Melody

like this

\(sin\theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}\\~\\ RHS=\frac{[cos\theta+isin\theta]-[cos(-\theta)+isin(-\theta)]}{2i}\\ RHS=\frac{[cos\theta+isin\theta]-[cos(\theta)-isin(\theta)]}{2i}\\ RHS=\frac{2isin\theta}{2i}\\ RHS=sin\theta\\~\\ RHS=LHS \qquad \qquad Q.E.D.\)

Well, how about sharing your solution with other people here. 

\((x+2\sqrt3)^7\\~\\ 7C3*(2\sqrt3)^4(x)^3 \)

That is the  x^3 tern.  You can take it from there.

Please just ask one question per post.

I am closing this post.

It would also be beneficial for you to talk about what you have tried, or thought about, for yourself.

Hi FantasticFlower  

We will welcome your questions especially if you tell us what you have attempted and you interact with the answerers.

If you are polite (which I think you will be) and you interact with answerers you will get a great reputation and your question will be responded to much more quickly and seriously.

I hope you learn lots and have fun here.      

This is the same one that I looked at the other day.

I know I did not answer it, but I did go to the effort to draw a comprehensive pic.   Which you completely ignored.

Anyway, it does not matter that I did anything at all.

This is a repost hence you need to guide people to the original question.

You need to include a hyperlink to the original and request that people answer on the original.

And for them to just leave a little note here saying that they have done so.

If you ever need an original question unlocked then send a polite note to a moderator asking them to open it for you.

Hi Technoblade,

You have already had your hand slapped by MPS and Why, and they are right, you should not take the credit for other people's answers.

BUT

I am pleased that you are showing an interest here. 

If you find an answer like you did here, post a link to it.  I'd be more than happy to give you a point for that.  

Finding previous answers is a skill in itself.  If you learn from that answer then better still. 

Don't be discouraged,  just don't plagiarize and you will fit in here just fine.   

Thanks very much guest, I am glad I could help   

Very impressive guest.

Let's see if I can find a quicker way...

1,2,2,3,3,4

A standard six-sided die is rolled 6 times.

You are told that among the rolls, there was one 1, 2 two's, two three 's, and one four.

How many possible sequences of rolls could there have been?

6!/(2!2!) = (2*3*4*5*6)/4 = 2*3*5*6 = 180 ways

or

How many ways to place the 2's     6C2 = 15

for each of those there are 4C2 ways to place the 3's = 6

then 2 ways to place the 4

and then only 1 way to place the 1

15*6*2=180 ways

So we have done it 3 ways in total and each time we got the answer 180 ways    

Hi xy   

Thanks for answering HSDx and guest     But, guest, my answer is for you to learn from as well    

What real value of t produces the smallest value of the quadratic t^2-9t-36?

maybe it will make more sense to you change the t to an x and then put it equal to y

\(y=x^2-9x-36 \)

Now you are being asked for the smallest value of x

     If you graph this you will get a concave up parabola.

    I know this becasue the x^2 makes it a parabola

    and the invisable number in front of the x^2 is  +1. 

    Since it is + the parabola will be concave up.

The lowest point of a concave up parabola will be the turning point. The x value for it which will be halfway between the roots.

roots are when y=0 so

\(0=x^2-9x-36\\ 0=(x-12)(x+3)\\ x=12\;\;or \;\;x=-3 \)

Halfway between the roots is  x= (12+-3)/2 = 4.5

That is the x value (actually the t value) that you want.     t=4.5

that is the value that will give the expression the smallest value.

If you wanted to find that smallest value then you would sub 4=4.5 back into the expression.