Latest Study WGU Applied-Algebra Questions | Latest Applied-Algebra Braindumps Files

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WGU Applied-Algebra Exam Syllabus Topics:

SectionObjectives
Topic 1: Linear Functions and Systems- Linear Equations
  • 1. Point-slope form
    • 2. Slope-intercept form
      - Systems of Equations
      • 1. Substitution method
        • 2. Elimination method
          Topic 2: Foundations of Algebra- Algebraic Expressions
          • 1. Evaluating expressions
            • 2. Order of operations
              • 3. Simplifying expressions
                - Equations and Inequalities
                • 1. Linear equations in one variable
                  • 2. Inequalities and interval notation
                    Topic 3: Exponential and Logarithmic Concepts- Exponential Functions
                    • 1. Growth and decay models
                      • 2. Exponential equations
                        - Logarithmic Basics
                        • 1. Logarithm properties
                          • 2. Conversion between exponential and logarithmic form
                            Topic 4: Functions and Graphs- Graphing Functions
                            • 1. Transformations of graphs
                              • 2. Coordinate plane interpretation
                                - Function Concepts
                                • 1. Domain and range
                                  • 2. Function notation
                                    Topic 5: Polynomial and Rational Expressions- Polynomials
                                    • 1. Operations with polynomials
                                      • 2. Factoring techniques
                                        - Rational Expressions
                                        • 1. Simplification
                                          • 2. Restrictions on variables

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                                            Latest Applied-Algebra Braindumps Files | Exam Applied-Algebra Simulator Fee

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                                            WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q50-Q55):

                                            NEW QUESTION # 50
                                            Based on collected data, the value of a painting can be modeled using the exponential function f(x)=240(1.12) x
                                            . In this case, x represents the number of years since 2005, and f(x) represents the value of the painting in dollars.
                                            Which value represents the average yearly rate of change of the painting's value from 2013 to 2015?

                                            Answer: B

                                            Explanation:
                                            This problem asks for an average yearly rate of change over a time interval. Since x represents years since
                                            2005, the year 2013 corresponds to x=8, and the year 2015 corresponds to x=10. Use the average rate of change formula:
                                            10#8
                                            f(10)#f(8)
                                            Evaluating the model gives f(8)=240(1.12)
                                            8
                                            #594.23 and f(10)=240(1.12)
                                            10
                                            #745.40. The total change is approximately 151.17 dollars over 2 years. Dividing by 2 gives about 75.59 dollars per year. Since the question asks for the average yearly rate, not the total increase, the correct answer is B.


                                            NEW QUESTION # 51
                                            The number of property sales in a region this year is expected to be 8 times the number of property sales in the region last year. The function H(x)represents the number of property sales this year, where xrepresents the number of properties sold last year.
                                            Which notation represents the number of property sales this year, given that the number of properties sold last year was 360?

                                            Answer: D

                                            Explanation:
                                            We are told that the number of property sales this year is expected to be 8 times the number of property sales last year.
                                            Let:
                                            x= " number of properties sold last year "
                                            and
                                            H(x)= " number of property sales this year "
                                            Since this year's sales are 8 times last year's sales, the function rule is:
                                            H(x)=8x
                                            The question says that last year's property sales were:
                                            x=360
                                            Substitute 360into the function:
                                            H(360)=8(360)
                                            H(360)=2,880
                                            So the correct notation is:
                                            H(360)=2,880
                                            This means: if 360 properties were sold last year, then 2,880 property sales are expected this year.
                                            Therefore, the correct answer is:
                                            # ( " C " )


                                            NEW QUESTION # 52
                                            A vehicle is traveling away from a town at a fixed rate. After 1 hours, the vehicle is 200 miles from the town.
                                            After 4 hours, the vehicle is 395 miles from the town.
                                            Which function represents the distance, d, between the vehicle and the town after thours?

                                            Answer: B

                                            Explanation:
                                            Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:
                                            d(t)=mt+b
                                            where:
                                            m= " rate of change "
                                            and
                                            b= " initial distance from the town "
                                            We are given two points:
                                            (1#200)
                                            and
                                            (4#395)
                                            These mean:
                                            After 1hour, the vehicle is 200miles away.
                                            After 4hours, the vehicle is 395miles away.
                                            First, find the rate of change:
                                            m=(395-200)/(4-1)
                                            m=195/3
                                            m=65
                                            So the vehicle is moving away from the town at a rate of:
                                            65 " miles per hour "
                                            Now the function has the form:
                                            d(t)=65t+b
                                            Use the point (1#200)to find b:
                                            200=65(1)+b
                                            200=65+b
                                            b=135
                                            Therefore, the function is:
                                            d(t)=65t+135
                                            Check using t=4:
                                            d(4)=65(4)+135
                                            d(4)=260+135
                                            d(4)=395


                                            NEW QUESTION # 53
                                            In the following graph showing the elevation profile, F(d), for a mountain biking course, the horizontal axis shows the number of miles since the start of the course and the vertical axis shows the elevation in feet.

                                            How can the concavity be described from d=10 to d=13.1?

                                            Answer: B

                                            Explanation:
                                            This question asks for a description of concavity and rate of change on a specific interval. From d=10 to part of the interval, the elevation is increasing, but the graph becomes less steep as it approaches a local maximum.
                                            That means the elevation increases slower and slower. After the peak, the graph begins to decrease, and the downward slope becomes steeper. This means the elevation decreases faster and faster. The correct statement must describe both parts of the interval in order: first increasing at a slowing rate, then decreasing at an increasing rate. The other answer choices either reverse the behavior or describe the wrong direction of change. Therefore, the correct answer is A.


                                            NEW QUESTION # 54
                                            The number of letters processed daily at a mail center is modeled by the decreasing exponential function shown in the graph.

                                            What is the long-term trend in the number of letters processed per day, based on the equation of the horizontal asymptote?

                                            Answer: A

                                            Explanation:
                                            The graph shows a decreasing exponential function.
                                            In Applied Algebra, a decreasing exponential function may approach a fixed value over time. This fixed value is called the horizontal asymptote.
                                            The horizontal asymptote represents the long-term value that the function gets closer and closer to, but does not necessarily cross or reach exactly.
                                            From the graph, the curve decreases quickly at first, then begins to level off near:
                                            y=1,000
                                            This means that as time continues, the number of letters processed per day approaches:
                                            1,000
                                            So the long-term trend is that the mail center will process about:
                                            1,000 " letters per day "


                                            NEW QUESTION # 55
                                            ......

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