Bosco

My phone number?    

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LOL, it was still Mt Everest.  It just hadn't been discovered yet.  

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Right.  A chalkboard can be green.  In fact,     

I've seen more green ones than black ones.    

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In Linguistics 101, the ratio of the number of juniors to the number of seniors is 3:2.  When six more juniors join the class, and one senior drops the class, the ratio of the number of juniors to the number of seniors becomes 2:3.  How many students are in the class after these changes?

I can tell you right now that increasing the number     

of juniors will not invert the ratio.  But I'll go through    

the motions for you anyway.     

                                                       J          3   

                                                      ––   =   ––  

                                                       S          2     

                                                   J + 6          2   

                                                   ––––   =   –––   

                                                   S – 1          3        

Cross multiply both                            2J  =  3S            (eq 1)  

                                                  3J + 18  =  2S – 2      (eq 2)      

Get either J or S in terms of the   

other.  It doesn't matter which, but  

J would be simpler.  From eq 1:            J  =  3S/2      

Substitute this in eq 2                (3) • (3S/2) + 18  =  2S – 2    

                                                            9S/2 + 18  =  2S – 2     

Multiply both sides by 2                          9S + 36  =  4S – 4    

Subtract 4S from both sides                   5S + 36  =  –4   

Subtract 36 from both sides                           5S  =  –40  

                                                                         S  =  –8  

You cannot have a negative number of seniors    

so something is wrong somewhere.  The error     

is in the initial proposition.  This problem has ...... no solution.    

In a different problem that I solved, the solution depended on accepting  

the concept of negative money.  Well, we *sort of* do have such a thing        

as that, it's called debt.  But I can't rationalize negative students.  LOL.    

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Franklin the fly starts at the point (0,0)  in the coordinate plane. At each point, Franklin takes a step to the right, left, up, or down. After 12 steps, how many different points could Franklin end up at?   

To find how many different points Franklin could end up at,    

that is, a maximum, we have to prohibit his going to any        

step where he landed, or could have landed, before.   

Step     Choices     Total  

   1            4            4                       Franklin could go in any of the four directions.  

   2            3            4 • 3 = 12          Only three choices because Franklin could go in any of the four  

                                                         directions except back to the place he was in the previous step.  

   3            3            4 • 3 • 3 = 36    

   4            3            4 • 3 • 3 • 3 = 108    

   5            3            4 • 3 • 3 • 3 • 3 = 324     See a pattern emerging.  Let's write this as a generality.  

   n            3            4 • 3^{(n – 1)}    

So, at step 12, Franklin could have ended up at 4 • 3¹¹  =  708,588 points    

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Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?  

Let the first integer be represented by x.   

Since the integers are consecutive, each following   

integer is one more than the integer before it.  

(x)  +  (x + 1)  +  (x + 2)  +  (x + 3)  +  (x + 4)  +  (x + 5)  +  (x + 6)  =  (21) • (x + 6)    

                                                             7x + 21  =  21x + 126    

                                                                 –14x  =  105     

                                                                       x  =  –7.5    

check answer    

(–7.5) + (–7.5 + 1) + (–7.5 + 2) + (–7.5 + 3) + (–7.5 + 4) + (–7.5 + 5) + (–7.5 + 6)  =  21 • (–7.5 + 6)    

(–7.5) +    (–6.5)    +    (–5.5)    +    (–4.5)    +    (–3.5)    +    (–2.5)    +    (–1.5)     =  21 • (–1.5)    

                                                                                                                      –31.5  =  –31.5   

_.

The sum of the interior angles of a polygon is (n – 2) • (180^o) where n is the number of sides.   

So the total of the interior angles is 360^o and opposite angles are equal, so set it up like so:    

                                         y + y + (2y – 51) + (2y – 51)  =  360    

                                                                        6y – 102  =  360  

                                                                                   6y  =  360 + 102    

                                                                                   6y  =  462           

                                                                                     y  =  77^o   

_.

Find the ordered triple (p,q,r) that satisfies the following system:  

p - 2q = 3  

q - 2r = -2 + q  

p + r = 9 + p  

Let's number those equations, just for identification:  

(1)  p – 2q  =  3  

(2)  q – 2r  =  –2 + q  

(3)  p + r  =  9 + p  

Per equation (2), r = 1  

Per equation (3), r = 9  

It can't be both at the same time, so the system is invalid.  

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I took a look at it.  Yeah, I remember that problem.  I can't solve it.  Sorry, wish I could help.    

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I've seen many of the questions on here a million times.  I've got to where I save answers I'd previously made and copy/paste them in the the new iteration of the math problem.  I don't think it's one person posting the same thing over and over and over.  Consider a math problem, take this one for example.  I think it's part of a math course – whether classroom, or online, or whatever – and students encounter it at different times.  Maybe they enroll at different times, or work at their own speed, but for whatever reason, they come to the question at different times.  Then they  bring it here.  

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