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3. Two boys and three girls are going to sit around a table with 5 different chairs. If the two boys want to sit together, in how many possible ways can they be seated?

We can anchor the boys in any two  adjacent chairs.....and they can be seated in 2!  =  2 ways in each of these chairs

And for each of these arrangements, the 3  girls can be seated in 3!  = 6 ways

So....the number of possible seating arrangements  is   2!  * 3! =  2 * 6   =    12 ways

2. The product of all digits of positive integer M  is 105  How many such M's are there with distinct digits?

I tried 6 for the problem above but it wasn't right... Any thoughts on why?

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I think the answer is not 6 because  1  can also be a digit of M.

105  =  3 * 5 * 7

So  5, 3, and 7 must be digits of M, while  1  might be a digit of M.

So the possiblites are:

357

375

537

573

735

753

In addition to:

the total number of possibilities   =   3! + 4!   =   6 + 6*4    =   6 + 24   =   30

5x^2 + 4  = 3x + 9                  

5x^2 - 3x - 5  = 0       use the Quadratic Formula                  

3 ±√[ 109]

___________   =   x       which implies that

        10

10x  =  3 ±√109          subtract 3 from both sides

10x - 3  =  ±√109       square both sides

So

(10x - 3)^2  = ( ±√109)^2  =

109

Just to clarify, is phi the answer?

Thank you so much, Guest!

Thank you very much! I really appreciate all your help :)

Note that   (10x - 3)²   =   (10x - 3)(10x - 3)   =   100x² - 60x + 9

Now let's look at the given equation:

5x² + 4  =  3x + 9

                                Subtract  3x  from both sides of the equation.

5x² - 3x + 4  =  9

                                Subtract  4  from both sides of the equation.

5x² - 3x  =  5

                                Multiply through by  20

100x² - 60x  =  100

                                         Add  9  to both sides of the equation.

100x² - 60x + 9   =   109

                                         And remember that  100x² - 60x + 9  =  (10x - 3)²

(10x - 3)²   =   109

:O Wow Cphill... Long solution... It's very nice!

The solutions to  \(4x^2 + 3 = 3x - 9\)  can be written in the form  \(x = a \pm b i,\)  where  \(a\)  and  \(b\)  are real numbers. What is  \(a + b^2\) ?  Express your answer as a fraction.

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\(4x^2 + 3 = 3x - 9\)

                                        Subtract  3x  from both sides and add  9  to both sides of the equation.

\(4x^2-3x + 12 = 0\)

                                        Now we can use the quadratic formula to solve for  x

\(x\ = \ \frac{3\pm\sqrt{3^2-4(4)(12)}}{2(4)}\ =\ \frac{3\pm\sqrt{-183}}{8}\ =\ \frac{3\pm\sqrt{183}i}{8}\ =\ \frac38\pm\frac{\sqrt{183}}{8}i\)

And now we have the solutions in the form   \(x = a \pm b i\)   so we can see...

\(a+b^2\ =\ \Big(\frac38\Big)+\Big(\frac{\sqrt{183}}{8}\Big)^2\ =\ \frac38+\frac{183}{64}\ =\ \frac{24}{64}+\frac{183}{64}\ =\ \frac{207}{64}\)_