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y - 8 = b  + 8                y ou give 8 and your b gains 8 to be equal 

      --- or ---                          y-b = 16 

y + 19  = 2 ( b-19)          or   y-2b = - 57 

                            subtract top equation from bottom to get     - b = -73    or        b = $ 73  

Oopps...I didn't read the entire question ...

  the three math books together can be arranged in   6 ways    ( 3!   or  3 P3) 

   then they act as one object of 5 objects     5 P 5 = 120 ways to arrange the books     ( 5!) 

Then  6 x 120 =   720 ways    

edit: The answer can't be 0 since it must be a positive integer. The second bolded number is the correct answer

There are 2 cases:

\(n = p_1 ^ 8\) or \(n = p_1 ^ 2 p_2 ^ 2\)

But wait! \(8 = 2^3\), so the second possibility won't work since it must not be divisible by a power of 2 greater than 4. Therefore, the smallest (and only) ice number with exactly 9 divisors is \(2^8 = \mathbf{256}\)

\(144_b = b^2 + 4b + 4 = (b + 2)^2\)

Since a perfect square can never be prime, thre are no bases that are prime

Try to do long division with \(1 \div 13\). If you do that you should get that \(\frac{1}{13} = 0.\overline{076923}\)

Therefore, \(\mathbf{(A, B, C) = (9, 2, 3)}\)

6 can be written with 6 dice in 2 ways:

6 * 1 * 1 * 1 * 1 * 1

2 * 3 * 1 * 1 * 1* 1

The first one has 6 ways to write it (6 spots for the 6 to go into)

The second one has 6 * 5 = 30 ways to write it (6 spots for the 3, 5 spots for the 2)

Therefore, there are 6 + 30 = 36 sequences of rolls

= 7! / (3! * 2! * 2!) = 5040/(6 * 2 * 2) = 5040/24 = 210 ways

Technically, the answer is 0 since sqrt[4](0) = 0, but if that's not it, then the second smallest integer is:

\(\sqrt[4]{1323*n*84*779*5*3*441}\)

\(1323 = 3^3 7^2\)

\(84 = 2^23^17^1\)

\(779 = 19^141^1\)

\(441 = 3^27^2\)

\(\sqrt[4]{1323*n*84*779*5*3*441} = 2^23^85^17^519^141^1\)

To make all exponents a multiple of four, n must be \(2^25^37^319^341^3 = \mathbf{81073047338500}\)

b + (b + 100) = 200      Given 

2b = 100

b = 50   cents 

The above possiblity presupposes the two members have different titles, such as president and vice president.   

In such case, for example, Janet & Frank would be different from Frank & Janet.  If there are no differentiations  

between the two members, however, the possibilities would be halved, and the total number would be 91 ways.   

_.   

Starting with the A moving one letter at a time vertically, horizontally, or diagonally, how many different paths spell ARCH?   

        A   

     RRR   

  CCCCC   

HHHHHHHH   

From the A, there are three ways to get to an R.  

From the R, there are three ways to get to a C.    

From the C, there are three ways to get to an H.   

3 x 3 x 3  =  27 paths   

There are 8 H's in the bottom row.  That 8th one is superfluous.   

_.   

A baking club wants to form an executive committee. There are $15$ people in the baking club, including Mark. In how many ways can the baking club form an executive committee with $2$ people, not including Mark?   

Excluding Mark, there are 14 members left to choose from.   

There are 14 choices for the first member of the committee.   

After the first is chosen, 13 choices are left for the second.    

14 x 13 = 182 ways to form the committee.    

_.

\(5x^2 - kx + 8 - 2x^2 + 25 = 0 \\ 3x^2 - kx + 33 = 0\)

After some simplification, I arrive at the following quadratic. In order to analyze when this quadratic has no real solutions, evaluate the discriminant, specifically when the discriminant is less than zero.

\(\Delta = b^2 - 4ac \\ \Delta = (-k)^2 - 4 * 3 * 33 \\ k^2 - 396 < 0 \\ k^2 < 396 \\ k < \sqrt{396} \approx 19.900\)

When the discriminant is less than zero, there are no real solutions, so k = 19 is the largest integer that makes the resulting quadratic have no real solutions.

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).    

The problem references a graph, but there is no graph provided.    

Taking into consideration only the three points that are specified,    

one parabola that goes through those three points is y = 2x² so    

a+b+c for that curve would be 2 + 0 + 0 = 2      

_.   

\(y=-5x^2-6x+10 <=> y'=-10x-6, y'=0 <=> x=-0.6\)

\(y= -5(-0.6)^2-6(-0.6)+10=11.8\)

so \((x,y)=(-0.6,11.8)\)

\(x^2+18x+5 = 1(x+9)^2+(-76)\)

\(x^2+15x+k\) with a=1,b=15 and c=k we want double root so \(\sqrt{b^2-4ac}=0 <=> \sqrt{15^2-4k}=0 <=> \sqrt{15^2} =\sqrt{4k}=0 <=> k=225/4\)