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thanks :)

I believe this is the Shrodinger equation in operator form for a quantum harmonic oscillator.

HAHAHA!!!!....your approach was better than mine....!!!!!

Note, KeyLime

The median of  a set of an even number n of ordered data values  is  the  average of the (n/2)th data value  and the (n + 2)/2 th data value

So...here....we want the  average of  the  (10/2)th data value  = 5th data value  and the (10+2)/2 = 6th data value

Since we want the average to be as small as possible....the 5th and 6th data values need to be the integers on either side of 10 that are closest to 10.....i.e.,  9 and 11   [ just as Rom said ]

Note that any other two values will produce an average  > 8.4

To see this....let's suppose that the 5th and 6th data values are  8  and 12

The median is  [ 8 + 12] / 2  = 10

And the elements < 8   need to be as small as possible...i.e, 1,2, 3 , 4

An d the elements  > 12 need to be as small as possible......i.e., 13, 14, 15, 16

Note that the average becomes   [ 1 + 2 + 3 + 4 + 8 + 12 + 13 + 14 + 15 + 16 ]  / 10   =  8.8

Which is > 8.4

Try this with the 5th and 6th data values  being 7 and 13

So....the elements need to be  1,2,3,4, 7, 13, 14, 15, 16, 17

Convince yourself that this average will also be > 8.4

Thanks CPhill!

Let the bottom left  vertex of the small square be located at (0,0)

Let the top right vertex of the large square be located at  (12, 9)

So....the equation of the line connecting these two points  is  y = (9/12) x  ⇒ y =  (3/4)x     (1) 

The right edge of the small square lies on the line x = 3      (2)

So....subbing  (2) into (1)  we have that

y = (3/4)(3)  =  9/4

Then....the height of the shaded area   =   9 - 9/4  =   27/4 cm

And the width of the shaded area  = 9  cm

And since the shaded area is triangular, its area = 

(1/2) width * height  =

(1/2)(9)(27/4)  = 

(9/2)(27/4)   =

243/8  cm^2   =

30.375 cm^2

But how did you get 9 and 11?? Never mind I just realized thanks

Oh I just found the answer:

The two triangles that are shaded blue are similar.

So we have the following ratios:

3/3+9 = x/9

x=9/4

So height of the black triangle is 9 - 9/4 = 27/4

So area of black triangle is (9 * 27/4)/2 = 243/8

EDIT: here's the image

As the numbers are distinct they can be ordered.

The median will be the average of the 5th and 6th elements so they must be 9 and 11

Then the 4 numbers less than 9 should be 1 2 3 4

and the 4 numbers greater than 11 should be 12 13 14 15

The average of all the numbers is

(10+20+54)/10 = 8.4

Here is the first one without using any trig

Look at the following diagram :

Again.....imagining that we have a cross-section.....

Let EF  be the radius of the sphere drawn to tangent DB

Let  AD be the cone height

We have triangles DAB and DEF

Angle DAB  = angle DFE = 90°

And angle ADF = angle EDF

So....by AA congruency....triangle ADB  is similar to triangle FDE

So.....DE / DB   = FE/AB

Let DB = the slant height of the cone = S

Let DE = cone height - 10  =  H - 10

FE = 10

AB = 15

So we have that

H-10    =   10

____        __

 S             15

H - 10           2

_____   =    ____

  S                 3

H - 10   = (2/3)S

H = (2/3)S + 10

So......using the Pythagorean Theorem

S^2 - [ (2/3)S + 10 ]^2  = 15^2

S^2 - [ (4/9)S^2 + (40/3)S + 100] =  225

S^2 - (4/9)S^2 -(40/3)S^2 - 100  =  225

(5/9)S^2  - (40/3)S - 325   = 0       multiply through by 9/5

S^2  - 24S  - 585  = 0       factor as

(S - 39) (S + 15) =  0

Since S is positive....then setting the first factor to 0  and solving for S  gives that S = 39

So....the height of the cone  = (2/3)(39) + 10   =  26 + 10   = 36

So....the volume of the cone is

(1/3)(15)^2* 36 * pi =

2700 pi  units^3

you're post is missing a key bit of info.

a=7 when .... what?

Very nice, Omi.....this one was somewhat difficult....!!!

\(\text{She has 52 cards to choose from, then 51, then 50}\\ N = 52\cdot 51\cdot 50\\ \text{The usual formula for what's known as a permutation of }n \text{ objects taken }r \text{ at a time is }\\ _nP_r = \dfrac{n!}{(n-r)!}\)

What is the number of centimeters in the length of the longer leg of the smaller triangle?

How to solve?

Hello there OldTimer     

It is nice to see you again.

\(Sin3xCos3x - Cos^2 2x + 1/2 = 0\\ 2Sin3xCos3x - 2Cos^2 2x + 1 = 0\\ sin(6x) - 2Cos^2 2x + 1 = 0\\ \qquad cos4x=cos^22x-sin^22x\\ \qquad cos4x=cos^22x-(1-cos^22x)\\ \qquad cos4x=2cos^22x-1\\ \qquad -cos4x=-2cos^22x+1\\ so\\ sin(6x) - 2Cos^2 2x + 1 = 0\\ sin(6x) - cos(4x)= 0\\\)

\(sin(6x) =cos(4x)\)

NOW

\(\qquad cos(\theta)=sin(\frac{\pi}{2}-\theta)=sin(\pi-(\frac{\pi}{2}-\theta))=sin(\frac{\pi}{2}+\theta)\\ \qquad \therefore\;\;\;\;cos(4x)=sin(\frac{\pi}{2}+4x)\)

\(sin(6x)=cos(4x)\\ sin(6x)=sin(\frac{\pi}{2}+4x)\\ 6x=\frac{\pi}{2}+4x+2\pi n \qquad n\in Z\\ 2x=\frac{\pi}{2}+2\pi n \qquad n\in Z\\ x=\frac{\pi}{4}+\pi n \qquad n\in Z\\\)