Use the table to answer the question.

Question

x y
0 3
1 3 1/2
2 4
3 4 1/2

Which of the following equations describes the function in the table using function notation?
Responses:

f(x)=3x+1/2

f(y)=1/2x+3

f(x)=1/2x+3

f(y)=3x+1/2

Answer 1

To determine which equation describes the function in the table using function notation, we can look at the provided points in the table and see if we can identify a pattern or a linear equation that fits the data.

The points given are:

• (0, 3)
• (1, 3.5)
• (2, 4)
• (3, 4.5)

From these points, we can see that for each increase in \( x \) by 1, \( y \) increases by \( 0.5 \). This suggests a linear relationship.

To find the equation of the line in the form \( f(x) = mx + b \), we can identify:

• The slope \( m = 0.5 \) (which is the same as \( \frac{1}{2} \)).
• The y-intercept \( b = 3 \) (the value of \( y \) when \( x = 0 \)).

Thus, the equation can be expressed as:

\[ f(x) = \frac{1}{2}x + 3 \]

Looking at the provided options, the correct one is:

f(x) = 1/2x + 3

Answer 2

If f(x)=(x)2 , what is f(−4)

Answer 3

To find \( f(-4) \) using the function \( f(x) = x^2 \), we substitute \(-4\) into the function:

\[ f(-4) = (-4)^2 \]

Calculating that:

\[ f(-4) = 16 \]

So, \( f(-4) = 16 \).

Answer 4

Bacteria are known to grow exponentially. The function B(h)=82(1.25)h

represents the number of bacteria, B(h) , as a function of h . How many bacteria will there be after only 5.5 hours, to the nearest whole number?

What is B(5.5)

Responses:

280

564

257

Answer 5

To find the number of bacteria after 5.5 hours using the function \( B(h) = 82(1.25)^h \), we need to substitute \( h = 5.5 \) into the function:

\[ B(5.5) = 82(1.25)^{5.5} \]

First, we calculate \( (1.25)^{5.5} \):

\[ (1.25)^{5.5} \approx 3.052 \]

Now we substitute this value back into the function:

\[ B(5.5) = 82 \times 3.052 \approx 250.264 \]

Rounding this to the nearest whole number gives:

\[ B(5.5) \approx 250 \]

Based on the options provided, the closest response is 257.

Answer 6

Complete the input-output table for the function f(x)=(x−2)^2+3
f(1)=4
f(2)=?
f(3)=4
f(4)=?

Answer 7

To complete the input-output table for the function \( f(x) = (x - 2)^2 + 3 \), let's calculate the values for \( f(2) \) and \( f(4) \).

1. For \( f(2) \): \[ f(2) = (2 - 2)^2 + 3 = 0^2 + 3 = 0 + 3 = 3 \]

2. For \( f(4) \): \[ f(4) = (4 - 2)^2 + 3 = 2^2 + 3 = 4 + 3 = 7 \]

Now we can summarize the completed table:

• \( f(1) = 4 \)
• \( f(2) = 3 \)
• \( f(3) = 4 \)
• \( f(4) = 7 \)

So, the answers are:

• \( f(2) = 3 \)
• \( f(4) = 7 \)

Answer 8

A dog groomer charges her clients by the weight of the dog. The following piecewise function represents the total charge, f(x)
, for a dog weighing x pounds

f(x)= 2.5x if 0≤x≤20
3.5x if 20 <x≤50
5x if x>50



If she grooms a dog that weighs 20 lbs, she will determine the amount to charge using the equation f(x)=2.5x

. What would be the point on the graph to show what she would charge for a 20 lb dog?

Responses
(20, 70)

(20, 50)

(20, 100)

Answer 9

To find the charge for a dog that weighs 20 lbs using the given piecewise function, we need to use the first piece of the function:

\[ f(x) = 2.5x \quad \text{for } 0 \leq x \leq 20 \]

Now, substituting \( x = 20 \) into the equation:

\[ f(20) = 2.5 \times 20 = 50 \]

This means the groomer would charge $50 for a 20 lb dog.

Thus, the point on the graph to represent this charge would be:

\[ (20, 50) \]

So the correct response is (20, 50).

Answer 10

On Friday, Kaitlyn went to the park to work out. She started with a slow jog at a constant rate for five minutes. She then increased her speed to a faster, constant pace for seven minutes. She took a three-minute break, then ran for another 7.5 minutes at a constant pace to finish her two-mile run. She rested for the remainder of the time. During what domain interval will she be at 1.25 miles?

Responses:

12≤s≤15

15≤s≤22.5

5≤s<12

22.5≤s≤25

Answer 11

Let's break down Kaitlyn's workout to find when she is at 1.25 miles.

1. First Segment (Slow Jog):

   • Duration: 5 minutes
   • Let's assume her speed in this segment is \( v_1 \).
   • In 5 minutes, she covers \( d_1 = v_1 \times 5 \) minutes.

2. Second Segment (Faster Pace):

   • Duration: 7 minutes
   • Let's assume her speed in this segment is \( v_2 \).
   • In 7 minutes, she covers \( d_2 = v_2 \times 7 \) minutes.

3. Break:

   • Duration: 3 minutes (no distance covered).

4. Final Segment (Running):

   • Duration: 7.5 minutes
   • Let's assume her speed in this segment is \( v_3 \).
   • In 7.5 minutes, she covers \( d_3 = v_3 \times 7.5 \) minutes.

Since we need to find the time during which she covers 1.25 miles, we need to calculate the distances covered in these segments until she reaches or exceeds 1.25 miles.

Total Routine

• Total Time: \( 5 + 7 + 3 + 7.5 = 22.5 \) minutes

• Since she runs a total of 2 miles, we can distribute this over the time she ran.

Finding when she reaches 1.25 miles:

Assuming her speeds allow her to cover the two miles as follows (this is a simplified calculation without specific speeds):

1. In the first 5 minutes, she jogs a small distance, ultimately contributing less than \( 0.625 \) miles since that would be half the distance after 10 minutes.

2. After 12 minutes (5 + 7), she may have covered possibly around \( 1.25 \) miles based on a typical pacing for joggers, depending on speeds. If she runs fast enough during the second segment, she could achieve this quickly, but since no specific velocities are provided, and we estimate her pace increases.

3. After the 3-minute break, only 7.5 minutes remain, contributing to her total running time.

Assuming continuous calculation, she hits 1.25 miles sometime before she hits 15 minutes in total and could very well reach it around 12 minutes after calculating approximately based on averages and assuming her second segment increased the distance covered significantly.

Results verify she doesn't cross 1.25 miles before 12 minutes but potentially crosses during that time:

Therefore, the likely interval where she is at 1.25 miles is:

12 ≤ s ≤ 15 (approximated run timings).

So, the correct answer is 12 ≤ s ≤ 15.