cryptoaops

No problem!

Also, I wanted to notify you that there should be no cheating on Alcumus or Homework as it is not allowed on AoPS.

Thank you for this. If people do need help, they can ask (sorry if im being rude).

\(2, 3, 5, x, 2x\) creates the largest median no matter what. \(x\) can be any integer you want it to be, but the median will always be \(5\)

First of all, the image is below not above 

The hexagon is composed of 6 isoceles triangles. In the sketch above, you just have to draw their apothem and then you get 6x2 right-angled triangles. So you have the same little right-angled triangles composing the whole hexagone. Each isoceles triangle has 6 little right-angled triangles and the total is \(\LaTeX\): \(6\text{isoceles}\triangle\times6\text{little right-angled triangle}=36\). You count \(\LaTeX\): \(4\text{little right-angled}\triangle\times2\text{shaded area}=8\text{shaded little right-angled triangles}\). So the ratio is: \(\frac{36}{8}=\frac{9\times4}{2\times4}=\boxed{\frac92}\). And \(\dfrac92=9:2\).

Do you mean \(=\) or \(\equiv\)? If you mean \(\equiv\) then I can help you.

From the second equation, since \(x \equiv 9 \pmod{20}\), we know \(x\) is odd. Then \(r\) can be one of \(1\), \(3\), and \(5\). From the third equation, \(x \equiv 4 \pmod{45}\). We see that \(x \equiv 1 \pmod{3}\) (since \(x = 45y+4\), for some integer \(y\)). Then we see that \(r\) can only be \(\boxed1\) (since if \(x \equiv 3 \mod{6}\), it will have remainder of \(0\) when divided by \(3\) and if \(x \equiv 5 \mod{6}\), it will have a remainder of \(2\) when divided by \(3\).

Well, let's take a deeper look.

It's in the \(5\times5\) square.

it's in all \(4\) of the \(4\times4\) squares.

it's in all \(9\) of the \(3\times3\) squares.

It's in \(4\) of the \(2\times2\) squares.

It's in \(1\) of the \(1\times1\) squares.

That would leave us with a total of \(\boxed{19}\) squares.

Not sure  about this... But  here's  my rational

 

We  know that  we  have  \(15\)  scores of \(10\).

 

Since  the  mode is  \(7\)  we  need to have  \(16\)  scores of \(7\)  (and no more)

 

The  median  is  \(9\), so the \(20\) and \(21st\)  scores  must be  \(9...\) to keep the average as small as possible.

 

And the other  \(3\)  scores  need to be  as small as possible so let them  \(=1\)

 

We  have

\(1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7 ,7, 7, 9 ,9, 9, 9, 9, 9,10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10\)

 

\(\text{Median}  = 9\)

\(\text{Mode} = 7\)

\(\text{Mean} = 7.975  =  8\)

Thank you!

Solution #1: \(2x^2 - 3x + 27 = 0 \)        

Let the roots be \(p\) and \(q\).

The sum of these roots \(=\frac32\).

The product of these roots \(=\frac{27}{2}\).

So

\(p + q   =  \frac32\) (1)

And

\(pq =\frac{27}{2}\)

\(2 pq =  27\) (2)

Square both sides of (1)

\(p^2 + 2pq  + q^2  = \frac94\) Sub (2) into this

\(p^2 + (27) + q^2 = \frac94\)

\(p^2 + q^2 =      \frac94 - 27 \)

\(p^2 + q^2 =   \frac94 - \frac{108}{4}\)

\(p^2 + q^2 =  -\frac{99}{4}\)

\(p^2 + q^2 =  \boxed{-24.75}\)

Solution #2: Use quadratic formula to find the two roots as  \(\frac34 + - 3.5968i\).

\((\frac34+ 3.5968i)^2 = \frac{9}{16} +\frac32 (3.5968i) - 12.9375\)

\((\frac34- 3.5968i)^2 = \frac{9}{16} -\frac32 (3.5968i) - 12.9375\)

Added together, \(\frac98 - 25.875 = \boxed{-24.75}\).