Answer:
It's a property of log() or ln() functions. It's states:

in other word, subtraction of log functions could be combined to a quotient in the log function.
A) 3n+4= 7n
B) 4n-5= -1n
You do a normal equation just add the N
The two pieces are continuous for every 
So, our only concern is to make sure that the pieces "glue" continuously at x=7.
To ensure this, we evaluate both pieces at x=7, and impose that the two values are equal.
The first piece evaluates to

The second piece evaluates to

So, we want

We move all terms involving c to the left hand side, and all the numerical constants on the right hand side:

M=-1
4m+2(m+1)=9m +5
4m+2m+2=9m+5
6m+2=9m+5
6m+2-6m= 9m+5-6m
2-5=3m+5-5
-3/3=3m/3
m=-1
Answer:
1/6
1/12
1/2
Step-by-step explanation:
There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner. So there are a total of 3×4=12 possible combinations. We can show that by making a grid:
![\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26R%26B%26G%5C%5CR%26RR%26BR%26GR%5C%5CB%26BR%26BB%26BG%5C%5CP%26PR%26BP%26GP%5C%5CY%26RY%26BY%26GY%5Cend%7Barray%7D%5Cright%5D)
Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB). So the probability is 2/12 = 1/6.
There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.
There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP). So the probability is 6/12 = 1/2.