Hi! Can you please show me how problem 8 is solved in test 2 section 3?

Sure! First, don’t get bogged down in the wall of text. You should read it, of course, but you should read it while remembering that the formula you’re going to need has already been given to you at the top of the question. A simplified version of the question is this:

nA = 360. If A > 50, what’s the greatest integer value of n?

I’ll suggest two ways to solve. First, you can backsolve. The question asks for the greatest number of sides, so start with the largest answer choice, 8, and see if you get a value of A that’s greater than 50 when you set n = 8.

8A = 360
A = 45

Nope, not quite. So try n = 7.

7A = 360
A = 51.4…

That works! So choice C is correct.

The mathier way is to use an inequality. First, get A by itself in the given equation:

A=\dfrac{360}{n}

Now, remembering that we know A must be greater than 50 and therefore \dfrac{360}{n} must be greater than 50, make an inequality.

50<\dfrac{360}{n}

50n<360

n<7.2

Since the question asked for the largest possible value of n, which can’t be a decimal because n represents a number of sides and a polygon can’t have a decimal number of sides, we go with 7.

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