Bosco

Find the vertex of the graph of the equation y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25.    

                                  y  =  –2x² + 8x – 15 – 3x² – 14x + 25    

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

set the first derivitive to zero    

                                   –10x – 6 = 0     ==>>     x = – 6 / 10  =  – 0.6    

plug this x back into the equation    

                                        y  =  –5•(–6/10)² – 6•(–6/10) + 10    

                                        y  =  –180 / 100  +  36 / 10  +  10    

                                        y  =  –180 / 100  +  360 / 100  +  1000 / 100    

                                        y  =  1180 / 100  =  59 / 5  =  11.8    

Thus, the vertex is at (–0.6, 11.8)    

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At a cafeteria, Mary orders two pieces of toast and a bagel, which comes out to $3.15. Gary orders a bagel and a muffin, which comes out to $3.50. Larry orders a piece of toast, two bagels, and three muffins, which comes out to $8.15. How many cents does one bagel cost?    

                                        T + B = 315 ¢    ==>>    T = 315 – B    

                                        B + M = 350 ¢    ==>>    M = 350 – B    

                                         T + 2B + 3M = 815 ¢   

                                         (315 – B) + 2B + [3 • (350 – B)]  =  815    

                                         315 – B + 2B + 1050 – 3B  =  815    

                                          – 2B + 1365  =  815    

                                          – 2B  =  – 550    

                                               B  =  275 ¢   

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(x+2)/(x) • (x+3)/(x+1) • (x+4)/(x+2) = 10    

cancel the (x+2)    

(1)/(x) • (x+3)/(x+1) • (x+4)/(1) = 10    

[ (1) • (x+3) • (x+4) ] / (x) • (x+1) • (1) ] = 10    

(x² + 7x + 12) / (x² + x) = 10    

x² + 7x + 12 = 10x² + 10x    

0 = 9x² + 3x – 12     this will factor    

(3x + 4) (3x – 3) = 0    

                      3x + 4 = 0     x will not be an integer, so disregard    

                      3x – 3 = 0    

                            3x = 3    

                              x = 1     therefore (x+2)/(x) = 3/1 = 3 is the value of the first fraction   

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Find all values of $x$ such that $x^2 - 5x + 4 = -x^2 - 25x - 14$. If you find more than one value, then list your solutions, separated by commas.  

x² – 5x + 4  =  – x² – 25x – 14    

2x² + 20x + 18  =  0    

  x² + 10x + 9  =  0    

(x + 1)(x + 9)  =  0    

x = –1 and x = –9    

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A pole of height 50 units has broken into two pieces, and the top of the pole is now at a point that is 25 units from the base of the pole. Find the difference between the lengths of the two pieces.    

The pole was 50 feet high.  The pole is now 25 feet high.    

That means the piece broken off is 50 – 25 = 25 feet long.    

The two pieces are the same length.  The difference is zero.    

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There are four cities: Capital City, Little Village, Mytown, and Yourtown.  The distance from Capital City to Little Village is $5$ miles.  The distance from Capital City to Mytown is $2$ miles.  The distance from Mytown to Yourtown is $3$ miles.  The distance from Little Village to Yourtown is $4$ miles.  Find the distance from Capital City to Yourtown, in miles.     

Little Village ----- 4 ----- Your Town ----- 1 ----- Capital City ----- 2 ----- My Town    

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x² + y²  =  4x + 12y is the equation of a circle.  

My approach will be to complete the squares     

of x and y, to obtain the radius of that circle.  

x² + y²  =  4x + 12y

(x² – 4x) + (y² – 12y)  =  0    

(x² – 4x + 4) + (y² – 12y + 36)  =  4 + 36    

(x – 2)² + (y – 6)²  =  40    

This draws a circle centered at (2, 6) with a radius sqrt(40).    

Never mind the center, all that interests us is the radius.    

By formula, the maximum x+y is the radius times sqrt(2).    

In this problem, the largest x+y is sqrt(40) times sqrt(2).    

sqrt(40) • sqrt(2)  =  sqrt(80)  =  4 • sqrt(5)    

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The Edit button to my above has disappeared, so I add the following:    

I am accustomed to the occasional ambiguously worded problem, but    

this time it must be questioned.  Exactly what is meant by “at least one    

of the boxes must be empty”?  Is it possible for both boxes to be empty? 

If both boxes are empty, the solution would contradict itself, because, in    

such a case, the balls would not have been distributed.    

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In how many ways can you distribute $8$ indistinguishable balls among $2$ distinguishable boxes, if at least one of the boxes must be empty?    

If all eight balls have to be used, and one of the two boxes has to remain empty,    

then you have to put all eight balls into the other box.  Only two ways to do that.    

Either put the balls in box A and leave box B empty, or vice versa.    

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Find the remainder when $1! + 2! + 3! + \dots + 100!$ is divided by $2$.    

Every factorial will have a "times 2" in it.  Every factorial that is, except 1!.    

All the factorials added together, except 1!, will sum to an even number.    

Therefore, the remainder when the total sum is divided by 2 will be that 1.    

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