Applied-Algebra Questions & Answers & Applied-Algebra Study Guide & Applied-Algebra Exam Preparation

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WGU Applied-Algebra Exam Syllabus Topics:

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

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WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q65-Q70):

NEW QUESTION # 65
An investment account accrues interest every year, and the value of the account is given by G(x). The graph of this function is shown.

What represents the value of the account after 4 years?

Answer: C

Explanation:
The graph represents the value of an investment account over time. The horizontal axis gives time in years, and the vertical axis gives the total account value in dollars. To find the value after 4 years, locate x=4 on the horizontal axis and read the corresponding G(x)-value on the graph. The plotted exponential growth curve is slightly above $1,000 at 4 years, closest to $1,052.87. Because the graph shows an increasing exponential trend, values such as $480.52, $584.62, and $711.28 are too low for the account value at x=4. The correct interpretation is that after 4 years, the investment account is worth approximately $1,052.87. Therefore, the correct answer is D.


NEW QUESTION # 66
The population of fish in a lake is changing according to the function P(t)=31t+438, where t is the number of months since the beginning of the year and P(t) is the fish population at time t. Which interpretation of the rate of change is correct?

Answer: A

Explanation:
The function P(t)=31t+438 is linear and follows the form P(t)=mt+b. In this form, m is the rate of change and b is the initial value. Here, the coefficient of t is 31, so the fish population changes by 31 fish per month.
Because 31 is positive, the population is increasing rather than decreasing. The number 438 is the initial fish population at the start of the year, not the rate of change. Therefore, the correct interpretation is that the number of fish in the lake is increasing at a constant rate of 31 fish per month. That matches option A.


NEW QUESTION # 67
The given function represents the price of a commodity, p, in dollars, based on the number of months, m, since the beginning of 2020.
p(m)=5m+5
What is the average rate of change of the price over the interval m=1to m=10?

Answer: C

Explanation:
The function is:
p(m)=5m+5
This is a linear function in the form:
p(m)=mx+b
For a linear function, the average rate of change over any interval is the same as the slope.
The slope is the coefficient of m:
5
So the price increases at a constant rate of:
$5 " per month "
We can also verify using the average rate of change formula:
(p(10)-p(1))/(10-1)
Find p(10):
p(10)=5(10)+5=55
Find p(1):
p(1)=5(1)+5=10
Now calculate:
(55-10)/(10-1)
=45/9
=5


NEW QUESTION # 68
After a water tank starts leaking, the amount of water in the tank is modeled by the exponential function that is graphed.

Which statement is justified considering the location of the horizontal asymptote?

Answer: C

Explanation:
The graph shows the water volume decreasing over time.
The vertical axis represents:
" Water volume in gallons "
The horizontal axis represents:
" Time in minutes "
The graph is an exponential decay curve. It decreases quickly at first, then levels off.
From the graph, the curve approaches a horizontal value near:
y=1,000
This means the horizontal asymptote is approximately:
y=1,000
Since the graph approaches 1,000 gallons, it will not decrease to 500 gallons according to this model.
The water volume does not change by a constant rate because exponential functions do not have constant additive change. Also, the graph decreases fastest at the beginning, not later.


NEW QUESTION # 69
The function
c(t)=-0.01t