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sin A =   BD / AB  =  BD / 24  = 1/3  →  BD = 8

sin C = BD / BC =  8 / BC = 1 / 2  →   BC  = 16

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

Simplify as

6x^2 - 21x -31 = 0

x =   [ 21 + sqrt [ 21^2 + 4(6)(31) ] ] / 12  =   [21 + sqrt (1185) ] / 12

(a + 2) / (a + 1)  =  b / 8

8(a + 2) /(a + 1)  = b

a      b

0    16

1    12

3    10

7      9

-2    0

-3    4

-5    6

-9    7

The answer is 5*pi/64.

I haven't figured out a or b, but here is the answer to part c:

1. Find the number of license plates with exactly four A's:

Choose 4 positions out of 6 for the A's: 6C4 = 15 ways

Fill the remaining 2 positions with any of the 26 letters: 26 * 26 = 676 ways

Total: 15 * 676 = 10,140 ways

2. Find the number of license plates with exactly two B's:

Choose 2 positions out of 6 for the B's: 6C2 = 15 ways

Fill the remaining 4 positions with any of the 26 letters: 26 * 26 * 26 * 26 = 456,976 ways

Total: 15 * 456,976 = 6,854,640 ways

3. Find the number of license plates with both four A's and two B's:

Choose 4 positions for the A's: 6C4 = 15 ways

Choose 2 of the remaining 2 positions for the B's: 2C2 = 1 way

Fill the remaining 0 positions: 1 way

Total: 15 * 1 * 1 = 15 ways

4. Find the total number of license plates:

Total = (Number with 4 A's) + (Number with 2 B's) - (Number with both)

Total = 10,140 + 6,854,640 - 15

Total = 6,864,765 ways

Therefore, there are 6,864,765 possible Aopslandian license plates that contain exactly four A's, or exactly two B's, or both.

The ratio CD/AB is equal to 12/5.

A quadrilateral has angles of $q$, $4q + 17^\circ $, $8q + 21^\circ$, and $15q + 148^\circ $. Find the largest angle of the quadrilateral, in degrees.    

q  +  (4q + 17^o)  +  (8q + 21^o)  +  (15q + 148^o)  =  360^o    

28q + 186^o  =  360^o    

28q  =  174^o    

q  =  174^o / 28  =  6.2^o     rounded to 6.2 here, but I didn't round it in my calculator    

by observation it's obvious that the largest angle is (15q + 148^o)    

(15 • 6.2^o) + 148^o  =  241.2^o    

_.    

Let x = AC   and  y = BC      and AB is the hypotenuse

CM = y/2

CN = x/2

Triangles  AMC  and BNC are both right 

AC^2 + ( BC/2) = AM

(AC/2)^2 + BC^2 = BN

So.....by the Pythagorean Theorem we have these two equations

x^2 + (y/2)^2   =1         

(x/2)^2 + y^2  = 1              simplify

x^2 + y^2/4  =1  →   4x^2 + y^2  = 4        (1)

x^2/4 + y^2  =1  →    -x^2 - 4y^2 = -4

Add the last two equations and we  get that

3x^2 - 3y^2  = 0

3x^2 = 3y^2

x^2 = y^2

Sub this  back into (1) for y^2

4x^2 + x^2  = 4

5x^2  = 4

x^2  = 4/5

and

y^2 = 4/5

AC^2 + BC^2  =  AB^2

4/5 + 4/5  = AB^2

AB =   sqrt [ 4/5 + 4/5 ]

AB = sqrt [ 8/5]  =  2sqrt [2/5] = (2/5)sqrt 10

What is the coefficient of x in (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 4x^2 + 8x + 7)?    

When you multiply two polynomials, each element in one is multiplied by each element in the other. 

(x^3 + x^2 + x + 1)(x^4 + 3x^3 + 4x^2 + 8x + 7)     x • 7    =  7x    

(x^3 + x^2 + x + 1)(x^4 + 3x^3 + 4x^2 + 8x + 7)     1 • 8x  =  8x    

                                         add them together                  15x    

Using Apollonius's Theorem we know that the median from the right angle is half the hypotenuse so the answer is 2x1 which is 2

Poltnomial  division

                       x^2

 x^2 - 2x -5   [  x^4 -2x^3 + kx^2  ]

                        x^4 - 2x^3 -5x^2

                       _______________

k - -5  = 0

k + 5  = 0

k = -5

We want to find the sum of f(0)+f(1)+f(2)+f(3)+f(4) where f(n) is the remainder of n^8+n^16 divided by 5

We start with 0 and get 0 so the remainder is 0

Then we plug in 1 to get 1+1 is 2 so the remainder is 2

For f(2) we know 2^8 is 256. 256 ends in a 6 so 2^16 which is (2^8)^2 ends in 6. 6+6=12 so 12/5 the remainder is 2

For f(3) we know 3^4 is 81 so the last diit of 3^8 is 1 then square 3^8 to get 3^16 and the last digit is also 1 so 1+1=2 so the remainder is also 2

For f(4) is the same as f(2) since it is bascically square all of f(2) so the remainder is still 2

Adding 0+2+2+2+2=8

To find the leading term you multiply the leading term of the functions. (-7x^4)(2x^8)=(-7)(2)(x^4)(x^8)=-14x^12

Leading term:-14x^12

The leading coefficient is -14 since that is the coefficient of the leading term

Leading coefficient: -14

Degree:12

To get the constant term you multiply -5 and 2 because they are the 2 constants

Constant term: -10

To find the coefficient of x^6 we multiply

(-7x^4)(18x^2)=-126x^6

(-3x^2)(-11x^4)=33x^6

(2)(5x^6)=10x^6

-126+33+10=-83

Coefficient of x^6: -83

To find the coefficient of x^2 you have to multiply

(x^2)x1=x^2

and

(x)(8x+7x)=(x)(15x)=15x^2

you add x^2+15x^2 to get 16x^2 so the answer is 16

Simplify as

x^2 + 18x + 3     complete the square on x

x^2 + 18x + 81 + 3 - 81

1(x + 9)^2 +(  - 78 )

(r-6)^2 + (r-6)^2  = r^2

2r^2 - 24r + 72  =  r^2

r^2 - 24r + 72 = 0     complete the square on r 

r^2 -24r + 144 = -72 + 144

(r -12)^2 = 72

r - 12  = sqrt [72]

r = 12 + 6sqrt 2

Ahh...Owinner. the sign is greater and EQUAL to, so -5 and -1/2 are actually valid solutions there.

the answer is

\((-\infty, -5]\cup(-1, -\frac{1}{2}]\)

Draw a line connecting B and E. Do you see any similar triangles? there should be 3.