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QTT2 Task 2 Passed – Finite Mathematics: Number Theory (WGU) Academic Year 2026

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QTT2 Finite Mathematics Task 2 (Number Theory) PASSED submission for Western Governors University (Academic Year 2026). Includes a clearly written, rubric-aligned example to help you understand expectations, structure your responses, and submit with confidence. Perfect for reviewing key number theory concepts and the exact format evaluators want. QTT2 Task, WGU QTT2, Task 2, Number Theory, Finite Math, Prime Numbers, Composite Numbers, GCD LCM, Divisibility Rules, Modular Arithmetic, Factorization Steps, Euclidean Algorithm, Congruence Problems, Rubric Guide, Passed Example, WGU Assessment, Math Performance

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QTT2
Finite Mathematics

TASK 2 – Passed
Number Theory
Western Governors University

, Part 1: Order of Operations

Figure 1:




Expression A:

2 3 The order of operation states that we must first
(−3) +14 ÷2*[(6−8) −(−4)]
3 2 solve all operations inside groupings. These will
−[|−2| −1+5 ] include parentheses, brackets, braces, and absolute
values. A fraction bar can also be considered a
grouping.

3
Looking at the grouping [(6 − 8) − (− 4)] in the
numerator, I would start by subtracting the 6 from 8
to get -2. Next, I would need to change the -4 to a
positive 4 and change the sign to addition. This is
due to the fact when you subtract a negative number
it becomes a positive. This would leave us with
3
[(− 2) + 4]. This grouping is not completely
solved as there are groupings inside the other
groupings. We would now need to solve the
exponent, which would give us − 8 + 4. To finish
out the grouping, we would solve − 8 + 4 to get
− 4.

The grouping |− 2| in the denominator, we would
solve is the absolute value. This would give us 2
2 The order of operation states that we would solve all
(−3) +14÷2*(−4)
3 2 exponents next.
−[(2 )−1+5 ] 2
In the numerator, we would solve (− 3) . Since the
-3 is in parentheses this will mean that we multiple
-3 by -3 to give us 9. If the negative was not in
parentheses it would not have been included in the
exponent.

In the denominator, we have two different
3 2
exponents to solve, (2 ) 𝑎𝑛𝑑 5 . These would equal
8 and 25, respectively.



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