Alan

Q(2+i) = a(2+i) + b = 2a+b + ai

Since Q(2+i) = P(2+i) we must have 2a+b + bi = 4 - 3i

Equate real and imaginary parts
Real:  2a + b = 4

Imag: b = -3

I'll leave you to finish this.

a^2 + bc = b^2 + ac

Rearrange as a^2 - b^2 = ac - bc

Factor as (a+b)(a-b) = (a-b)c

Simplify, a + b = c

Can you take it from here?

Use the calculator on the Home page here.  Hover over the log button and you will see lg2 appear.

Base 2 palindrome:   10001       = 17 in base 10

Base 16 palindrome:   11           = 17 in base 10        = 10000 in base 2

Probability that no dice show a 6:  p1 = (5/6)^5

Probability that exactly one die shows a six:  p2 = 5*(1/6)*(5/6)^4

Probability that at least two dice show a six: p = 1 - p1 - p2 ≈ 0.196

I think there are 6 such values

Rearrange the L1 equation as \(y=-\frac{5}{8}x-\frac{13}{8}\)

The slope of L2 must be \(m=\frac{8}{5}\) so its equation will be \(y=\frac{8}{5}x+b\)

We know it goes through point (10, -10), so  \(-10=\frac{8}{5}\times10+b\)

Can you take it from here?

You could also tackle it this way:

(x + y)³ = x³  + 3x²y + 3xy² + y³ =  x³ + y³ + 3xy(x + y)

So  9³ =   x³ + y³ +3*10*9

x³ + y³ = 729 - 270 = 459

asinus has it right as far as he went, so:

Expected win = (0 + 1 + 1 - 1 + 1 - 1)/6 = $1/6