Math 8 – Bid Is Non Negotiable (Read First Before Bidding!!!)
Read first before bidding. Accept Bid as is or don’t Bid at all. I’m not increasing so don’t ask and don’t waste my time
Lesson 4.
2
Introduction
Course Objectives
This lesson will address the following course outcomes:
· 2
0
. Translate problems from a variety of contexts into mathematical representation and vice versa (linear, exponential, simple quadratics).
· 2
3
. Determine the exponential function for a situation when given an initial value and either the growth/decay rate or a second function value. Interpret the initial value and growth rate of an exponential function. Include compound interest as one application.
· 25. Use functional models to make predictions and solve problems.
Specific Objectives
Students will understand
· the differences and similarities between exponential growth and decay.
· the differences between linear and exponential growth
Students will be able to
· write an equation for an exponential decay model.
In this lesson, you will connect the exponential mathematics of compounding to related applications, such as automobile depreciation and spread of disease.
Depreciation
Problem Situation
1
: Understanding Depreciation
Depreciation is a process of losing value, opposite to that of accruing interest. For example, new automobiles lose 15% to 20% of their value each year for the first few years you own them.
Suppose you purchase a a
$
2
6
,000 automobile that depreciates 15% per year.
#1 Points possible: 5. Total attempts: 5
What will the value of the car be when it is 1 year old?
$
#2 Points possible: 5. Total attempts: 5
Think back to the last lesson, and how you came up with the general formula for the balance of the 5year CD after t years.
Using the same idea, find a formula for the depreciated value (V) of this car after t years.
V =
Hint: If you get stuck, we’ll offer some hints
#3 Points possible: 5. Total attempts: 5
Using the formula you just created, predict the value of the car after 5 years. Round to the nearest cent if needed.
$
#4 Points possible: 5. Total attempts: 5
If you were to graph the value of the car over time, which of the following graphs reflects what you would expect to see?
In the last lesson, you looked at compound interest which is an example of exponential growth. In the previous problem, depreciation is an example of exponential decay. Note the similarities and differences to the growth model.
· Both have a vertical intercept that represents the starting value and is in the equation.
· The base of the exponent in the growth model is >1. The base of the exponent in the decay model is between zero and one.
· Both the compound interest formula (exponential growth) and the decay model have the same basic form, y=Caxy=Cax , with a starting value (C) that is multiplied by a factor raised to a variable power. The base of the exponential is 1+growth rate1+growth rate for growth, and 1−decay rate1decay rate for decay. Note that the general compound interest formula looks different, but the (1+rk)(1+rk) factor can be simplified down to one number.
· In growth, rate of change starts slow and increases. In decay situations, the rate of change initially represents a steep negative decline but becomes less steep (but still negative) as time goes on.
· Both are based on multiplicative (or relative) change.
Spreading Disease
Problem Situation 2: A Spreading Disease
During 2014, there was an Ebola breakout in West Africa. On April 1, 2014, there had been 130 cases reported. A month later there had been 234 cases reported.
#5 Points possible: 12. Total attempts: 5
Find the absolute and relative change between the two reported values.
Absolute change:
Relative change: As a decimal: . As a percent: %
#6 Points possible: 5. Total attempts: 5
Let C represent the number of cases after t months, so t = 0 corresponds to April 2014.
Find a formula for a linear model based on the data given, assuming the trend continues.
C =
Hint: If you get stuck, we’ll offer some hints
#7 Points possible: 5. Total attempts: 5
Let C represent the number of cases after t months, so t = 0 corresponds to April 2014.
The relative change you found two questions earlier is the percent increase, or percent growth rate, during the month. Use it to find a formula for an exponential model based on the data given, assuming the trend continues.
C =
Hint: If you get stuck, we’ll offer some hints
#8 Points possible:
10
. Total attempts: 5
What does each model predict the number of cases to be after 6 months? Round to the nearest whole number.
Linear model: cases
Exponential model: cases
#
9
Points possible: 10. Total attempts: 5
What does each model predict the number of cases to be after 2 years (24 months)? Round to the nearest whole number.
Linear model: cases
Exponential model: cases
Summary of Exponential Equations
In this lesson and the previous one you learned about exponential models. Exponential equations are used when a quantity is growing or decreasing by the same percent every time interval.
Exponential Equations
Exponential equations have the form y=P(1+r)ty=P(1+r)t
· y is the output variable
· P is the initial value, or vertical intercept (yintercept)
· r is the percent growth rate (relative growth rate) written as a decimal
· If the quantity is growing, r will be positive
· If the quantity is decreasing, then r will be negative
· t is time
A special case of the exponential equation is the compound interest equation, which looks like A=P(1+rn)ntA=P(1+rn)nt , where n is the number of compounds in a year.
To find an exponential equation from given information:
· The initial value will be given. Identify it
· Is the percent growth rate given?
· If yes, identify it
· If no, then look for a second output value, and find the relative (percent) change
Example 1: A population is currently 30,000 and is growing by 4.2% per year.
· Initial population is P=30000
· Growth rate is r= 4.2% = 0.042
· Equation will be y=30000(1+.042)t=30000(1.042)ty=30000(1+.042)t=30000(1.042)t
Example 2: A car was purchased for $12000 and its value decreases by 8% each year
· Initial value is P=12000
· Growth rate is r= 8% = 0.08. Notice it is negative because it’s decreasing in value
· Equation will be y=12000(1−0.08)t=12000(0.92)ty=12000(10.08)t=12000(0.92)t
Example 3: A biologist started with 200 bacteria in a dish, and in 1 hour the amount grew to 300. Find an exponential equation for the number of bacteria.
· Initial value is P=200
· We don’t know the growth rate, so we’ll find the relative change from 200 to 300: 300−200200=100200=0.5300200200=100200=0.5 . So the growth rate is 50% = 0.5.
· Equation will be y=200(1+0.5)t=200(1.5)ty=200(1+0.5)t=200(1.5)t , where t is in hours.
HW 4.2
#1 Points possible: 5. Total attempts: 5
Which of the following was one of the main mathematical ideas of the lesson?
· Cars lose value quickly after purchase.
· Multiplying a number by a second number between 0 and 1 will give you a larger number than you started with.
· All exponential models share certain characteristics such as the general shape of the graph, increase or decrease by a percentage, and having a vertical intercept.
· Depreciation can be modeled with an exponential equation.
#2 Points possible: 8. Total attempts: 5
One way to describe a linear model is that it is based on additive change while an exponential model is based on multiplicative change. Complete the sentences below illustrating why these terms apply to each model.
In the linear table below, each time x increases by 1, we .
In the exponential table below, each time x increases by 1, we .
y = 2x+1
x
y
0 11
3 25
3
7
4
9y = 3(2)x
x
y
0
3
1
62
12
3
24
4
48
#3 Points possible: 12. Total attempts: 5
Select the letter of each graph next to the equation that best matches it.
y = 1,000(0.95)x y = 200 + 11x y = 5(1.1)x
#4 Points possible: 15. Total attempts: 5
Certain drugs are eliminated from the bloodstream at an exponential rate. Doctors and pharmacists need to know how long it takes for a drug to reach a certain level to determine how often patients should take medications. Answer the following questions about this situation.
a. Select the correct statement:
· More of the drug will be eliminated in the first hour than in the second hour.
· Less of the drug will be eliminated in the first hour than in the second hour.
· The same amount of the drug will be eliminated in the first hour as in the second hour.
b.
Write an exponential equation for the following situation. The drug dosage is 500 mg. The drug is eliminated at a rate of 5.2% per hour. Use D = the amount of the drug in milligrams and t = time in hours.
c. How much of the drug is left after 6 hours? Round to the nearest milligram.
milligrams
#5 Points possible: 18. Total attempts: 5
In Lesson 1.2, you looked at historical data of the world’s population. Scientists use models to project population in the future. While these models have limitations, they help leaders plan ahead for the resources people will need. The World Bank estimates that that the world population growth rate in
2010
was 1.1%. The U.S. Census Bureau estimated the world population in 2010 to be about 6.9 billion people.
2
a. This is an exponential situation because the population is increasing each year by a percentage of the previous year. Complete the table below based on the growth rate of 1.1%. Some of the entries are done for you. Round projected populations to 2 decimal places.
0
1
billion
billion
Calendar Year 
Number of Years after 2010 
Projected Population 

2010 
6.90 billion 

2011 
billion 

2012 

2013 
b.
c. Write an exponential equation for the world population growth after 2010. Let P = the projected population in billions and t = the number of years after 2010. Hint: Think about how you calculated the entries in the table. If you started with 6.9 billion each time, how could you calculate each population? Think back to your work in the lesson.
d. Use your model to predict the population in 2020.
billion
#6 Points possible: 10. Total attempts: 5
The Center for Disease Control (CDC) published the following information about respiratory disease in children. Respiratory disease is an illness that affects a person’s ability to breathe and use oxygen.3
In 2005, approximately one fourth of the 2.4 million hospitalizations for children aged < 15 years were for respiratory diseases, the largest category of hospitalization diagnoses in this age group. Of these, 31% were for pneumonia, 25% for asthma, 25% for acute bronchitis and bronchiolitis, and 19% for other respiratory diseases, including croup and chronic disease of tonsils and adenoids.
a. Based on this information, how many children were hospitalized for pneumonia in 2005?
children
b. Which of the following pie charts accurately represents the data for children hospitalized for respiratory diseases?
#7 Points possible: 15. Total attempts: 5
A ball is placed at the top of ramp and released. It’s height above ground over time is shown in the table below.
0 19
6
1
Time 
Height 

10  
a. Plot the points from the table.
Height(cm)
Clear All Draw: 
Time (sec) 
b.
c. Is this a linear model?
Why or why not?
Is this an exponential model?
Why or why not?
Lesson 4.3
Introduction
Course Objectives
This lesson will address the following course outcomes:
· 24. Translate exponential statements to equivalent logarithmic statements, interpret logarithmic scales, and use logarithms to solve basic exponential equations.
Specific Objectives
Students will understand
· the relationship between a logarithm and an exponent.
Students will be able to
· use logarithms to solve an exponential equation.
In Lesson 4.1 and Lesson 4.2, you created increasing exponential models for compound interest and an exponential decay model for depreciation. In those types of situations, we might be interested in how long it will take our account to grow to a certain amount, or when the value of a car will drop below a certain level. To do that, we need to be able to solve exponential equations.
When we solve an equation like 3x=123x=12 , we use division to solve the equation, since division “undoes” multiplication: 3×3=1233×3=123 , so x=4x=4 . Likewise we use multiplication to “undo” division, addition to “undo” subtraction, and so on.
If we have an exponential equation, it would be helpful to have something to “undo” the exponential so we could solve exponential equations. The logarithm is that tool.
Common Log
Common Logarithm
The common logarithm, written log(x), undoes the exponential 10x. This means that
log(10x)
=
x, and likewise 10log(x) = x.
This also means
the statement 10a = b is equivalent to the statement log(b) = a.
For example, since 102 = 100, then log(100) = 2.
log(x) is read as “log of x”, and means “the logarithm of the value x”. It is important to note that this is not multiplication – the log doesn’t mean anything by itself, just like √ doesn’t mean anything by itself; it has to be applied to a number.
#1 Points possible: 20. Total attempts: 5
For each of the following, write the number as a power of 10, then use that to evaluate the log.
=
= log( )
=
= log( )
=
=
= log( )
=
a) log(1,000) 
= log(103) 
=  
b) log(100) 
= log( ) 

c) log(10) 

d) log(1) 

e) log(0.1) 
= log(110)log(110) = log(10−1)log(101) 

f) log(0.01) 
It is helpful to note that from the last problem that the number we’re taking the log of has to get 10 times bigger for the log to increase in value by 1.
#2 Points possible: 15. Total attempts: 5
For each of the following, write the number as a power of 10, then use that to evaluate the log.
a) log(100,000) =
b) log(11000)log(11000) =
c) log(0.01) =
Of course, most numbers cannot be written as a nice simple power of 10. For those numbers, we can evaluate the log using a scientific calculator with a log button. Here a couple videos showing how to use your calculator:
·
A graphing calculator (like a TI84)
[+]
·
A scientific calculator (like a TI30x)
[+]
#3 Points possible: 5. Total attempts: 5
Evaluate log(150) using a calculator. Give at least 3 decimal places.
With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.
Example: Solve 10x = 1000:
Taking the log of both sides gives log(10x) = log(1000)
Since the log undoes the exponential, log(10x) = x. Similarly log(1000) = log(103) = 3.
The equation simplifies then to x = 3.
#4 Points possible: 5. Total attempts: 5
Solve 3(10x)=6,0003(10x)=6,000 . Give at least 3 decimal places.
This approach allows us to solve exponential equations with powers of 10, but what about problems like 2 = 1.03n from earlier, which have a base of 1.03? For that, we need the exponent property for logs.
Exponent Property
Properties of Logs: Exponent Property
log(Ar)=rlog(A)log(Ar)=rlog(A)
#5 Points possible: 5. Total attempts: 5
Rewrite each expression using the exponent property of logs.
a. log(1.03x)=log(1.03x)=
b. log(35x)=log(35x)=
This property will allow us to solve exponential equations. An exponential equation is one where the variable is in an exponent.
Solving exponential equations with logarithms
1. Isolate the exponential. In other words, get it by itself on one side of the equation. This usually involves dividing by a number multiplying the exponential.
2.
Take the log of both sides
of the equation.
3. Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.
4. Divide as needed to solve for the variable.
Example: Solve
200(1.03x) = 8000
:
200(1.03x) = 8000 
Divide both sides by 200 
1.03x = 40 
Take the log of both sides 
log(1.03x) = log(40) 
Use the exponent property of logs on the left 
x log(1.03) = log(40) 
Divide by log(1.03) 
x=log(40)log(1.03)x=log(40)log(1.03) 
If appropriate, we can use a calculator to approximate the answer 
x ≈ 124.8 
#6 Points possible: 5. Total attempts: 5
Solve 40(3x)=6,00040(3x)=6,000 . Give at least 3 decimal places.
#7 Points possible: 5. Total attempts: 5
The value of a car is depreciating following the formula V=18,000(0.82t)V=18,000(0.82t) , where V is the value and t is years since purchase. Determine when the value of the car will fall to $12,000. Round your final answer to 2 decimal places (but make sure you keep more during the calculations).
years after purchase
#8 Points possible: 5. Total attempts: 5
The population of a town was 72 thousand in 2010, and has been growing by 8% each year. When will the population reach 160 thousand if the trend continues? Give at least 1 decimal place.
years after 2010
Hint: If you have trouble, we’ll provide some hints
HW 4.3
#1 Points possible: 5. Total attempts: 5
Which of the following was one of the main mathematical ideas of the lesson?
· Logarithms are used in carbon dating to determine how old artifacts are
· Logarithms are used to undo exponents
· Logarithmic scales are better than linear scales
· Logarithms can only solve exponential equations where the base of the exponential is 10
#2 Points possible: 5. Total attempts: 5
Evaluate using your calculator, rounding to 3 decimal places:
a. log(71) =
b. log(6814) =
c. log(0.00315) =
d. log(0.72) =
#3 Points possible: 5. Total attempts: 5
Evaluate each of the following, to at least 3 decimal places.
a. log(1,000,000) =
b. log(11000)log(11000) =
c. log(70) =
#4 Points possible: 5. Total attempts: 5
Evaluate each of the following, rounding to 3 decimal places if needed:
a. 10001001000100 =
b. log(1000)log(100)log(1000)log(100) =
Notice the results are different. This question is to help you notice that log is not something that is multiplying the numbers – it is an operation being applied to the number. This means you cannot “cancel” the logs.
#5 Points possible: 5. Total attempts: 5
Solve, to 3 decimal places, the equation 2(10x)=4802(10x)=480
x =
#6 Points possible: 5. Total attempts: 5
Solve for k: 200(1.05)k=16,790200(1.05)k=16,790
k=k=
Give an exact answer, or a decimal approximation accurate to at least 3 decimal places.
#7 Points possible: 5. Total attempts: 5
Solve for x. Round your answer to the nearest hundredth.
54x=8954x=89
#8 Points possible: 5. Total attempts: 5
The value of a car is depreciating according to the model V=11000(0.82)tV=11000(0.82)t , where V is the value of the car t years after purchase.
Solve, using logarithms, for when the value of the car will reach $4,000. Give your answer to the nearest tenth.
years after purchase
Lesson 4.4
Introduction
Course Objectives
This lesson will address the following course outcomes:
· 2
0
. Translate problems from a variety of contexts into mathematical representation and vice versa (linear, exponential, simple quadratics).
· 24. Translate exponential statements to equivalent logarithmic statements, interpret logarithmic scales, and use logarithms to solve basic exponential equations.
· 25. Use functional models to make predictions and solve problems.
Specific Objectives
Students will understand
· That log scales reveal magnitudes and relative change
Students will be able to
· Solve applied problems using logarithms
· Read and place values on a logarithmic scale
In the last lesson, you learned how to solve exponential equations. In this lesson we will explore some applications of that skill. We will also look at how logarithms are useful for representing numbers that vary greatly in size.
Carbon Dating
Problem Situation 1: Carbon Dating
When ancient artifacts are found, scientists determine their age using a technique called radioactive dating. Carbon14 is a radioactive isotope that is in living organisms. All living organisms have approximately the same ratio of carbon14 to carbon12. After the organism dies, the carbon14 decays to nitrogen14, changing the ratio of carbon14 to carbon12. The fraction of the original carbon14 remaining in an organism after t thousand years is modeled by A=0.8861tA=0.8861t .
#1 Points possible: 12. Total attempts: 5
Suppose an object, such as a tree, has 1 unit of carbon14 in it at the time it dies. Complete the table for the fraction of carbon14 remaining after the given numbers of years since death. Give your values to 3 decimal places.
Years since death
Carbon14 Remaining
0
1.000
2000
4000
6000
8000
Hint: Be careful with how t is defined in the equation. t represents thousands of years, so t = 1 is one thousand years.
#2 Points possible: 5. Total attempts: 5
Textbooks will commonly report that the “halflife” of carbon14 is
5730
years. Based on your calculations, what do you think “halflife” means?
· After 5730 years, half the carbon14’s life will have passed, after which it will grow back to its original amount
· Carbon14 has a total life of
11460
years, after which it will be gone
· The amount of carbon14 remaining decreases by half in 5730 years, then after another 5730 years there will be half of that remaining, or 25% of the original
· The amount of carbon14 remaining decreases by half in 5730 years, then after another 5730 years there will be none remaining
#3 Points possible: 12. Total attempts: 5
Just using the idea of halflife, complete the table.
Years since deathCarbon14 Remaining
01.000
5730 
11460 
17190 
22920 
#4 Points possible: 5. Total attempts: 5
An artifact is found, and scientists determine that it contains 20% of its original carbon14.
Estimate the age of the artifact, using the results from the previous problem.
years old
#5 Points possible: 5. Total attempts: 5
An artifact is found, and scientists determine that it contains 20% of its original carbon 14.
Determine the approximate age of the artifact using algebra and the equation A=0.8861tA=0.8861t given at the start of the problem situation. Give your answer as a number of years, to the nearest year.
years old
Log Scales
Problem Situation 2: Log Scales
When numbers are very small or very large, or a set of values varies greatly in size, it can be hard to represent those values. Consider the distance of the planets of our solar system from the sun:
Planet 
Distance (millions of km) 
Mercury 
58 
Venus 
108 
Earth 
150 
Mars 
228 
Jupiter 
779 
Saturn 
1430 
Uranus 
2880 
Neptune 
4500 
Placed on a linear scale – one with equally spaced values – these values get bunched up.
However, representing each value as a power of 10, and using a scale spaced by powers of 10, the values are more reasonably spaced.
#6 Points possible: 5. Total attempts: 5
The scale above shows the dwarf planet Pluto. Use the scale to estimate the distance in kilometers from the sun to Pluto. Round to the nearest million km.
million km
#7 Points possible: 5. Total attempts: 5
Ida is a large asteroid in asteroid belt, approximately 429 million kilometers from the sun. Plot on the logarithmic scale below where Ida belongs.
101
.5
102
102.5
103
103.5
104
Clear All Draw: Dot
Vertical Scale
Sometimes you will see a logarithmic scale on the vertical axis of a graph. A common example is in stock charts. Both stock charts below show the Dow Jones Industrial Average, from 1928 to 2010.
Both charts have a linear horizontal scale, but the first graph has a linear vertical scale, while the second has a logarithmic vertical scale.
In 1929, the stock market value dropped from 380 to 42. In 2008, the stock market value dropped from 14,
100
to 7,500.
#8 Points possible: 16. Total attempts: 5
Compute the absolute and relative change in each market drop. Give your answers to 1 decimal place. Since we’re looking for the size of the drop, give your answers as positive values.
Relative Change: %
1929: 
Absolute change:

Relative Change: % 

2008: 
Absolute change: 
#9 Points possible: 10. Total attempts: 5
Which graph better reveals the drop with the largest absolute change?
· The first graph, with linear scale
· The second graph, with log scale
Which graph better reveals the drop with the largest relative change?
· The first graph, with linear scale
· The second graph, with log scale
HW 4.4
#1 Points possible: 5. Total attempts: 5
Polluted water is passed through a series of filters. Each filter removes 90% of the remaining impurities from the water. If you have 10 million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to below 500 particles per gallon? You can only use a whole number of filters.
filters
#2 Points possible: 5. Total attempts: 5
India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If the population continues following this trend, during what year will the population reach 2 billion?
#3 Points possible: 5. Total attempts: 5
If $1000 is invested in an account earning 3% compounded monthly, how long will it take the account to grow in value to $1500? Round to the nearest month.
years, months
#4 Points possible: 5. Total attempts: 5
In Lesson 4.2, you found a model for a spread of a disease.
Suppose that another outbreak has been detected. On Sept 1, there were 60 reported cases. A week later, there were 75 reported cases.
If the disease continues spreading exponentially, how many weeks after Sept 1 will 850 cases be expected? Round to the nearest week.
weeks
#5 Points possible: 5. Total attempts: 5
In chemistry, the acidity of a liquid is commonly measured in pH. pH is important outside of chemistry to many things, including keeping swimming pools maintained, and measuring the acidity of rain. The pH scale is related to the concentration of hydrogen ions in the liquid, a number that tends to be very small, so the pH scale was introduced to make the numbers easier to work with. For a given concentration of hydrogen ions, H, measured in Molars, the pH is determined by pH = log(H)
Purified tap water has a hydrogen ion concentration of 107 M. Find the pH of water.
#6 Points possible: 5. Total attempts: 5
The number line below shows a value on a log scale.
105
104
103
102
101
100
101
102
103
104
105
What is, approximately, the value of the point shown on the number line? x =
#7 Points possible: 5. Total attempts: 5
Plot the number 0.0001 on the log scale below.
105
104
103
102
101
100
101
102
103
104
105
Clear All Draw: Dot
#8 Points possible: 5. Total attempts: 5
Plot the number 0.0032 on the log scale below.
105
104
103
102
101
100
101
102
103
104
105
Clear All Draw: Dot
Lesson 4.5
Introduction
Course Objectives
This lesson will address the following course outcomes:
·
20
. Translate problems from a variety of contexts into mathematical representation and vice versa (linear, exponential, simple quadratics).
· 21. Describe the behavior of common types of functions using words, algebraic symbols, graphs, and tables. Include descriptions of the dependent and independent variables.
Specific Objectives
Students will understand that
· a variable is a symbol that is used in algebra to represent a quantity that can change.
· many variables can be present in a scenario or experiment, but some can be held fixed in order to analyze the effect that the change in one variable has on another.
· verbal, numerical, algebraic, and graphical representations provide different yet consistent information on quantifiable situations.
Students will be able to
· evaluate an expression.
· informally describe the change in one variable as another variable changes. verbally describe useful conclusions after exploration of a quantifiable situation.
· create a table of values from a given formula.
· create a graph using the table of values.
· compare the features of a quadratic graph and a linear graph.
You have now explored linear and exponential models in some depth. In this lesson, we will explore another type of model.
Intuition
Problem Situation: Calculating the Braking Distance of a Car
Experts agree that driving defensively saves lives. Knowing how far it takes your vehicle to come to a complete stop is one aspect of safe driving. For example, when you are going only 45 miles per hour (mph), you are traveling about 66 feet every second. This means that to be a safe driver, you need to drive “in front of you” (i.e., you need to know what is going on ahead of you so that you can react accordingly). In this lesson, you will learn more about what it takes to drive defensively by examining the braking distance of a vehicle. Braking distance is the distance a car travels in the time between when the brake is applied and when it comes to a full stop. For this problem, we will be looking at minimum braking distance when the brakes are fully applied and the car is skidding.
#1 Points possible: 5. Total attempts: 5
What are some variables that might affect the braking distance of a car?
Come up with your own list, and after submitting it, compare it to ours.
#2 Points possible: 5. Total attempts: 5
For this lesson, you will focus on examining how speed affects braking distance.
Make a prediction on what will happen to braking distance if you were to double the speed of the car before it applies the brakes.
· The braking distance would be shorter
· The braking distance would be the same
· The braking distance would be longer, but less than twice as long
· The braking distance would be twice as long
· The braking distance would be more than twice as long
Formula
The formula for the braking distance of a car is d
=
V
22
g(f
+
G)d=V22g(f+G)
V = initial velocity of the car (feet per second). That is, the velocity of the car when the brakes were applied, when t = 0.
d = braking distance (feet)
G = roadway grade (percent written in decimal form). Note: There are no units for this variable.
f = coefficient of friction between the tires and the roadway (0 < f < 1). Note: Good tires on good pavement provide a coefficient of friction of about 0.
8
to 0.85.
g = acceleration due to gravity (32.2 ft/sec2). This is a constant.
Since g is a constant, this formula has four variables. To understand the relationships between the variables, you will hold two of them fixed. That leaves you with two variables—one that will affect the other. Since you want to see how speed affects braking distance, you will hold the other two variables, f and G, fixed.
#3 Points possible: 5. Total attempts: 5
Let f = 0.8 and G = 0.05. Write a simplified form of the formula for the braking distance d using these values for the two variables. Your formula should involve the remaining variable V.
d =
Do not round the value in the denominator. If you have trouble, you will get a hint after two attempts.
While the original equation involved four variables, this simplified formula just involves two, allowing us to compare how changing the initial velocity changes braking distance.
#4 Points possible:
16
. Total attempts: 5
For each of the velocities in the table, given in miles per hour, first convert them to feet per second. Then, use your simplified formula from above to determine the braking distance. Give answers to two decimal places.
Velocity 
Velocity 
Braking Distance 
10 

20  
40 

80 
#5 Points possible: 12. Total attempts: 5
Suppose the speed doubles from 10mph to 20mph. Complete the two sentences below, giving answers to one decimal place.
The braking distance at 20mph is ft longer than the braking distance at 10mph
The braking distance at 20mph is times as long as the braking distance at 10mph
Suppose the speed doubles from 40mph to 80mph. Complete the two sentences below, giving answers to one decimal place.
The braking distance at 80mph is ft longer than the braking distance at 40mph
The braking distance at 80mph is times as long as the braking distance at 40mph
#6 Points possible: 5. Total attempts: 5
Now we can return to a question we asked you to make a prediction about earlier.
Based on the pattern from your calculations, what will happen to braking distance if you were to double the speed of the car before it applies the brakes?
· The braking distance would be shorter
· The braking distance would be the same
· The braking distance would be twice as long
· The braking distance would be three times as long
· The braking distance would be four times as long
· The braking distance would be five times as long
#7 Points possible: 5. Total attempts: 5
Now try to extend that idea to answer to complete the sentence below.
If the speed were to triple, the braking distance would be times as long
#8 Points possible: 5. Total attempts: 5
Plot the data from your table of values you calculated earlier, with velocity in miles per hour on the horizontal axis, and braking distance on the vertical axis.
Clear All Draw:
#9 Points possible: 5. Total attempts: 5
What best describes the shape of the graph of the data?
· Linear
– line shaped
· Curving upwards
#10 Points possible: 5. Total attempts: 5
Look at your simplified formula from earlier. What family of equations does this formula belong to?
· Linear
· Exponential
· Quadratic
· None of these
HW 4.5
#1 Points possible: 5. Total attempts: 5
Which of the following was one of the main mathematical ideas of the lesson?
· Braking distance is affected by many factors.
· When using variables, it is only important to know what numbers to substitute in for them.
· When using variables, it is important to know what they represent and what units should be used with them.
· A subscript is a label on a variable.
#2 Points possible: 12. Total attempts: 5
In the lesson, you investigated the relationship between velocity and braking distance. You will now investigate the relationship between the coefficient of friction and the braking distance.
Recall that the formula for the braking distance of a car is d=V22g(f+G)d=V22g(f+G)
a. Which of the variables in the formula represents a constant?
· f
· d
· V
· G
· g
b. To investigate the relationship between the coefficient of friction and the braking distance, you need to hold the other variables fixed. Let G = 0.02. Which of the following is a correct interpretation of the value G = 0.02?
· The grade of a road is 2%, which is a vertical increase of 2 feet for every 1 foot of horizontal increase.
· The grade of a road is 0.02%, which is a vertical increase of 0.02 feet for every 1 foot of horizontal increase.
· The grade of a road is 2%, which is a vertical increase of 2 feet for every 100 feet of horizontal increase.
· The grade of a road is 0.02%, which is a vertical increase of 0.02 feet for every 100 feet of horizontal increase.
c. Let V = 72 mph and use the value of G in Part (b). Which of the following expressions represents the simplified form of the formula using these values?
· d=
11
,
15
1.3664.4f+1.288 ftd=11,151.3664.4f+1.288 ft
· d=11,151.3664.4f+8,657.89 ftd=11,151.3664.4f+8,657.89 ft
· d=11,151.3664.4f+0.02 ftd=11,151.3664.4f+0.02 ft
#3 Points possible: 18. Total attempts: 5
Use the formula you found in part c of the previous question.
a. Create a table of values for f and d (in feet). Use the values of f given in the table. Perform one of the calculations on paper showing the units. You may then use technology to complete the table. Round answers to the nearest tenth.
f 
d (feet) 
0.30 

0.50 

0.70 

0.90 
b.
c. The four values of f correspond to the coefficient of friction for four road conditions: an icy road, a very good road with great tires, an asphalt road with fair tires, and a wet road with fair tires. Match the coefficients of friction to the appropriate conditions by looking at the braking distance required.
i. Icy road, f =
ii. Very good road with great tires, f =
iii. Asphalt road with fair tires, f =
iv. Wet road with fair tires, f =
d. In the table, the coefficient of friction, f, is increasing at a constant rate, since each value is 0.2 more than the previous value. How is d changing as f increases at a constant rate?
. The stopping distance is decreasing
. The stopping distance is constant
. The stopping distance is increasing
#4 Points possible: 5. Total attempts: 5
In statistics, the formula E=1.96⋅√ˆp(1−ˆp)nE=1.96⋅p^(1p^)n is used to calculate the margin of error E (at a 95% confidence level) for a survey of n people where ˆpp^ is the proportion of the people who responded a particular way.
In a particular survey of
13
00 people, the proportion who favored stiffer penalties for drunk driving was 71%, so ˆp=0.71p^=0.71 . Determine the margin of error, E. Report the answer rounded to 3 decimal places.
Margin of error:
#5 Points possible: 20. Total attempts: 5
Newton’s law of gravitational force (measured in Newtons) between two objects r meters apart with masses m1 kilograms and m2 kilograms is given by the formula F=Gm1m2r2F=Gm1m2r2 , where G is a constant approximately equal to 6.674
×
10−116.674×1011 . The earth and the moon are about 384,000 kilometers apart. The mass of the moon is about 73,480,000,000,000,000,000,000 kg, and the mass of the earth is about 5,972,200,000,000,000,000,000,000 kg.
a. The size of the numbers in this question make them hard to work with. Rewrite them using scientific notation. Don’t forget to check the units, and make any necessary conversions.
The distance between the earth and the moon:
meters
Mass of the moon:
kilograms
Mass of the earth:
kilograms
b. Calculate the gravitational force. Give your answer in scientific notation. Note: Since the number 384,000 only has three significant digits (numbers before trailing zeros), it is appropriate to round your final answer so that it also has three significant digits (2 decimal places, in scientific notation).
Newtons
#6 Points possible: 20. Total attempts: 5
The tuition at a daycare center is based on family income. A reduced tuition has a subsidy. There are three levels of tuition:
· Full subsidy – the family does not pay any tuition
· Partial subsidy – the family pays part of the tuition
· No subsidy – the family pays the full tuition
The data for the daycare center, showing how many students there are for each age level and tuition level, is given below. Answer the questions below. Round to the nearest whole percent.
15
Total
Tuition Level 

Full Subsidy 
Partial Subsidy 
No Subsidy 
Total 

Age 
3 yearolds 
17 
13  8  
4 yearolds 
22 
14 
15  
5 yearolds 
16  11  
a. Complete the last column and last row.
b. What percentage of 3 yearolds received a full or partial subsidy? %
c. What percentage of those who receive no subsidy are 5 years old? %
d. What percentage of the students are 3 years old? %
e. The daycare center’s funding for one term comes from federal funding for the subsidy and the tuition paid by families based on the formula below. Find the funding for the center.
· Funding = 1,530F + 1,750P + 1,875N where
· F = number of children receiving a full subsidy
· P = number of children receiving a partial subsidy
· N = number of children receiving no subsidy
Funding for the center = $
#7 Points possible: 10. Total attempts: 5
In Lesson 2.1, you used a formula that was written as steps in a form to calculate taxes for different people. Formulas are often written in this way. One example is the Expected Family Contribution (EFC) Formula, which is used to determine if a college student is eligible for financial aid. The EFC has many different sections that each use different calculations. One section of the 201011 form is shown below.
+
Student’s Contribution from Assets 

Cash, savings, and checking 

Net worth of investments 
+  
Net worth of business and/or investment farm 

Net worth (sum of lines 45 through 47) 

Assessment rate 
× 
0.20 
Students Contribution from Assets 
= 
a. Calculate the Student’s Contribution from Assets given the following information, to the nearest dollar.
Cash: $500
Savings: $1,240
Investments: $0
Business: $0
Checking: $732
Student’s Contribution from Assets: $
b. Write a formula (equation) that summarizes the calculation in this form using the following variables:
C = Cash including savings and checking
Ni = Net worth of investment
Nb = Net worth of business or farm
S = Student’s contribution from assets
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