geno3141

Since the height of the cone is the same as the hieght of the scoop of ice cream, the height of the

cone =  2r,  where r is the radius of both the cone and the sphere of ice cream.

The volume of the sphere:  V  =  (4/3)·pi·radius³  =  V  =  (4/3)·pi·r³​ 

The volume of the cone:  V  =  (1/3)·pi·radius²·(height)   =  (1/3)·pi·r²·(2r)  =  (2/3)·pi·r³

The volume of the sphere is twice the volume of the cone, so it can fill two cones.

I believe that the number 2 is found only once; in the line 1  2   1.

Unless what is meant the number of times a digit is found (which can be part of a longer number) ...

If you look at the "first terms", the 101, 93, ... 5, these have a difference of 8.

We can use this difference to express not only the "first terms" (101, 93, ... 5) but also the "second terms"

(97, 89, ... 1):

If n = 0:     8n + 5  =  5        8n + 1  =  1

...

If n = 11,   8n + 5  =  93      8n + 1  =  89

If n = 12,  8n + 5  =  101     8n + 1  =  97

Also note that if you multiply the expressions together, you get a difference of squares:

  (101 + 97)(101 - 97)  =  101² - 97²  =  792

  (93 + 89)(93 - 89)      =  93² - 89²     =  729

  ...

  (5 + 1)(5 - 1)  =  5² - 1²  =  24

If we rewrite the numbers using the  8n + 5  and  8n + 1  forms;

    [ (8n + 5) + (8n + 1) ] ·[ (8n + 5) - (8n + 1) ]  

=    (8n + 5)² - (8n + 1)² 

=    64n² + 80n + 25) - (64n² + 16n + 1)

=    64n + 24    [this clearly indicates that it will be an arithmetic progression with a common difference of 64]

First term:  n = 0     --->   value  = 24

Last term:  n = 12   --->   value = 792         [there are 13 term]

Sum  =  (13)(24 + 792) / 2  =  5304

tan(30^o)  =  5 / x

           x  =  5 / tan(30^o)

If you complete the square on  x² - 3x

                                                 x² - 3x + 9/4                 (divide -3 by 2 and square the result)

                                                 ( x - 3/2 )²

So:  x² - 3x + 1  =  ( x² - 3x ) + 1

                          =  ( x² - 3x + 9/4 ) + 1 - 9/4                 (if you add 9/4 to a side, you must also subtract it)

                          =  ( x - 3/2 )² - 5/4

Left for you to finish ...

csc(x)  =  1 / sin(x)

--->   csc(x) / sin(x)  =  [ 1 / sin(x) ] / sin(x)  =  1 / sin²(x) 

cot(x)  =  cos(x) / sin(x)

tan(x)  =  sin(x) / cos(x)

--->   cot(x) / tan(x)  =  [ cos(x) / sin(x) ] / [ sin(x) / cos(x) ]  =  cos²(x) / sin²(x)

So, we now have:  [ 1 / sin²(x) ] - [ cos²(x) / sin²(x) ]

--->                         [ 1 - cos²(x) ] / sin²(x)

--->                                  sin²(x) / sin²(x)  =  1 

Fermat's Last Theorem (which was proven by Andrew Wiles) states that there are no three posiitve integers greater than 2

for which the equation aⁿ + bⁿ = cⁿ is true.

The restriction that the values must be positive disallows situations such as this:

  if a = -1, b = 1, and c = 0, then (-1)³ + (1)³ = (0)³  [which is true].

However, in the expression a¹⁰⁰ + b¹⁰⁰  =  c¹⁰⁰, because of the even exponents, if you could find any negative value that could be used, its corresponding positive value could also be used; but this is not allowed under Fermat's Last Theorem.

This is what I did:

1)  I looked at triangle(BAP) and used the Law of Cosines to find angle(BAP).

     I set the length of AB = x.

     BP²  =  AP² + AB² - 2·AP·AB·cos( BAP )

        3²  =  1² + x² - 2·1·x·cos( BAP )

          9  =  1 + x² - 2x·cos( BAP )

         ( 8 - x² ) / ( -2x )  =  cos( BAP )

          cos( BAP ) =  ( x² - 8 ) / (2x )

2)  Then, I looked at triangle(DAP) and used the Law of Cosines to find angle(DAP).

     Since it's a square, AD = x.

     DP²  =  AP² + AD² - 2·AP·AD·cos( DAP )

        7²  =  1² + x² - 2·1·x·cos( DAP )

         49  =  1 + x² - 2x·cos( DAP )

         ( 48 - x² ) / ( -2x )  =  cos( DAP )

          cos( DAP ) =  ( x² - 48 ) / (2x )

3)  Since angle(DAP) = 90^o - angle(BAP)   --->   cos( DAP )  =  sin( BAP )

                      --->   sin( BAP )  =  ( x² - 48 ) / (2x )

4)  Since  sin²( BAP ) + cos²( BAP )  =  1

                [ ( x² - 48 ) / (2x ) ]²  +  [ ( x² - 8 ) / (2x ) ]²  =  1

           [ ( x⁴ - 96x² + 2304 ) / ( 4x² ) ]  +  [ ( x⁴ - 16x² + 64 ) / ( 4x² ) ]  =  1

multiplying both sides by 4x²:

            ( x⁴ - 96x² + 2304 )  +  ( x⁴ - 16x² + 64 )  =  4x²

                 2x⁴ - 116x²  + 2368  =  0

But this doesn't have any real solutions?

Where did I mess up?

1)  Find BC (use the Law of Cosines)

2)  Find angle(DCB) (use either Law of Sines or Law of Cosines)

3)  Find angle(ACE)

4)  Find angle(CAE)

5)  Find AE.

Find the vertex:                                    f(x)  =  -4x² - 12x + 9

Subtract 9 from both sides:             f(x) - 9  =  -4x² - 12x

Factor out the -4:                            f(x) - 9  =  -4(x² + 3x)

Complete the square:                f(x) - 9 - 9  =  -4(x² + 3x + 9/4)

                                                      f(x) - 18  =  -4(x + 3/2)²

Vertex:  (-3/2, 18)