Bosco

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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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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          Area = 144 units²    

          If one of the angles is 90^o   

          then all of the angles are 90^o   

          and this rhombus is a square.  

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The coefficient of y⁴ in the expansion will be 81.   

Combine terms of (2y – 5 + y – 6)⁴  to  (3y – 11)⁴   

There will be only one term raised to the fourth power   

and that's the "3y" term ...  3 raised to the fourth is 81.   

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A = 100 units²     

Perimeter 40 and diagonal 10 • sqrt(2) indicate that this     

figure is the special case of rectangle, which is a square.    

Each side is 10, therefore the area is 10 • 10 = 100.     

Deleted.  I misread the question.

                 7!             5040  

₇P₇  =  –––––––  =  ––––––  =  5040       

             (7 – 7)!           1   

You might question zero in the denominator,   

but zero factorial is defined as 1 so it's okay.  

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               AB + CD  

Area  =  —–—–—–  •  height   

                     2   

That's just the formula for the area of a trapezoid.   

Without specific values, I can't take it any further.  

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A_{big 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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