Bosco

If you let the first machine operate 2 times, it will have added 10 pieces,   

then you let the second machine operate 3 times, it will remove 9 pieces.   

This would leave 1 piece in the vat.   

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Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 28/9 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?   

                                                                                              (x + 6)    28   

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

                                                                                                  1         9   

                                28x + 168   

             7x + 21  =  –––––––––   

                                      9    

          63x + 189  =  28x + 168       

                     35x  =  – 21   

                                 – 21            3  

                         x  =  ––––  =   – –––      

                                   35             5    

                   The solution is not an integer, so the    

                   problem, as written, is not satisfied.       

                   Assuming the problem can be solved,   

                   I hope someone will show me my error.    

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Seven years ago, Grogg's dad was $9$ times as old as Grogg. Four years ago, Grogg's dad was $6$ times as old as Grogg. How old is Grogg's dad currently?   

(D – 7)  =  (9)(G – 7)    ..............—>   D – 7  =  9G – 63     (eq 1)   

(D – 4)  =  (6)(G – 4)    ..............—>   D – 4  =  6G – 24     (eq 2)   

subtract (2) from (1)                                – 3  =  3G – 39   

                                                                  36  =  3G   

                                                            Grogg  =  12    

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Laverne starts counting out loud by 5's.  She starts with 2.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds 20, Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running?   

Laverne's           Shirley's   

Counting          Summation   

     2                           2   

     7                           9    

   12                         21    

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What is the largest positive integer $n$ such that $1457$, $1797$, $709$, $15$, $24$, $197$, $428$ all leave the same remainder when divided by $n$?   

First, let's eliminate as many divisors as we can.  Consider 15 and 24.   

The divisor n has to leave a remainder, so factors can be eliminated.   

n cannot be 3 or 5, nor 2, 4, 6, 8, or 12 .... leaving 7, 9, 10, 11, 13, and 14.   

Remember, the remainder has to be the same.  Try them out on 15 and 24 first.  

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

Well, I have run out of divisors and ideas at the same time.   

Either the problem has no solution, or I'm missing something.   

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What is the fifth smallest prime number greater than 100?    

The first five primes greater than 100 are 101, 103, 107, 109, and 113.   

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Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.   

                                        2x² + 3x + 8x – x² + 4x + k   

Combine like terms           x² + 15x + k    

In math, the term "double root" means "repeated root" ... i.e., both roots are the same   

Examples:    x² + 2x + 1    factors to (x+1)(x+1) and both roots are –1   

                    x² + 6x + 9    factors to (x+3)(x+3) and both roots are –3   

                    x² – 8x + 16  factors to (x–4)(x–4) and both roots are +4   

One way to determine that a quadratic has a double root is when its discriminant equals zero   

The discriminant is the expression underneath the radical in the quadratic equation   

When a quadratic is posed as ax² + bx + c the discriminant is b² – 4ac   

In x² + 15x +k in this problem, the discriminant is 15² – (4)(1)(k)   

Multiply it out, set equal to 0, and solve for k       225 – 4k = 0   

                                                                                    –4k  =  –225   

                                                                                        k  =  56.25    

The problem doesn't ask for the repeated root, but it's the square root of k.    

The factors of the quadratic are (x + 7.5)(x + 7.5) and the repeated root is –7.5   

check answer   

                                      x² + 15x + 56.25 = 0   

                                     (–7.5)² + (15)(–7.5) + 56.25 = 0   

                                       56.25 – 112.5 + 56.25 = 0    

                                                                       0 = 0   

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The parabola y = ax^2 + bx + c is graphed below. Find a + b + c. (The grid lines are one unit apart.)    
The parabola passes through (-2,8), (0,0), and (2,8).   

There is no graph shown.  Relying only on the three points provided,    

one parabola that passes through (–2,8), (0,0), and (2,8) is y = 2x²    

If that is correct, then a + b + c  =  2 + 0 + 0  =  2.   

Those three points, and the vertex at (0,0), indicate that the parabola    

is symmetrical to the y-axis and that the equation has a double root.  

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Oh, my stars.  I misread the problem.  I thought it said x² – mx + 24 = 0,  

CPhill solved it correctly at https://web2.0calc.com/questions/help-help_55   

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