Henry is taking a medicine for a common cold. The table below shows the amount of medicine f(t), in mg, that was present in Henry's body after time t:

t (hours) 1 2 3 4 5
f(t) (mg) 282 265.08 249.18 234.22 220.17

Greg was administered 200 mg of the same medicine. The amount of medicine in his body f(t) after time t is shown by the equation below:

f(t) = 200(0.88)t

Which statement best describes the rate at which Henry's and Greg's bodies eliminated the medicine?

a. Henry's body eliminated the antibiotic faster than Greg's body.
b. Henry's body eliminated the antibiotic at the same rate as Greg's body.
c. Henry's body eliminated the antibiotic at half of the rate at which Greg's body eliminated the antibiotic.
d. Henry's body eliminated the antibiotic at one-fourth of the rate at which Greg's body eliminated the antibiotic.

To compare the rates at which Henry's and Greg's bodies eliminate the medicine, we can look at the slopes of the functions representing the amount of medicine in their bodies over time.

For Henry, the amount of medicine in his body decreases gradually over time. To find the rate of elimination, we can look at the change in the amount of medicine divided by the change in time. Taking two consecutive data points from the table:

Slope of Henry's function at t = 2 is (265.08 - 282) / (2 - 1) = -16.92
Slope of Henry's function at t = 3 is (249.18 - 265.08) / (3 - 2) = -15.9

Similarly, for Greg, the rate of elimination is given by the exponent in the equation. From the given equation, we can see that the rate is 0.88.

Comparing these rates, we can conclude that Henry's body eliminated the antibiotic at a rate that is approximately 0.88 / 15.9 = 0.055 times the rate at which Greg's body eliminated the antibiotic.

Therefore, the statement that best describes the rate at which Henry's and Greg's bodies eliminated the medicine is:

c. Henry's body eliminated the antibiotic at half of the rate at which Greg's body eliminated the antibiotic.

Which of the following is equal to the square root of the cube root of 6?

a. 6 to the power of one sixth
b. 6 to the power of one third
c. 6 to the power of two thirds
d. 6 to the power of three halves

The square root of the cube root of 6 can be expressed as (6^(1/3))^(1/2).

To simplify this, we can apply the power of a power rule, which states that when raising an exponent to another exponent, we multiply the exponents.

Therefore, (6^(1/3))^(1/2) is equal to 6^((1/3)*(1/2)) = 6^(1/6).

So, the square root of the cube root of 6 is equal to 6^(1/6).

Therefore, the correct option is:

a. 6 to the power of one sixth

The functions f(x) and g(x) in the table below show Kim's and Ben's savings respectively, in dollars, after x years. Some values are missing in the table.

x(years) 1 2 3
g(x) = 3x
Kim's savings in dollars 3 9
f(x) = 3x + 5
Ben's savings in dollars 8 11

Which statement best describes Kim's and Ben's savings in the long run?

a. The savings will increase by the same factor for both.
b. The savings will increase by the same amount for both.
c. Kim will save more than Ben.
d. Ben will save more than Kim.

To determine the savings in the long run, we can look at the behavior of the functions as x approaches infinity.

For Kim's savings, the function f(x) = 3x + 5 increases linearly as x increases. As x approaches infinity, f(x) will continue to increase indefinitely. Therefore, Kim's savings will increase without limit in the long run.

For Ben's savings, the function g(x) = 3x also increases linearly as x increases. In the long run, as x approaches infinity, g(x) will also continue to increase indefinitely. However, it increases at a slower rate compared to f(x) since there is no constant term (like the "+5" in f(x)).

Therefore, in the long run, Kim's savings will increase more than Ben's savings.

The correct statement is:

c. Kim will save more than Ben.

The table represents the amount of money in a bank account each month.

Month Balance ($)
1 3,600.00
2 3,960.00
3 4,356.00
4 4,791.60
5 5,270.76

What type of function represents the bank account as a function of time? Justify your answer.

a. This is an exponential function because there is a common difference in the balance between the months.
b. This is an exponential function because there is a common ratio in the balance between the months.
c. This is a linear function because there is a common difference in the balance between the months.
d. This is a linear function because there is a common ratio in the balance between the months.

To determine the type of function that represents the bank account as a function of time, we need to analyze the pattern of change in the balances between the months.

In this case, the balance is increasing each month, but the increments are not constant. There is a difference between the balances from one month to the next:

Second month - First month: 3,960 - 3,600 = 360
Third month - Second month: 4,356 - 3,960 = 396
Fourth month - Third month: 4,791.60 - 4,356 = 435.60
Fifth month - Fourth month: 5,270.76 - 4,791.60 = 479.16

As we can see, the differences between the balances are not constant. This suggests that the function representing the bank account as a function of time is not a linear function.

However, if we calculate the ratios between the balances from one month to the next, we can observe a pattern:

Second month / First month: 3,960 / 3,600 ≈ 1.1
Third month / Second month: 4,356 / 3,960 ≈ 1.1
Fourth month / Third month: 4,791.60 / 4,356 ≈ 1.1
Fifth month / Fourth month: 5,270.76 / 4,791.60 ≈ 1.1

The ratios between the balances are approximately equal to 1.1.

This suggests that the function representing the bank account as a function of time is an exponential function, with a common ratio of 1.1. Each month, the balance is approximately 1.1 times the previous month's balance.

Therefore, the correct answer is:

b. This is an exponential function because there is a common ratio in the balance between the months.

Peter made custard every weekend. The more times he made custard, the more practice he gained. The time he took to make the same amount of custard every week decreased with practice. The function below shows the number of hours f(x) Peter took to make custard after he had made it x times:

f(x) = 2(0.6)x

Which graph best represents the function?

a. graph of an increasing exponential function going through six tenths on the y axis
b. graph of a decreasing exponential function going through one on the y axis
c. graph of a decreasing exponential function going through two on the y axis
d. graph of an increasing exponential function going through one on the y axis

To determine the best graph representation of the function f(x) = 2(0.6)^x, we need to analyze the behavior of the function as x increases.

For this function, as x increases, the value of (0.6)^x decreases. This means that the function is decreasing, as the exponent becomes increasingly negative.

Additionally, the initial value of the function is 2, which means at x=0, the function has a value of 2.

Therefore, the best graph representation would be a graph of a decreasing exponential function that goes through the point (0, 2) on the y-axis.

The correct answer is:

c. graph of a decreasing exponential function going through two on the y-axis.

Jessa bought her home for $125,000 in 2010. Property values have increased 15% every year since she has owned the home. Which of the following equations can be used to represent the price of the home x years after 2010?

y = 125,000(1.15)x
y = 125,000(1.25)x
y = 125,000(0.85)x
y = 125,000(0.75)x

To represent the price of the home x years after 2010, we need to account for the annual increase of 15%.

Since the price is increasing, we need to multiply the initial price of $125,000 by a factor greater than 1. The factor will be 1 plus the annual increase rate.

Therefore, the correct equation to represent the price is:

y = 125,000(1.15)^x

The correct answer is:

a. y = 125,000(1.15)^x

Jack invested some money in a bank at a fixed rate of interest compounded annually. The equation below shows the value of his investment after x years:

f(x) = 300(1.02)x

What was the average rate of change of the value of Jack's investment from the third year to the fifth year?

6.43 dollars per year
8.24 dollars per year
12.86 dollars per year
14.26 dollars per year

To find the average rate of change of the value of Jack's investment from the third year to the fifth year, we need to calculate the change in value and divide it by the change in time.

The value of Jack's investment in the third year can be found by substituting x = 3 into the equation:

f(3) = 300(1.02)^3 = 300(1.061208) ≈ 318.36

Similarly, the value in the fifth year can be found by substituting x = 5:

f(5) = 300(1.02)^5 = 300(1.04008) ≈ 312.02

The change in value from the third year to the fifth year is:

318.36 - 312.02 ≈ 6.34

The change in time from the third year to the fifth year is 5 - 3 = 2.

To find the average rate of change, we divide the change in value by the change in time:

6.34 divided by 2 ≈ 3.17

Therefore, the average rate of change of the value of Jack's investment from the third year to the fifth year is approximately $3.17 per year.

The closest answer option is:

a. 6.43 dollars per year (which is not accurate)

Therefore, none of the given answer options are correct.

The following table shows the expressions that represent the sales of two companies:

Company A B
Sales 110(0.70)x 430(1.90)x

Which company has the largest percent of increase?

Company A
Company B
Both Company A and Company B
Neither Company A nor Company B

To determine which company has the largest percent of increase, we need to compare the growth rates of the two expressions.

For Company A, the expression is 110(0.70)x, while for Company B, the expression is 430(1.90)x.

To determine the growth rate, we need to look at the exponent in each expression. The higher the exponent, the greater the growth rate.

Comparing the exponents in the two expressions:
- For Company A, the exponent is x.
- For Company B, the exponent is also x.

Since the exponents are the same for both companies, the growth rates are equal.

Therefore, the correct answer is:

c. Both Company A and Company B.

Both companies have the same rate of increase, as indicated by the x exponent in their respective expressions.

Graph h(x) = 38(0.51)x. What is the constant percent rate of change of f(x) with respect to x? Does the graph show growth or decay?

49% growth
49% decay
51% growth
51% decay

To determine the constant percent rate of change of the function h(x) = 38(0.51)^x with respect to x, we can look at the base of the exponent, which is 0.51.

The constant percent rate of change is equal to 1 minus the base of the exponent, expressed as a percentage.

So, the constant percent rate of change of h(x) with respect to x is 1 - 0.51 = 0.49, or 49%.

Since the rate of change is less than 1 (0.49), the graph of h(x) represents decay, as the values of the function decrease as x increases.

Therefore, the correct answer is:

49% decay.

The number of bacteria in a sample increased by a factor of 1.02 every minute from the time shown in the graph. The function shows the number of bacteria in the sample f(x) after x minutes from 9:05 a.m:

f(x) = 10(1.02)x

exponential graph showing Time in Minutes on the x axis and Number of Bacteria on the y axis going through zero comma ten and approximately fifty comma twenty seven

Which of the following is a reasonable domain for the function from 9:05 to 9:55 am?

0 ≤ x ≤ 10
5 ≤ x ≤ 55
0 ≤ x ≤ 50
All positive integers

To determine the reasonable domain for the function from 9:05 am to 9:55 am, we need to consider the time interval shown on the graph.

At 9:05 am, the time starts at x = 0. As we can see from the graph, the number of bacteria is approximately 10.

By 9:55 am, the time has passed for 50 minutes. From the graph, the number of bacteria is approximately 50,270.

Based on this information, a reasonable domain for the function from 9:05 am to 9:55 am would be:

0 ≤ x ≤ 50

This range includes the time from the start at 9:05 am (x = 0) to 50 minutes later at 9:55 am (x = 50).

Therefore, the correct answer is:

0 ≤ x ≤ 50