this.

Think about temperature.  In the freezer it would be a negative temperature and in a fire it would be a high positive temperature so you can think of positive numbers as being 'hot' numbers and negative numbers as being 'cold' numbers.

So  - - means take away cold.     If you take away cold then it must get HOTTER!   So - - = +

Lets think of the other possibilities 

- + means take away heats.  If you take away heat it must get colder So - + = -

+ - means add cold. If you add cold it must get colder So + - = -

+ + means add heat.  If you add heat it must get hotter. So + + = +

Now that you understand why this is true it is easy.  I'll say it in a different way.

If the + and - signs are SO close together that you can put a circle around them without anything else getting in the way then.  

If the signs are the same replace them with a +

If the signs are different replace them with a -

This is another post that you may find helpful.

CPhill is correct - He's rarely wrong.  I just want to look at it myself.

Think about this for a moment

$$|a|=|t|$$

This would be true if x=t and it would also be true if x=-t 

The same thing applies to your question:

$$\begin{array}{rlll}
|3y+6|&=&|3y+3|\\
(3y+6)&=&\pm (3y+3)\quad &\mbox{now factorise both sides}\\
3(y+2)&=&\pm 3(y+1)\quad &\mbox{next divide both sides by 3}\\
y+2&=&\pm (y+1)




\end{array}$$

Now we seperate this into 2 equations

$$\begin{array}{rlllrl}
y+2&=&+(y+1)\quad &\mbox{and}\qquad y+2&=&-(y+1)\\
y+2&=&y+1\quad &\mbox{and}\qquad y+2&=&-y-1\\
y&=&y-1\quad &\mbox{and}\qquad 2y&=&-3\\
y&=&\mbox{no soln}\quad &\mbox{and}\qquad y&=&\frac{-3}{2}\\
\end{array}$$

So the answer is y=-1.5

Sunday 1/6/14

Answerers must always remember that many/most people who use this forum have never heard of imaginary or complex numbers!  This poster has been trying to figure out our answer to this question for quite some time!

Hi anonymous,

Now I finally understand what your problem was. (Thank you Alan)  I am so glad that you have been persistent in your questioning.  This is what I want everyone to do!  If you do not understand our answer or you think we are wrong just keep at us until you are happy.  Remember though to try and work out WHY we don't understand your question or your problems.  That way we can help each other.

In this case higher mathematician, including me, forgot that most  student have never heard of imaginary numbers.  I am so glad that you have reminded me of this divide.  

Thanks, I am genuinely grateful to you.

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I just want to talk about it a little more:

x^2+7x+53=11/3

Think about this in this way (Alan has done it a bit different but visually this might be easier for you to understand.)

$$x^2+7x+53=\frac{11}{3}$$

I am going to split this up into two bits.   I'm going to let y = the left side and y=right side. Like this;

$$y=x^2+7x+53\mbox\qquad { and }\qquad y=\frac{11}{3}$$

Now I am going to graph these two functions.  Where they cross each other will be where  $$x^2+7x+53=\frac{11}{3}$$

NOTE:  This is how you always solve simultaneous equations when doing it graphically.

You can see from the graph that it is impossible for

  $$y\quad \mbox{to equal}\quad x^2+7x+53\mbox\quad {and}\quad \frac{11}{3}\quad \mbox {both at the same time.}$$

So as Alan said, there are no real solutions, the answer in the set of REAL numbers is {} (an empty set.)

There are however complex number solutions - These are imaginary numbers! They are the answers that we were so determined to give you.  Sorry.

Why didn't you do this yourself on the site calc anonymous?