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how many languages are u fluent in or did u use google translate? Melody

https://translate.google.com/#view=home&op=translate&sl=auto&tl=en&text=%D8%AA%D9%85%20%D8%B3%D8%AD%D8%A8%20%D8%AD%D8%B1%D9%81%20%D9%85%D9%86%20%D8%A7%D9%84%D8%A8%D8%B7%D8%A7%D9%82%D8%A7%D8%AA%20%D9%88%D8%AA%D9%85%D8%AA%20%D8%A7%D8%AF%D8%A7%D8%B1%D8%A9%20%D8%A7%D9%84%D9%82%D8%B1%D8%B5%20%D8%A7%D9%84%D8%AF%D9%88%D8%A7%D8%B1.%20%D9%85%D8%A7%20%D8%A7%D8%AD%D8%AA%D9%85%D8%A7%D9%84%20%D8%A7%D9%86%20%D9%8A%D9%83%D9%88%D9%86%20%D9%83%D9%84%20%D9%85%D9%86%D9%87%D9%85%D8%A7%20%D8%AD%D8%B1%D9%81%D8%A7%20%D9%85%D8%AA%D8%AD%D8%B1%D9%83%D8%A7%20%D8%9F

25 combanations, so total is 3628800....... simplify \(\frac{25}{3628800}\)which equals \(\frac{1}{145152}\)

Thank you!

Fair enough.

Here is a good site to give you the basic definitions.

Thank you very much sir!!

thanks, guest!

Yes, that is much better.  Be careful to proofread your questions.   

Have you worked out what is wrong yet?   I will leave you with that.

ALSO

You had no hope of following guests logic as he did not have any. He had no idea what he was talking about.

Here is my take:

Consider y=(x+a)(2x+p)     

The roots for this are -a and -p/2    So (-p/2, 0)   is a point on this parabola

Now consider.

  \(y=x^2+px+q\\ \text{The axis of symmetry for this is }\frac{-p}{2}\\ \text{since this is a concave up parabola and the roots are real}\\ \text{It follows that when } x=\frac{-p}{2} \text{ the y value is negative or 0.}\\\)

When I add these two parabolas together the resulting y value when x=-p/2  MUST be equal to or less than 0.

The resultant parabola is concave up since the x^2 coefficient is positive (+3 to be precise).

Hence it must cross the x in 1 or 2 places and hence have 2 real roots (they can be equal roots)

Here is a graph that may help you understand.

im sorry.

Let \(p\) and \(q\) be real numbers such that the roots of \(x^2 + px + q = 0\)are real. Prove that the roots of \(x^2 + px + q + (x + a)(2x + p) = 0\) are real, for any real number \(a\). 

Brilliant! Thank you very much Alan.

Looking at the first few numbers in the sequence we see that the remainders cycle in the order 1, 2, 2, 4, 3, 2 with period 6. 

So we just need to find the remainder when 1000 is divided by 6. This is 4.  The numbers above correspond to remainders of 0, 1, 2, 3, 4, 5 respectively, so the remainder when a₁₀₀₀ is divided by 5 is 3.

The five numbers: 1, 2, 3, 4, 5, give: 5! = 120 permutations. Since each permutation sums up to: 1+2+3+4+5=15. So 15 x 120 =1,800 - the sum of all 120 arrangements.

"The cubic polynomial x^3 + ax^2 + b has three roots 2, 4, and r.  Find a and b."

Your question is not complete.

I can guess at the bits that are missing but I shouldn't have to do that.

Please check it properly.

No I don't know, I've been looking to get some explanation on YouTube and the net on how to start but couldn't find anything helpful. 
 

I think of a number.

mmm I chose 8

I multiply it by 2  so that is    2*8

I subtract 9 so that is           2*8-9

Now I do the same thing only I let the number I choose be x.   Try it.

(You have to just pretend that x is a number)

If you have the pic in front of you then the question is probably done.

I'd draw one of the triangles and find the 3 centres, then I'd probably start on the other triangle.

But I have not done it so maybe I would have to change my mind,

Have you made any start at all?

Do you even know what those three centres are?

Do not answer with just 'yes'

I am asking you to interact with me properly.