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| Section | Objectives |
|---|---|
| Polynomial and Rational Expressions | - Polynomials
|
| Functions and Graphs | - Graphing Functions
|
| Linear Functions and Systems | - Linear Equations
|
| Exponential and Logarithmic Concepts | - Exponential Functions
|
| Foundations of Algebra | - Algebraic Expressions
|
>> Actual Applied-Algebra Test <<
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NEW QUESTION # 80
The function N(t)models the number of subscribers to a virtual newsletter over time. The graph of N(t)is shown. The horizontal axis represents the number of years since the newsletter started, and the vertical axis represents the number of subscribers, in thousands.
How is the number of subscribers changing over time based on the graph?
Answer: A
Explanation:
The graph shows an increasing exponential curve.
The number of subscribers is going up as time passes, so the function is increasing.
Also, the graph becomes steeper as time increases. This means the rate of increase is getting larger.
So the number of subscribers is:
" increasing faster and faster "
This is a key feature of exponential growth.
NEW QUESTION # 81
A vehicle is travelling away from a rest stop. The function
D(t)=48t+23
represents the distance from the vehicle to the rest stop at time t, where t is measured in hours and D is measured in miles.
What is the value of D(1.7)?
Answer: D
Explanation:
The function is:
D(t)=48t+23
This is a linear function because it has the form:
D(t)=mt+b
where:
48
is the rate of change, and:
23
is the starting distance from the rest stop.
We need to find:
D(1.7)
That means substitute:
t=1.7
into the function.
D(1.7)=48(1.7)+23
First multiply:
48(1.7)=81.6
Now add 23:
D(1.7)=81.6+23
D(1.7)=104.6
So the vehicle is:
104.6 miles
from the rest stop after 1.7 hours.
NEW QUESTION # 82
The graph shows the number of people waiting in a virtual queue to buy tickets for an event.
What does the horizontal asymptote mean?
Answer: C
Explanation:
The graph shows the queue size changing over time.
The horizontal axis represents:
" Time in minutes "
The vertical axis represents:
" Queue size "
The graph is a decreasing exponential curve. It starts near 750people and decreases quickly at first. Then the curve begins to flatten near:
y=100
A horizontal asymptote is the horizontal line that the graph approaches as time continues.
Here, the graph approaches the horizontal line:
y=100
This means that as time passes, the number of people in the virtual queue gets closer and closer to 100.
It does not mean the queue decreases to 0, because the graph levels off near 100, not near 0.
Therefore, the correct interpretation is:
" As time passes, the number of people in the queue approaches 100. "
NEW QUESTION # 83
The traffic flow on a roadway after 8:00 a.m. is modeled by the graphed exponential function.
Which conclusion is valid, based on the horizontal asymptote?
Answer: B
Explanation:
The graph shows an exponential decay model for traffic flow. A horizontal asymptote represents the long- term value the function approaches. In this case, the graph decreases but levels off above zero. Because the horizontal asymptote is above the horizontal axis, the model predicts that the traffic flow approaches a positive number rather than decreasing all the way to zero. Exponential functions do not change by a constant additive rate; that is a property of linear functions. Instead, exponential models change by a constant factor.
Therefore, statements about constant rate are not valid here. The valid conclusion based on the asymptote is that the traffic flow will not decrease to 0 vehicles per hour. The correct answer is B.
NEW QUESTION # 84
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: C
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 # 85
......
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