Bosco

How many integers from 1 to 9 are divisors of the five-digit number 24,519?   

They all are.  But you probably mean evenly divisible.  

This an easy one.  All you have to do is try them all to find out. 

You don't even have to try the even numbers – i.e, 2, 4, 6, & 8 – so that cuts the  

effort in half.   

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11(-5)+4(7)    

                                       11(–5) + 4(7)    

                                        –55 + 28   

This is the same as            28 – 55     so, whichever way is easier for you to visualize.   

Answer is                          –27    

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what is -2(bx - 5) = 16    

                                                                –2(bx – 5)  =  16    

                                                                –2bx + 10  =  16    

                                                                –2bx  =  6    

(Assuming you want to solve for x)   

                                                                                6          

                                                                      x  =  ––––  

                                                                              –2b    

                                                                                   3         

                                                                      x  =   – –––  

                                                                                    b    

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Find the number of non-congruent right triangles, where all the sides are positive integers, and one of the legs is $6$.  

It's talking about right triangles, therefore c² = a² + b²     

The problem requires that all sides be integers, i.e., no fractions.   

There's only one triangle that I can find to satisfy both conditions.  That's 6, 8, 10.   

BTW, a right triangle with integer sides is called a Pythagorean Triple.      

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Guest GA, did you sit on a hot frypan and burn out your brain? 

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Find the value of $c$ so that the quadratic equation $4x^2 + 2x + c = 3x^2 - 8x$ has exactly one solution in $x$.   

Get everything on the left                                4x² – 3x² + 2x + 8x + c  =  0    

and combine like terms                                      x²          + 10x      + c  =  0      

The quadratic is now in the form ax² + bx + c = 0         

For it to be a square, then b² – 4ac must equal 0     

                                                                          10² – (4)(1)(c)  =  0       

                                                                           100  –  4c         =  0    

                                                                                              4c  =  100    

                                                                                                c  =  25      

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