Applied-Algebra Valid Braindumps Ppt - Applied-Algebra Certified Questions

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WGU Applied-Algebra Exam Syllabus Topics:

Section	Objectives
Topic 1: Linear Functions and Systems	- Linear Equations 1. Slope-intercept form 2. Point-slope form - Systems of Equations 1. Substitution method 2. Elimination method
Topic 2: 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 3: Polynomial and Rational Expressions	- Polynomials 1. Operations with polynomials 2. Factoring techniques - Rational Expressions 1. Simplification 2. Restrictions on variables
Topic 4: Foundations of Algebra	- Equations and Inequalities 1. Inequalities and interval notation 2. Linear equations in one variable - Algebraic Expressions 1. Order of operations 2. Evaluating expressions 3. Simplifying expressions
Topic 5: Functions and Graphs	- Function Concepts 1. Domain and range 2. Function notation - Graphing Functions 1. Transformations of graphs 2. Coordinate plane interpretation

>> Applied-Algebra Valid Braindumps Ppt <<

WGU Realistic Applied-Algebra Valid Braindumps Ppt Free PDF Quiz

Eliminates confusion while taking the WGU Applied-Algebra certification exam. Prepares you for the format of your WGU Applied-Algebra exam dumps, including multiple-choice questions and fill-in-the-blank answers. Comprehensive, up-to-date coverage of the entire WGU Applied-Algebra Certification curriculum.

WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q63-Q68):

NEW QUESTION # 63
The function P(t)represents the yearly profit, in millions of dollars, for a streaming service since opening. The graph of P(t)is shown.

What is the correct interpretation of the maximum value?

A. Approximately 22 years after opening, a maximum yearly profit of approximately $1,410,000was earned.
B. Approximately 22 years after opening, a maximum yearly profit of approximately $235,000was earned.
C. Approximately 10.5 years after opening, a maximum yearly profit of approximately $1,410,000was earned.
D. Approximately 10.5 years after opening, a maximum yearly profit of approximately $235,000was earned.

Answer: C

Explanation:
The graph represents yearly profit as a function of time:
P(t)= " yearly profit "
where:
t= " years since opening "
The graph is a downward-opening curve, so its maximum value occurs at the highest point, also called the vertex.
From the graph, the highest point occurs at approximately:
t=10.5
So the maximum yearly profit happened approximately:
10.5 " years after opening "
The vertical axis measures yearly profit in millions of dollars. From the graph, the maximum value is approximately:
1.41
Since the units are millions of dollars:
1.41×1,000,000=1,410,000
So the maximum yearly profit was approximately:
$1,410,000
Therefore, the correct answer is:
# ( " C " )

NEW QUESTION # 64
Consider the function I(m)=41m²#490m+30, which represents the daily income, I, in dollars, of a store based on the number of months, m, since the beginning of the year. What is the difference between I(7) and I(9)?

A. Approximately 332 months
B. Approximately $2
C. Approximately 2 months
D. Approximately $332

Answer: D

Explanation:
This is a polynomial function evaluation problem. Since I(m) represents daily income in dollars, the difference between I(7) and I(9) must also be in dollars. First evaluate I(7)=41(49)#490(7)
+30=2009#3430+30=#1391. Then evaluate I(9)=41(81)#490(9)+30=3321#4410+30=#1059. The difference is
|#1059#(#1391)|=332. Because the question asks for the difference between the two outputs, the correct unit is dollars, not months. So the difference between I(7) and I(9) is approximately $332. Therefore, the correct answer is C.

NEW QUESTION # 65
As sacks are unloaded off a wagon, the total weight of the wagon and sacks changes. Each sack has the same weight. After 3 sacks are removed, the total weight of the cart and remaining sacks is 116 pounds. After 6 sacks are removed, the total weight is 101 pounds.
What is the weight of each sack?

A. 18 pounds
B. 5 pounds
C. 32 pounds
D. 81 pounds

Answer: B

Explanation:
This situation can be modeled using a linear relationship because each sack has the same weight.
We are given:
After 3sacks are removed, the total weight is 116pounds.
After 6sacks are removed, the total weight is 101pounds.
From 3 sacks removed to 6 sacks removed, the number of removed sacks increases by:
6-3=3
During that time, the total weight decreases from 116pounds to 101pounds:
116-101=15
So removing 3 additional sacks decreases the total weight by 15 pounds.
Now divide to find the weight of one sack:
15/3=5
So each sack weighs:
5 " pounds "
Check:
If 3 more sacks are removed and each sack weighs 5 pounds, the total weight should decrease by:
3(5)=15
116-15=101
This matches the given information.

NEW QUESTION # 66
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?

A. (0#12)
B. (0#16)
C. (0#8)
D. (8#16)

Answer: D

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 # 67
The number of registered voters in a district is modeled by a decreasing exponential function, where x represents the number of years since 2005 and f(x) represents the number of registered voters.

Which interval is associated with the slowest average decrease in the number of registered voters?

A. 2005 to 2008
B. 2007 to 2014
C. 2008 to 2023
D. 2015 to 2025

Answer: D

Explanation:
A decreasing exponential function falls quickly at first and then levels off over time. The "slowest average decrease" occurs where the graph is flattest, because the output changes by the smallest amount per year.
Earlier intervals, such as 2005 to 2008, occur when the graph is steepest, so those intervals have a faster average decrease. Later intervals occur after the exponential decay has slowed. From the graph, the interval from 2015 to 2025 is farthest to the right and has the least steep downward trend. Therefore, the number of registered voters decreases most slowly during that interval. This is why the correct answer is D.

NEW QUESTION # 68
......

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