Bosco

250 in base 10 is FA in base 16.    

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A school orders $99$ textbooks, all for the same price.    

When the bill for the total order comes, the first and last    

digits are obscured. What are the missing digits?    
_18,486.7_    

What is the criterion for a correct answer?  If we can get       

a price per book that comes out even?  What about that      

the price of a book is a plausible amount?  You decide.       

                                               518,486.76     

                                              ––––––––––  =  5,237.24     

                                                      99              

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(5 x 6 x 7 x 8) + 1  =  1680 + 1  =  1681 apples          

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Two integers are relatively prime when there are no common factors other than 1.   

Determine the number of positive integers less than 191 that are relatively prime to 191.   

Since 191 is a prime number to start with, then, not counting 1, wouldn't that mean that       

there are 189 integers less than 191 and relatively prime to 191?  Or would you count 1?        

Not sure of the definition.     

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Dang, I made a careless mistake.  I wonder how many of you caught it.   

I overlooked the minus in front of 23x.  Too late to edit my mistake, grrr.   

The lines should look like this:   

        (x³ + x² + x + 1)(x⁴ – 8x³ + 17x² – 23x + 14)  gives us   14x     

        (x³ + x² + x + 1)(x⁴ – 8x³ + 17x² – 23x + 14)  gives us  –23x     

                                                              Add them together  –9x        

                         So the coefficient the problem is asking for is  –9.     

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        (x³ + x² + x + 1)(x⁴ - 8x³ + 17x² - 23x + 14)    

        Every term in the first expression will be multiplied with every term in the second expression.    

        Which term(s) times which term(s) will result in x with no exponent....   

        (x³ + x² + x + 1)(x⁴ - 8x³ + 17x² - 23x + 14)  gives us 14x     

        (x³ + x² + x + 1)(x⁴ - 8x³ + 17x² - 23x + 14)  gives us 23x     

                                                             Add them together 37x        

                    So the coefficient the problem is asking for is 37.     

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There is no parabola shown below, but just from the information that        

is provided, an equation that satisfies the three points is y = 2x².   

That would make a + b + c  =  2 + 0 + 0  =  2     

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

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

The easiest way – the easiest way I know, anyway – is     

to set the first derivitive equal to zero, and solve for x.   

                                                ^dy/_dx  = –10x – 6   

                                                      0  =  –10x – 6     

                                                   10x  =  –6   

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

                                                                          plug it back into the original equation     

                                                        y  =  –5(–0.6²) – 6(–0.6) + 10      

                                                        y  =  –1.8 + 3.6 + 10   

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

                                                        The vertex is located at (–0.6 , 11.8)    

                                                         Verified the answer with Desmos Graphing Calculator.    

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                                               2x^2 + 3x + 8x - x^2 + 4x + k   

Combine like terms                 x² + 15x + k   

When the quadratic is in the format like the above,    

to have a double root (also called a repeated root),    

the constant is the square of half the coefficient of x.    

Coefficient of x is 15. 

Half of 15 equals 7.5. 

7.5 squared is 56.25.               So, k = 56.25   

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x² – mx + 24 = 10   

x² – mx + 14 = 0   

the coefficient of x is the sum of the factors of 14   

the integer factors of +14 are (+2, +7) and (–2, –7)   

                                         and (+1, +14) and (–1, –14)   

So m could equal +9 or –9 or +15 or –15   

Therefore, m could have 4 possible values    

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p - 2q = 3               (eq 1)   

q - 2r = -2 + q         (eq 2)   

p + r = 9 + p            (eq 3)   

By eq 2  r = 1

By eq 3  r = 9   

Both can't be true so there is no solution.   

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