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Let p be a prime. What are the possible remainders when p is divided by 17?  Select all that apply.    

        17/17 = 1 R 0      

      103/17 = 6 R 1        73/17 = 4 R 5        43/17 = 2 R 9          47/17 = 2 R 13    

        19/17 = 1 R 2        23/17 = 1 R 6        61/17 = 3 R 10        31/17 = 1 R 14    

        37/17 = 2 R 3        41/17 = 2 R 7        79/17 = 4 R 11         83/17 = 4 R 15    

        89/17 = 5 R 4        59/17 = 3 R 8        29/17 = 1 R 12         67/17 = 3 R 16    

So, we can say, shown empirically, that all the possible     

remainders are every integer from 0 to 16, inclusive.    

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Let x and y be integers.  If x and y satisfy 37x + 82y = -172 and x + y = -1, then find x.    

                                            37x + 82y = –172    

                                                x   +   y =     –1    ==>>    y = –x – 1     

substitute (–x – 1) for y   

in the other equation    

                                            37x + 82(–x – 1)  =  –172    

                                            37x – 82x – 82     =  –172    

                                                              – 45x  =  – 90     

                                                                      x  =  2    

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The answer is x = 50, y = -47.

Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.  

combine like terms and   

arrange as ax² + bx + c                              x² + (k – 9) • x + 16    

For the quadratic to be the square of a binomial, the "b" term    

has to be twice the value of the square root of the "c" term.    

                            k – 9  =  (2)(+4)    ==>>    k  =  ( +8 + 9 )    

                            when  k = 17    ==>>   x² + (17 – 9) • x + 16  =  (x + 4)²    

                            when  k =   1    ==>>   x² + (  1 – 9) • x + 16  =  (x – 4)²    

   sum of both values of  k  =  18       

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All members of our painting team paint at the same rate. If $10$ members can paint a $1000$ square foot wall in $10$ minutes, then how long would it take for $5$ members to paint a $1000$ square foot wall, in minutes?    

This is a special case of a rate x time problem,    

so it's possible to take a short cut in this case.    

With half as many painters, and with all else equal,   

the job would take twice the time.  So, 20 minutes    

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   Evaluate $a^3 - \dfrac{1}{a^3}$ if $a - \dfrac{1}{a} = 0$.   

                               1                            1   

       What is  a³ –  –––    when    a  –  ––– = 0    

                               a³                           a   

        For a – (1/a) to equal 0  the only way is when a = +1     

         If so, then a³ – (1/a³) = 0 as well   

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You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $5$ pieces of candy per scoop to the vat, and another machine that can remove exactly $3$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?    

Let the 1st machine cycle two times to add         10 pieces    

Let the 2nd machine cycle three times to remove   9 pieces    

                                                                                1 piece remains    

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Real numbers a and b satisfy    
a + ab = 250    
a - ab = -240    
Enter all possible values of a, separated by commas.    

given the two equations    

                                                     a + ab  =    250    

                                                     a – ab  =  –240    

simply add the two equations    

                                                    2a         =      10    ==>>    a = 5          

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Find the discriminant of the quadratic 5x^2 - 2x + 8 - 2x^2 + 6x + 2.  

Collect and combine like terms    

& arrange them as ax² + bx + c               5x² – 2x + 8 – 2x² + 6x + 2     

                                                                3x² + 4x + 10        

The discriminant is b² – 4ac                    D  =  (4)² – (4)(3)(10)    

                                                                 D  =  16 – 120    

                                                                 D  =  – 104    

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7 P 7 =   7!  = 5040  ways 

     or

7 choices for first book 

6 choices for second book 

5 choices for third book 

.

.

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1 choice  for last book                   7x6x5x4x3x2x1 - 7! = 5040 ways 

(x-1) = 0 produces a root of x =1 

(x+1) = 0 produces a root of -1 

  multiply them together :

(x-1)(x+1) =0

x^2  +  (0) x  - 1  = 0       or simply   x^2 -1 = 0 

The squares with two or more vertices in the set {A, B, C, D} are:

ABCD itself

The squares formed by taking adjacent vertices of ABCD (ABDC, ACBD, ADBC, BCDA, BDCA, CDAB)

Therefore, there are 7 squares that have two or more vertices in the set {A, B, C, D}.

The number $100$ has four perfect square divisors, namely $1,$ $4,$ $25,$ and $100.$   
What is the smallest positive integer that has exactly $2$ perfect square divisors?   

I reckon that would be four ... it has two perfect square divisors, namely one and four.   

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To solve the equation tan(2x) + tan(3x) = sec(3x), we can first use the identity sec(x) = 1/cos(x) to rewrite the equation as:

tan(2x) + tan(3x) = 1/cos(3x)

Next, we can use the identity tan(x) = sin(x)/cos(x) to rewrite the equation again:

sin(2x)/cos(2x) + sin(3x)/cos(3x) = 1/cos(3x)

Now, we can multiply both sides of the equation by cos(2x)cos(3x) to eliminate the denominators:

sin(2x)cos(3x) + sin(3x)cos(2x) = cos(2x)

Using the double angle and angle addition formulas, we can simplify the left side of the equation:

sin(x)cos(x) = cos(2x)

Finally, we can divide both sides of the equation by cos(x) and use the identity tan(x) = sin(x)/cos(x) to get:

tan(x) = 1/cos(x)

Therefore, the equation is equivalent to tan(x) = sec(x).

To find the smallest positive solution to this equation, we can use the unit circle or a calculator. We can see that the solution is x = π/4.

Therefore, the smallest positive solution to the equation tan(2x) + tan(3x) = sec(3x) is x = π/4 radians.

Let's visualize the problem:

Imagine a right-angled corner of a room. Place a circular table in this corner so that one edge of the table touches both walls.

Point P: A point on the edge of the table that is 8 units away from one wall and 9 units away from the other.

Radius: The distance from the center of the table to any point on its edge (like point P).

Solution:

Form a right triangle:

The distance from point P to the first wall (8 units) is one leg of a right triangle.

The distance from point P to the second wall (9 units) is the other leg of the right triangle.

The hypotenuse of this right triangle is the diameter of the table, which is twice the radius.

Use the Pythagorean theorem:

a² + b² = c²

8² + 9² = c²

64 + 81 = c²

145 = c²

c = √145

Find the radius:

Since the diameter is twice the radius, the radius is half of √145.

Radius = (√145) / 2

Therefore, the radius of the table is (√145) / 2 units.