CPhill

The second one requires a little thinking

If the center of the circle lies on the line  y = 6x + 2  and the circle is tangent to the x axis  at (-2, 0)  then the the x coordinate of the center of the circle  is at  x = -2

So.....the y cordinate of the center  must be y = 6(-2) + 2  =  -10

So....the center  is  ( -2, -10)

And since this circle is tangent to the x axis......the radius must = 10

So.....the equation is   (x + 2)^2 + (y + 10)^2  = 100

See the graph here :   https://www.desmos.com/calculator/8abh7hhozi

The first  isn't too difficult 

We have the form   (x - h)^2  + (y - h)^2  = r^2

The center  is ( x , h)    and the radius  =  sqrt (r^2) = r

So....we have

(x - 2)^2 + (y + 3)^2  = 80   writing this in a slightly different form we have

(x - 2)^2  + ( y - (-3) )^2  = 80

So.....the center  is   (2, -3)    and the radius  = sqrt (80)  =  sqrt (16 *5)  = 4sqrt (5)

Egyptian Fractions have the form :  1 / positive integer 

1/3 =  1/a + 1/b

Let z  = 3

a,b  must be > 3

So let a  =  z + m

And let b =  z + n

So we have

1/ z  =    1/ [z + m]  +  1/[;z + n]

1/z  =  [ 2z + m + n] / [(z + m) (z + n)]      cross-multiply

(z + m) (z + n)  =  z (2z + m + n ]

z^2 + mz + nz + mn =  2z^2 + mz + nz

z^2 + mn = 2z^2

z^2  = mn       so

3*2 = mn

9 = mn

So the possibilites for m,n  are

m   n

1    9

3    3

So....the possible fractions are

1/ [3 + 1]   +  1/ [ 3 + 9]  =     1/4 + 1/12

1/[3 + 3] + 1/[3 + 3]  =  1/6 + 1/6

Only the first is what we need

So

1/4  + 1/12 

Yep...if A,B  can be the same integers, you are correct

[ I assumed that they had to be different ]

5 consecutive heads can occur as follows

HHHHHTT

THHHHHT

TTHHHHH

6 consecutive heads can occur as follows:

HHHHHHT

THHHHHH

7 consecutive heads can occur as follows :

HHHHHHH

So....the probability  is 

[ 3 + 2 + 1 ] / 2^7  =

6 /128 =

3 / 64

If A,B  can be the same integers....you are correct, HSM  !!!

[ I assumed that they must be unique ]

Whatever.....our answers only differ by 1, so  it probably depends upon our assumptions  ....LOL!!!

25AB

The sum of the digits must be a multiple of 9......so.....

2 + 5  +  A  + B  =   9M

A + B   =  9M - 7

If  M =  1                    M  =  2                M  = 3

A + B  =  2                A + B  =  11         A + B  = 20 

A , B  =                     A , B =                 No possibilities  for A, B when  M > 2

0, 2                           2, 9

and the reverse        3, 8

                                 4, 7

                                 5, 6

                                 and the reverses

So....by my count.....we have 5(2) =  10 such numbers

That sounds like NO FUN, CU  !!!

If   3AA1  is divisible by 9, the sum of its digits must be divisible by 9

The means  that 

3 + 1 + 2A  = 9M

4 + 2A  = 9M

2A  = 9M - 4

M must be an even number  > 0

So

If

M  = 2    A  = 7

And 3771 / 9  =  419

(3/2)x^2 + 11x + c  = 0

Using the discriminant, we have that

(-11)^2 -  4(3/2)c   = 7

121 -6c =  7      subtract 121 from each side

-6c  =  -114   divide both sides by -6

c =  -114/-6

c = 19