Take $1.29 times 3(number of nights) and add $.50 = $4.37
You can solve this problem through factoring.
First, you have the equation,

Then, you can factor the numerator.

You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with 
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
Answer: 9/7
Step-by-step explanation:
(3a^4 * b^-2) * (2a^7 * b^-1)
= 6*a^(4+7)*b^(-2-1)
= 6*a^11*b^-3
.7777777778 the calculator says, only because it ran out of room :)