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Since  sqrt (7)   and  1 - sqrt (6)  are roots  then  so  are  -sqrt (7) and  1 + sqrt (6)

So....we  have a 5th degree polynomial

This polynomial  is

a ( x + 3) ( x- sqrt 7) ( x + sqrt 7) ( x - 1  + sqrt 6)  ( x - 1 - sqrt (6)

Simplifying this  we  gat

a (x^5 - x^4 - 18x^3  - 22x^2  + 77x  + 105)

And  since  we  know   that   P(-1)  =  8, we  can solve  this  for  "a"

8  =  a ( -1 + 3) (- 1 - sqrt 7)  (-1 + sqrt 7) ( -1 - 1 + sqrt (6))  ( -1 - 1 - sqrt (6))   simplify

8   = a  (24)

a =  (8/24)  =   1/3

The polynomial  is    (1/3)  ( x^5 + x^4 - 18 x^3 - 22 x^2 + 77 x + 105)

So.....P(0)  =     105 / 3    =   35

I got \(2,-3,2+i\sqrt{3}, 2-i\sqrt3\) but the website said it was wrong

Very nice, Melody......I could  prove this with coordinates, but  your solution is more elegant  (I just didn't draw  enough lines.....LOL!!!!)

2n^5  +25n^2 + 32n - 15

________________________

6n^4 + 2n^3  -11n^2-2n+ 17

You are correct.....the limit does not exist

If  we  divide  every term by  n^4   and then  let  n  approach infinity.....we will be  left with

2n  + 0 + 0   +  0

________________  =     (2/6)n    =     n / 3     =  inf  / 3   =    the limit does not exist

  6  + 0 + 0 + 0 + 0

This will alwas happen  when  we  have  a higher power polynomial /  a lower power polynomial

Just plug in the points. 

(y−1=-3(x−2))

=^._.^=

Nice work, or you could simplify. 

11!(1 + 12 + 12*13)

11!(169)

11! * 13 * 13

=^._.^=

(-3)^2 = 9

-3^2 = -9

The reason for this is pemdas, a calculator will first calculate the exponent (3^2), and then the negative sign. 

-3^2 = -9 

-(3^2) = -9

-(9) = -9

To fix this problem, just add a parentheses. 

(-3)^2 = 9

=^._.^=

Find the largest prime divisor of 11! + 12! + 13! ==6,745,939,200 = 2^8 * 3^4 * 5^2 * 7 * 11 * 13^2

13 is the largest prime divisor.

Ohhhh. :))

Yes of course it can. 

The people on here will help you understand those problems. 

=^._.^=

The web2 calc of course.     

It depends what help you want.  It won't just give you answers without you understanding anything.

x*x*x = x   (number of men * working hours a day * number of days = number of articles)

y*y*y = something

y = (x) *y/x

x*(y/x)*x*(y/x)*x*(y/x) = x*(y/x)*(y/x)*(y/x)

y*y*y = y^3/x^2

This problem quite tricky. 

=^._.^=

Ok I have simplified it and made it a 2x by 2y rectangle (which is a special case parallelogram)

This is a rough outline:

Triangle DPC and triangle EPB are similar.  DC=2x   and  EB=x    so the little one is half the size of the big one.

So  JP=2*IP

so 3*IP= 2y

IP = 2y/3

so  PG= y-  2y/3  = y/3

Now look at similar triangles  AF and i forgot to give it a name .. . ok look at the big green triangle and the similar little grey one.

PG is y/3  and  A? is  y  so the big one is 3 times as big as the little one.

So AP:AF = 2:3

Which calculator?

=^._.^=

(x-2)(x+3)(x^2-4x+7)

That's the factored form. 

Can you take it from here?

=^._.^=

I really appreciate the help!

Thanks this really helped!

I don't know but i have an interactive graph here:

I have no idea, sorry.

I have opened the question though, so other people can answer here. 

If it starts with an even

There are 4 first digit possibilities (not 5 becasue it can's start with 0)  , 4 third digits 3 fifth digits etc  so that is    4*4!  *   (now for the odd digits)  4!

If it starts with an odd then there will be  5!*4! different ones

Now add them together.

4*4!*4! + 5!*4!

Oh, I only just saw the 'same guest here' bit.

It is really good that you are discussing your own problem.  We really like that.

Why don't you become a member and get known to us.  

Members who make a good impression get preferential treatment. 

Hi Guest, your logic sounds good to me.   

It can definitely be a square or a rhombus.  

I think the translation condition means that opposite sides must be parallel and equal length.

The second condition means that the diagonals must intersect at right angles.

there are no other quadrilaterals that meet both these conditions.

Let p(x) = x^(2), q(x) = 2x^(3)+ 1, and r(x) = 7x^(4) - 5x^(3) - 6x - 2. Find p(q(x)) + r(x).

\( p(x) = x^2, \quad q(x) = 2x^3+ 1,\quad and \quad r(x) = 7x^4 - 5x^3- 6x - 2. \\Find \qquad p(q(x)) + r(x).\\ \\~\\ p(q(x)) + r(x)=(2x^3+ 1)^2\quad +7x^4 - 5x^3- 6x - 2\)

You can finish it.

Thank you CPhill,

here is a nice youtube video

How To prove that 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...))))