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Answer D .

tan A = 2/3.

There are 40 ways of making a change for a $20 bill as follows:

       $1   2   5 10

1  = (20, 0, 0, 0)
2  = (18, 1, 0, 0)
3  = (16, 2, 0, 0)
4  = (14, 3, 0, 0)
5  = (12, 4, 0, 0)
6  = (10, 5, 0, 0)
7  = (8, 6, 0, 0)
8  = (6, 7, 0, 0)
9  = (4, 8, 0, 0)
10  = (2, 9, 0, 0)
11  = (0, 10, 0, 0)
12  = (15, 0, 1, 0)
13  = (13, 1, 1, 0)
14  = (11, 2, 1, 0)
15  = (9, 3, 1, 0)
16  = (7, 4, 1, 0)
17  = (5, 5, 1, 0)
18  = (3, 6, 1, 0)
19  = (1, 7, 1, 0)
20  = (10, 0, 2, 0)
21  = (8, 1, 2, 0)
22  = (6, 2, 2, 0)
23  = (4, 3, 2, 0)
24  = (2, 4, 2, 0)
25  = (0, 5, 2, 0)
26  = (5, 0, 3, 0)
27  = (3, 1, 3, 0)
28  = (1, 2, 3, 0)
29  = (0, 0, 4, 0)
30  = (10, 0, 0, 1)
31  = (8, 1, 0, 1)
32  = (6, 2, 0, 1)
33  = (4, 3, 0, 1)
34  = (2, 4, 0, 1)
35  = (0, 5, 0, 1)
36  = (5, 0, 1, 1)
37  = (3, 1, 1, 1)
38  = (1, 2, 1, 1)
39  = (0, 0, 2, 1)
40  = (0, 0, 0, 2)
 

Sorry, that's incorrect.

Take a look at:

It will be of the form 

y=a(x+2)(x+3)

plug in (5,10) to find a

The area of triangle ACE is 120.

There are exactly three quadratics that Babette could have been thinking of.

(j + 20) = 3/4 (m - 26)

(j - 30) = 1/4 (m + 30)

4j + 104 = 3m - 78 - (1)

4j - 120 = m + 30 - (2)

Subs (1) from (2) we get

224 = 2m - 108

2m = 332

m = 166

Putting value of m in eq^m (1)

(j + 26) = 3/4 (166 - 26)

(j + 26) 3/4 (140)

(j + 26) = 3 * 35

j + 26 = 105

j = 79

The order is 20.

Angle YZX = 72 degrees.

The difference is 15*pi.

So let blue marbles in bag = x & green marbles in bag = y

1/6x = 3/8y

x = 9/4y --(i)

Now, 48 green marbles are removed.

x = 9 (y - 48) -(ii)

9/4y = 9 (y - 48) {x term eqn (i)}

y = 4y - 192

3y = 192

y = 64 marbles

x = 9/4 * 64

   = 9 * 16

x = 144 marbles

green marbles = 64

blue marbles = 144

AB + AC = 18.

The Answer is wrong

534 , 543 , 634 , 643 , 734 , 743 , 834 , 843 , 934 , 943 , Total =  10 such numbers.

We can write

3a + 4b  = 7

9a + 12b = k

We have infinite solutions when one equation is a multiple of the other.....so.....

Note that  multiplying the first equation through by 3 we get.....  

3 (3a + 4b)  =  3 * 7

9a + 12b  =  21  =  k

Let the quadratic be of the form   ax^2 + bx + c

Let us suppose  that the quadratic is monic  ( a =1)   

By Vieta....

The sum of the roots  =  -b  =    (3 + sqrt 3) / 5   + ( 3 - sqrt 3) / 5   =  3/5 + 3/5  = 6/5

So b =  -6/5

And the product of the roots  =  c  =  (3 + sqrt 3) / 5   *   (3 - sqrt 3)  / 5 =  [ 9  - 3 ] /25  =  6/25

So....the monic polynomial   is

x^2  - (6/5)x  + ( 6 / 25 )

If we wanted  integer coefficients  we also could have

25x^2 - 30x + 6    .......  many other answers are also possible