Applied-Algebra Exam Dumps - WGU Courses and Certificates Questions and Answers

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 "

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.

The graph shows:

P(t)= " weekly profit in thousands of dollars "

where:

t= " years since opening "

The graph is a downward-opening curve, so it has a maximum value at its highest point. In Applied Algebra, the highest point of this kind of polynomial graph is called the vertex.

To answer the question, we need the t-value of the vertex because the question asks when the store had its maximum weekly profit.

From the graph, the highest point occurs between:

t=5

and

t=6

More specifically, it is approximately:

t=5.5

So the store had its maximum weekly profit approximately:

5.5 " years after opening "

The graph shows the temperature of an object decreasing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve decreases and approaches the horizontal axis as time increases.

A horizontal asymptote is a horizontal line that the graph approaches but may not reach. Horizontal lines have equations of the form:

y= " constant "

From the graph, the curve approaches:

y=0

So the horizontal asymptote is:

y=0

The options x=0and x=500are vertical lines, not horizontal asymptotes.

Therefore, the correct answer is:

▭ ( " B " )

A logistic function often increases quickly in the middle and then begins to level off as it approaches a maximum carrying capacity. From year 7 to year 10, the graph is still increasing, so the bison population is going up. However, the curve is becoming flatter during this interval. A flatter curve means the slope is getting smaller, so the population is increasing at a slower rate. This behavior is described as “increases slower and slower.” It is not decreasing, because the graph continues to move upward. It is also not increasing faster and faster, because the slope is not becoming steeper in this interval. Therefore, the correct interpretation is answer B.

This problem requires reading a value from a graph. The horizontal axis gives time in minutes, and the vertical axis gives the number of viewers. To solve it, find 432 on the vertical axis, locate where the graph reaches that value, and then move down to the horizontal axis. From the exhibit, the graph reaches 432 viewers at minute 1. That means the ordered pair is approximately (1, 432), so the input is 1 minute and the output is 432 viewers. The other listed minutes correspond to different viewer counts on the graph. In graphical depictions, matching the correct input-output pair is the key step. Therefore, the correct answer is B.

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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.

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 "

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. "

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