Bosco

Catherine rolls a standard 6-sided die six times. If the product of her rolls is 200, then how many different sequences of rolls could there have been? (The order of the rolls matters.)  

The prime factorization of 200 is 2 x 2 x 2 x 5 x 5 but  

that's only five factors and Cat rolled six times so we  

need to add a 1 to the rolls. 

She had to roll these numbers:  1, 2, 2, 2, 5, 5     

How many sequences of these could have occurred.  

The permutation of six things taken six at a time.  

                            ₆P₆ = 720

Oh, she could have rolled 1, 1, 4, 2, 5, 5  etc.      

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In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

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     A_{large square}  =  6.25 units²  

     The base of each triangle is 1 unit. 

     There are 2¹/₂ bases going across.  

     So the side of the square is 2¹/₂ units.  

     Two and a half squared is six and a quarter.  

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Let X be a point inside rectangle ABCD.  If AX = 3, BX = 3, and CX = 3, then find DX.   

                                        AX² + CX²  =  BX² + DX²         

                                         3²  +  3²  =  3²  +  DX²     

                                         DX²  =  3²  

                                          DX  =  3    

i copied the following from the internet.  

In Euclidean geometry, the British Flag Theorem says that if a point P is chosen inside a rectangle ABCD, then the sum of the squares of the Euclidean distances from P to two opposite corners of the rectangle equals the sum to the other two opposite corners. 
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Let "r" = the amount added to each successive term.  

a₁ + r = a₂   and   a₂ + r = a3   therefore a₃ = a₁ + r + r    

given that a₁ + a₃ = 5   therefore (a₁) + (a₁ + 2r) = 5                 (eq 1)  

a₁ + r = a₂   and   a₂ + r = a₃   and   a₃ + r = a₄   therefore a₄ = a₁ + r + r + r   

given that a₂ + a₄ = 6   therefore (a₁ + r) + (a₁ + 3r) = 6            (eq 2)   

subtract (eq 2) minus (eq 1)   

(eq 2) ...............................................  2a₁ + 4r = 6    

(eq 1) ...............................................  2a₁ + 2r = 5    

                                                                     2r = 1         

                                                                       r = 0.5      

substitute r back into (eq 2)               2a₁ + (4)(0.5) = 6   

                                                                       2a₁ = 4  

                                                                         a₁ = 2     

check answer  

                          a₁ = 2.0 , a₂ = 2.5 , a₃ = 3.0 , a₄ = 3.5    

                          a₁ +  a₃ = 5                a₂ +  a₄  = 6   

                         2.0 + 3.0 = 5               2.5 + 3.5 = 6    

Let  
f(x) = \sqrt{x}  
Find the largest three-digit value of x such that f(x) is an integer.  

31²  is    961   

32²  is  1024   

Therefore, x = 961   

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In any polygon, the sum of the interior angle and the exterior angle at a vertex is 180^o.  

In this polygon, the value of the interior angle is twice that of the exterior angle,    

therefore, the interior angle is 120^o and the exterior angle is 60^o.   .  

In any regular polygon, the sum of all the exterior angles is 360^o     

irrespective of the number of vertexes.  

Since one exterior angle of this polygon is 60^o     

there must be six of them to total of 360^o. 

We shall presume that there are the same number of sides that there are vertexes.   

Therefore, this polygon has six sides.    

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  2x^2 + 3x + 8x - x^2 + 4x + k has a double root    

  combine like terms                          (2x² – x²) + (3x +8x + 4x) + k  

                                                           x² + 11x + k    

  to have a double root,    

  i.e., both roots are the same,   

  the discriminate must equal zero        b² – 4ac = 0    

                                                            (11)² – (4)(1)(k)  =  0    

                                                             121 – 4k  =  0    

                                                                     –4k  =  –121    

                                                                          k  =  121 / 4    

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1.   At the vertex of a polygon, the sum of the exterior angle and the interior angle is 180^o.   

      Since the interior angle is 12 times the exterior angle, the exterior angle has to be 15^o.  

      The sum of all exterior angles of a polygon is 360^o, irrespective of the number of sides.  

       Therefore, the polygon has 360 / 15 = 24 vertices.   

       We shall presuppose that a polygon has the same number of sides as it has vertices,    

       Therefore, this polygon has 24 sides.    

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                                       y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25    

group like terms              y  =  (–2x² – 3x²) + (8x – 14x) + (–15 + 25)    

combine like terms           y  =  –5x² – 6x + 10    

first derivitive                    ^dy/_dx  =  –10x – 6    

set equal to zero                    0  =  –10x – 6    

                                           10x  =  –6   

                                               x  =  – 0.6    this is the x-coordinate of the vertex

sub x into original                   y  =  [ –5 • (–0.6)² ]  –  [6 • (–0.6) ]  +  10    

                                               y  =  (–5 • 0.36)  –  (–3.6)  +  10    

                                               y  =  –1.8 + 3.6 + 10    

                                               y  =  11.8      this is the y-coordinate of the vertex

the vertex is at                        (–0.6 , 11.8)    

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                                                 x^2-mx+24 = 10   

subtract 10 from both sides      x² – mx + 14  =  0   

integers that multiply to 14        (+7)(+2)   

                                                  (–7)(–2)   

                                                  (+1)(+14)   

                                                  (–1)(–14)  

possible values of m                   –9   

                                                   +9   

                                                  –15   

                                                  +15   

total number of

possible values of m                      4    

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                                        5x^2 - 2x + 8 - 2x^2 + 6x + 2 - 2x^2 + 4x - 1   

can be regarded as        (5x² – 2x² – 2x²) + (–2x + 6x + 4x) + (8 + 2 – 1)   

combine like terms            x² + 8x + 9   

discriminant is                   8² – (4)(1)(9)  =  64 – 36  =  28   

The discriminant is the expression under the radical   

in the quadratic formula.   

When a quadratic is stated in the form ax² + bx + c  

the discriminant is b² – 4ac.  

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