Answer:
Colin has <em>8 sheets </em>left for his third class.
Step-by-step explanation:
Given that:
Total Number of pieces of papers = 
Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad
Writing the equation:

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

So, the answer is:
Colin has <em>8</em> <em>sheets </em>left for his third class.
A. (5,7)
b. (5,5)
c. (5,3)
d. (-2,5)
e. (-4,5)
f. (-6,5)
Between 90 and 180 degrees
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