New Launch Applied-Algebra Dumps [2026] - WGU Applied-Algebra Exam Questions

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WGU Applied-Algebra Exam Syllabus Topics:

SectionWeightObjectives
Topic 1: Graphing and Functions25%- Coordinate plane and plotting points
- Slope and equations of lines
- Function notation and evaluation
- Linear functions and their graphs
Topic 2: Systems of Equations and Inequalities15%- Solving by substitution and elimination
- Graphical solutions
- Applications of systems
Topic 3: Algebraic Expressions and Operations20%- Simplifying and evaluating expressions
- Operations with polynomials
- Order of operations
- Variables, constants, and coefficients
Topic 4: Exponents, Radicals, and Quadratic Relationships15%- Properties of exponents
- Simplifying radical expressions
- Solving quadratic equations
- Basic quadratic graphs
Topic 5: Linear Equations and Inequalities25%- Real-world applications
- Solving single-variable equations
- Solving and graphing inequalities

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WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q58-Q63):

NEW QUESTION # 58
The logistic function f(x), whose graph is shown, models the number of people who have created a website account, where xrepresents the number of days since the website started and f(x)represents the number of people who have created an account.

What is one range of values for which the graph is concave down?

Answer: C

Explanation:
For a logistic graph, concavity changes at the inflection point.
Before the inflection point, the graph is usually concave up. This means the function is increasing faster and faster.
After the inflection point, the graph is concave down. This means the function is still increasing, but it is increasing slower and slower.
From the graph, the curve changes concavity around:
x=8
After x=8, the graph begins to level off as it approaches its upper limit.
So one interval where the graph is concave down is:
(8#16)
Therefore, the correct answer is:
# ( " A " )


NEW QUESTION # 59
The value of an investment portfolio after x years is modeled by the function g(x)=3100(1.059) x
.
What is the value of the portfolio when x=10?

Answer: A

Explanation:
This problem uses an exponential growth model. The initial portfolio value is 3100, and the growth factor is
1.059. Since the growth factor is greater than 1, the portfolio increases over time. To find the value after 10 years, substitute x=10 into the function: g(10)=3100(1.059)
10
First evaluate the exponent, then multiply by 3100. This gives g(10)#5499.4758. Rounded to the nearest cent, the value is $5,499.48. The answer must be written as money because the function models the value of an investment portfolio. Among the choices, only option B matches the correctly evaluated exponential expression. Therefore, the correct answer is B.


NEW QUESTION # 60
The number of people auditioning for a game show is expected to be 7 more than the number of people who auditioned last year. The function A(t) can be used to model the situation, where t represents the number of people who auditioned last year and A(t) represents the number of people expected to audition this year.
Which quantity represents the number of people expected to audition this year, given that 440 people auditioned last year?

Answer: A

Explanation:
In this function-notation problem, the input represents last year's number of people and the output represents this year's expected number. Because this year's number is 7 more than last year's, the rule is A(t)=t+7. The question says 440 people auditioned last year, so the correct input is 440. Substituting gives A(440)
=440+7=447. This means if 440 people auditioned last year, then 447 are expected to audition this year. The notation must place 440 inside the parentheses because that is the input value, and 447 must be written as the output value. The other options reverse the input and output or subtract instead of adding. Therefore, option C is correct.


NEW QUESTION # 61
The function d(x)=14+65xrepresents the distance, in meters, from a tower to an object at time x, in seconds.
What is the value of d(1.5)?

Answer: D

Explanation:
We are given the function:
d(x)=14+65x
This is a linear function because it has the form:
d(x)=mx+b
where 65is the rate of change and 14is the starting value.
We need to find:
d(1.5)
That means substitute x=1.5into the function:
d(1.5)=14+65(1.5)
Now multiply:
65(1.5)=97.5
Then add:
d(1.5)=14+97.5
d(1.5)=111.5
So the distance from the tower to the object at 1.5seconds is:
111.5 " meters "


NEW QUESTION # 62
Which statement correctly interprets u(8), given that u(t)=0.21t
3
#4.09t
2
+18.89t+69.5 represents the number of units, u, in an inventory warehouse t hours after 9:00 a.m.?

Answer: C

Explanation:
The function u(t) gives the number of units in the inventory warehouse t hours after 9:00 a.m. The expression u(8) means the value of the inventory function 8 hours after 9:00 a.m. Since 8 hours after 9:00 a.m. is 5:00 p.
m., the time must be 5:00 p.m., not 10:00 a.m. Now evaluate the function: u(8)=0.21(8)
3
#4.09(8)
2
+18.89(8)+69.5. This gives approximately 66.38, which rounds to about 66 units. Therefore, the correct interpretation is that there are approximately 66 units in the inventory at 5:00 p.m.


NEW QUESTION # 63
......

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