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As follows;

Find the sum of the digits of
\(777,777,777,777^2 - 222,222,222,223^2\).

Formula: \(a^2-b^2 = (a-b)*(a+b)\)

\(\begin{array}{|rcll|} \hline && \mathbf{ 777,777,777,777^2 - 222,222,222,223^2 } \\ &=& (777,777,777,777 - 222,222,222,223)*(777,777,777,777 + 222,222,222,223) \\ &=& 555,555,555,554*1,000,000,000,000\\ &=& \mathbf{555,555,555,554,000,000,000,000 }\\ \hline \end{array}\)

In trapezoid ABCD, sides AB and CD are parallel, A = 2D and C = 3B + A.  Find B.

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∠A + ∠D = 180º

2D + D = 180 

∠A = 120º

∠D = 60º

∠B + ∠C = 180º

B + 3B + 2D = 180

4B = 60

B = 15

Hi, wouldn't AD = AB as well?

a = a

b = a + 2

c = a + 4

b^2 - a^2 = 348

(b+a)(b-a) = 348

(a+2+a)(a+2-a) = 348

Solve for a and you'll have b and c. :))

=^._.^=

(a-4)(b-4)

ab - 4(a + b) + 16

Remember vietas can help you find sum and product of roots. 

=^._.^=

I'm going to assume that A is the center of the circle. 

AC = AB since they're both the radius. 

AD + EC = AB

Can you figure it out from here?

=^._.^=

.....You'll need to find the inverse functions

for f(0)  you need inverse of    3-x    which is 3-x         f^-1(0) = 3

for 6 you need to find when    -x^3  + 2x^2 + 3x  = 6       or when   f(?) = 6         this occurs at     ? = 2

3+2 = 5

 x = 32 degrees.

The smallest possible value of f(t) is 45.

[ABMO]/[ABCDEFGH] = 13/80.

Here's  an  approach  to  the  same problem......only  the  numbers  are changed.......see if it helps

 https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_11

Rewrite  as

x(x + 2)  +  n =  0

x^2  + 2x   +  n  = 0

If we have no real solutions, the discriminant  is  <  0       ....so.....

2^2 - 4n < 0

4  - 4n < 0

4 < 4n

n >  1

So... no  real  solutions  exist    when  n =  [ 2, 10 ]

So         9 /10   proobability of no real solutions

Just set the equation = to -5

2x+7 = - 5

2x = -12

x = -6      this is the first part of the piecemeal equation and - 6 is less than - 2      so first answer = -6

Now for the second part of the equation

-x^2 -x+1 = - 5

-x^2 -x + 6 = 0

x^2 + x-6 = 0

(x-2)(x+3) = 0       for this x can be = to 2   or -3 (throw out...because this part of the equation is only for x >=2)

    so the two possibilities are x = - 6 or 2

By Vieta  ....

The  sum  of  the roots  =  k/2  ⇒   k =  2 *(sum of the roots)

And  the  product of  the roots  =  28 /2   =   14

Either

7 * 2   =  14         and  the   sum of roots  = 2(7 + 2)  =  18  =   k

or

14 * 1  = 14          and  the  sum of  the roots  = 2 ( 14 + 1)   = 30  =  k

or

-7 * - 2  =  14        and  the  sum  of  the roots  = 2(-7 + -2)  =  -18  = k

or

-14 * -1  =  14        and the  sum of  the roots  = 2 9-14 + -1)  = -30 = k

The  sum of all possible k's =   0

I will assume these are only positive numbers

A = 16     B = 96      Difference = 80

If the numbers can be negative

A = -16     B = 96     Difference = 112

a^2 - b^2  =  (a + b) ( a - b)  =  (-1) (a - b)

So

a^2  - b^2  + 3a   + b  =

(-1)(a - b)  + 3a  + b  =

-a + b + 3a  +  b  =

2a + 2b  =

2 (a + b)  =

2 (-1)   =

-2

The largest possible value of f(12) is 148.

See my answer  under   your original post.....

B - A = 12.

f(-2)  =  (-2)^2  - 2   =  2

f (2)  =  2(2) - 20  =  -16

So

2 - 16 =  - 14

So   we need  f(a) to  =   14

Note  that  if  a  =  -4    then f (a)  =  f (-4)  = (-4)^2  - 2  =  14

And  if  a  =  17  then   2(17) - 20  =  14

So

a = 17    or  a  = - 4

And  their  +  difference  =  21

f(f(f(f(1 + i)))) = -64.

The possible values of P are 3, 4, 5, 6, 7, 8, 9.