20.5: Alternating Current versus Direct Current
Learning Objectives
By the end of this section, you will be able to:
- Explain the differences and similarities between AC and DC current.
- Calculate rms voltage, current, and average power.
- Explain why AC current is used for power transmission.
Alternating Current
Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure \(\PageIndex{1}\) shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world.
Figure \(\PageIndex{2}\) shows a schematic of a simple circuit with an AC voltage source. The voltage between the terminals fluctuates as shown, with the AC voltage given by \[V = V_{0} sin 2 \pi ft, \label{20.6.1}\] where \(V\) is the voltage at time \(t\), \(V_{0}\), \(V_{0}\) is the peak voltage, and \(f\) is the frequency in hertz. For this simple resistance circuit, \(I = V/R\), and so the AC current is
\[I = I_{0} sin 2 \pi ft, \label{20.6.2}\]
where \(I\) is the current at time \(t\), and \(I_{0} = V_{0} / R\) is the peak current. For this example, the voltage and current are said to be in phase, as seen in Figure \(\PageIndex{1b}\).
Current in the resistor alternates back and forth just like the driving voltage, since \(I = V / R\). If the resistor is a fluorescent light bulb, for example, it brightens and dims 120 times per second as the current repeatedly goes through zero. A 120-Hz flicker is too rapid for your eyes to detect, but if you wave your hand back and forth between your face and a fluorescent light, you will see a stroboscopic effect evidencing AC. The fact that the light output fluctuates means that the power is fluctuating. The power supplied is \(P = IV\). Using the expressions for \(I\) and \(V\) above, we see that the time dependence of power is \(P = I_{0}V_{0}sin^{2}2\pi ft\), as shown in Figure \(\PageIndex{3}\).
Making Connections: Take-Home Experiment—AC/DC Lights
Wave your hand back and forth between your face and a fluorescent light bulb. Do you observe the same thing with the headlights on your car? Explain what you observe. Warning: Do not look directly at very bright light.
We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average power consumption of 60 W, for example. As illustrated in Figure 3, the average power \(P_{ave}\) is \[P_{ave} = \frac{1}{2}I_{0}V_{0}. \label{20.6.3}\] This is evident from the graph, since the areas above and below the \(\left(1/2\right)I_{0}V_{0}\) line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current \(I_{rms}\) and average or rms voltage \(V_{rms}\) to be, respectively,
\[I_{rms} = \frac{I_{0}}{\sqrt{2}}\label{20.6.4}\]
and
\[V_{rms} = \frac{V_{0}}{\sqrt{2}}.\label{20.6.5}\]
where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, its mean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now, \[P_{ave} = I_{rms}V_{rms}, \label{20.6.6}\] which gives
\[P_{ave} = \frac{I_{0}}{\sqrt{2}} \cdot \frac{V_{0}}{\sqrt{2}} = \frac{1}{2}I_{0}V_{0}, \label{20.6.7}\]
as stated above. It is standard practice to quote \(I_{rms}\), \(V_{rms}\), and \(P_{ave}\) rather than the peak values. For example, most household electricity is 120 V AC, which means that \(V_{rms}\) is 120 V. The common 10-A circuit breaker will interrupt a sustained \(I_{rms}\) greater than 10 A. Your 1.0-kW microwave oven consumes \(P_{ave} = 1.0 kW\), and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive circuit.
To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average values are used for AC. Thus, for AC, Ohm’s law is written
\[I_{rms} = \frac{V_{rms}}{R}.\label{20.6.8}\]
The various expressions for AC power \(P_{ave}\) are
\[P_{ave} = I_{rms}V_{rms},\label{20.6.9}\]
\[P_{ave} = \frac{V^{2}_{rms}}{R},\label{20.6.10}\]
and
\[P_{ave} = I^{2}_{rms}R.\label{20.6.11}\]
Example \(\PageIndex{1}\): Peak Voltage and Power for AC
(a) What is the value of the peak voltage for 120-V AC power?
Strategy
We are told that \(V_{rms}\) is 120V and \(P_{ave}\) is 60.0 W. We can use \(V_{rms} = \frac{V_{0}}{\sqrt{2}}\) to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.
Solution
Solving the equation \(V_{rms}=\frac{V_{0}}{\sqrt{2}}\) for the peak voltage \(V_{0}\) and substituting the known value for \(V_{rms}\) gives \[V_{0} = \sqrt{2} V_{rms} = 1.414\left(120 V\right) = 170 V.\]
Discussion
This means that the AC voltage swings from 170 V to \(-170 V\) and back 60 times every second. An equivalent DC voltage is a constant 120 V.
(b) What is the peak power consumption rate of a 60.0-W AC light bulb?
Solution
Peak power is peak current times peak voltage. Thus, \[P_{0} = I_{0}V_{0} = 2\left(\frac{1}{2} I_{0} V_{0} \right) = 2P_{ave}.\] We know the average power is 60.0 W, and so \[P_{0} = 2\left(60.0 W\right) = 120 W.\]
Discussion
So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.
Why Use AC for Power Distribution?
Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than to build numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized. High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See Figure 4.) For safety reasons, the voltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC is used in most large power distribution systems.
Example \(\PageIndex{2}\): Power losses are less for high-voltage transmission
(a) What current is needed to transmit 100 MW of power at 200 kV?
Strategy
We are given \(P_{ave} = 100 MW\), \(V_{rms} = 200 kV\), and the resistance of the lines is \(R = 1.00 \Omega\). Using these givens, we can find the current flowing (from \(P = IV\) ) and then the power dissipated in the lines (\(P = I^{2}R\)), and we take the ratio to the total power transmitted.
Solution
To find the current, we rearrange the relationship \(P_{ave} = I_{rms}V_{rms}\) and substitute known values. This gives \[I_{rms} = \frac{P_{ave}}{V_{rms}} = \frac{100 \times 10^{6} W}{200 \times 0^{3} V} = 500A .\]
(b) What is the power dissipated by the transmission lines if they have a resistance of \(1.00 \Omega\)?
Solution
Knowing the current and given the resistance of the lines, the power dissipated in them is found from \(P_{ave} = I^{2}_{rms}R.\). Substituting the known value gives \[P_{ave} = I^{2}_{rms}R = \left(500 A\right)^{2} \left(1.00 \Omega\right) = 250 kW.\]
(c) What percentage of the power is lost in the transmission lines?
Solution
The percent loss is the ratio of this lost power to the total or input power, multiplied by 100: \[% loss = \frac{250kW}{100 MW} \times 100 = 0.250 %.\]
Discussion
One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.
It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.
PHET EXPLORATIONS: GENERATOR
Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.
Figure \(\PageIndex{5}\): Generator
Summary
- Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant.
- The voltage source of an alternating current (AC) system puts out \(V= V_{0} \sin{2\pi} ft\), where \(V\) is the voltage at time \(t\), \(V_{0}\) is the peak voltage, and \(f\) is the frequency in hertz.
- In a simple circuit, \(I = V/R\) and AC current is \(I = I_{0} \sin{2\pi}ft\), where \(I\) is the current at time \(t\), and \(I_{0} = V_{0}/R\) is the peak current.
- The average AC power is \(P_{ave} = \frac{1}{2} I_{0}V_{0}\).
- Average (rms) current \(I_{rms}\) and average (rms) voltage \(V_{rms}\) and \(I_{rms} = \frac{I_{0}}{\sqrt{2}}\) and \(V_{rms} = \frac{V_{0}}{\sqrt{2}}\), where rms stands for root mean square.
- Thus, \(P_{ave} = I_{rms}V_{rms}\).
- Ohm's law for AC is \(I_{rms} = \frac{V_{rms}}{R}\).
- Expressions for the average power of an AC circuit are \(P_{ave} = I_{rms}V_{rms}\), \(P_{ave} = \frac{V_{rms}^{2}}{R}\), and \(P_{ave} = I_{rms}^{2}R\), analogous to the expressions for DC circuits.
Glossary
- direct current
- (DC) the flow of electric charge in only one direction
- alternating current
- (AC) the flow of electric charge that periodically reverses direction
- AC voltage
- voltage that fluctuates sinusoidally with time, expressed as V = V 0 sin 2 πft , where V is the voltage at time t, V 0 is the peak voltage, and f is the frequency in hertz
- AC current
- current that fluctuates sinusoidally with time, expressed as I = I 0 sin 2 πft , where I is the current at time t, I 0 is the peak current, and f is the frequency in hertz
- rms current
- the root mean square of the current, \(I_{rms}=I_0/\sqrt{2}\), where I 0 is the peak current, in an AC system
- rms voltage
- the root mean square of the voltage, \(V_{rms}=V_0/\sqrt{2}\), where V 0 is the peak voltage, in an AC system


6.13: Stakeholders
- Compare and contrast different type of stakeholders in business reports
When writing any professional document, it is important to identify the potential stakeholders. A stakeholder is anyone who will be affected by the contents of what you write. How you choose to word your document—or even the choice to write the document—can become an ethical matter for stakeholders. It is crucial to consider your main objectives before writing. For example, a report on consumer focus groups that were generally negative toward the company’s new ad campaign must take into consideration issues such as how upper management will receive the news and what information will be most helpful to the team working on the campaign moving forward.
According to Paul Anderson, the author of Technical Communication: A Reader-Centered Approach , there are three types of stakeholders [1] :
- Direct
- Indirect
- Remote
The direct stakeholders are those initially impacted by what you write. For instance, if you are writing about opening a new waste disposal site, the stakeholders clearly include the company you are writing the report for. However, disposal companies that might use this waste site in the future are also considered direct stakeholders because they will be in the same situation as the current company. Their future business will be impacted based on whether your proposal is accepted or declined.
The indirect stakeholders are those that are not impacted until a later time. Using the previous example of the waste disposal site, citizens in the area would be indirect stakeholders. The stakeholders don’t necessarily need to be people: the nearby ecosystems would be indirect stakeholders of this same proposal. If toxic waste was dumped there, it would harm the animals, rivers, and plant life nearby.
Finally, the remote stakeholders are not affected until far into the future. One example, following our hypothetical waste disposal site, is future generations. While it may seem far-fetched, historically, there have been instances where toxic or poisonous materials have been disposed of incorrectly and the run-off that went into lakes and streams caused birth defects. While this is remote, it must be considered when writing a document. This category of remote stakeholders brings to the forefront one major difference between reports and other types of business communication: reports can have long lifespans and be revisited far into the future.
- Anderson, Paul V. T echnical Communication: A Reader-Centered Approach . Mason, OH: Cengage Learning, 2007.
Contributors and Attributions
- Stakeholders. Authored by : Susan Kendall. Provided by : Lumen Learning. License : CC BY-SA: Attribution-ShareAlike
- Ethics and Technical. Provided by : WikiBooks. Located at : en.wikibooks.org/wiki/Professional_and_Technical_Writing/Ethics#Legal_Issues_and_Communication. Project : Professional and Technical Writing. License : CC BY-SA: Attribution-ShareAlike


20.S: Carboxylic Acids and Nitriles (Summary)
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Concepts & Vocabulary
20.0 Chapter Objectives and Introduction to Carboxylic Acids
- Carboxylic acids are defined as molecules that have a hydroxyl group bonded to a carbonyl group (RCOOH).
- Fatty acids are naturally occurring carboxylic acids that are important components of lipids.
- Fatty acids can either be saturated (all carbon-carbon bonds are single bonds) and unsaturated (one or more carbon-carbon double bonds).
- Several other functional groups can be prepared from carboxylic acids and are considered derivatives (acyl halides, anhydrides, esters, amides and nitriles).
20.1 Naming Carboxylic Acids and Nitriles
- Carboxylic acids are named following IUPAC rules with ending -oic acid.
- When carboxylic acids are named as an attachment to a ring, add -carboxylic acid to the name of the cyclic compound.
- Nitriles are named following IUPAC rules with the ending -nitrile or -carbonitrile.
20.2 Structure and Properties of Carboxylic Acids
- Carboxylic acids have trigonal planar shape around the carbonyl carbon and due to resonance structures, all the atoms of the functional group are co-planar.
- Conjugation in carboxylic acids increases acidity of the hydroxyl hydrogen compared to alcohols. The conjugate base (carboxylate) is stabilized by resonance.
- Carboxylic acids form hydrogen bonded dimers yielding higher boiling points than for other molecules of similar size.
- Small carboxylic acids are often soluble in water due to hydrogen bonding. Larger carboxylic acids are often slightly soluble in water, though when converted to a carboxylate salt by treating with base, become water soluble.
20.3 Biological Acids and the Henderson-Hasselbalch Equation
- Carboxylic acids are fully protonated in acidic solutions.
- Carboxylic acids are fully dissociated in basic solutions.
- The Henderson-Hasselbalch equation can be used to relate pH, pKa and ratio of concentrations of acid and conjugate base.
20.4 Substituent Effects on Acidity
- Withdrawing groups (through inductive or resonance effects) can stabilize carboxylates, increasing acidity of carboxylic acids.
- Donating groups (through inductive or resonance effects) destabilize carboxylates, decreasing acidity of carboxylic acids.
- Nearly all substituents in ortho positions increase acid strength whether donating or withdrawing, this is called the ortho -effect.
20.5 Preparing Carboxylic Acids
- Carboxylic acids can be prepared through oxidation of several functional groups including: 1 o alcohols, aldehydes, alkyl arene side chains, and alkenes.
- Carboxylic acids can also be prepared by carboxylation of Grignard reagents and hydrolysis of nitriles.
20.6 Reactions of Carboxylic Acids: An Overview
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There are four main categories of carboxylic acid reactions:
- Acid/base reaction (salt formation)
- Nucleophilic acyl substitution
- Reduction
- Substitution alpha to the carbonyl
- Nitriles have linear geometry due to the sp hybrid orbitals on carbon and nitrogen.
- Nitriles are less basic than amines.
- Nitriles can be prepared from aldehydes, ketones, or amides.
- Nitriles react similarly to carbonyl compounds where nucleophiles will attack the carbon end of the nitrile.
- Nitriles can be hydrolyzed to carboxylic acids under basic or acidic conditions.
- Nitriles can be reduced to amines by hydride reduction.
- Nitriles can be converted to aldehyde by reaction with DIBALH.
- Nitriles can be converted to ketones by reaction with organometallic compounds.
20.8 Spectroscopy of Carboxylic Acids and Nitriles
- IR of carboxylic acids typically show a very strong and broad OH stretch from about 2500 to 3300 cm -1 as well as a strong carbonyl stretch around 1710 cm -1 .
- 1 H NMR of carboxylic acids show the OH proton between 10-12 ppm as well as hydrogens on carbon adjacent to the carbonyl around 2-3 ppm.
- 13 C NMR of carboxylic acids show the carbonyl carbon at 160-180 ppm.
- Mass spectra of carboxylic acids often have fragments for OH (loss of 17) and COOH (loss of 45).
- The most identifying characteristic of IR of nitriles is the CN triple bond stretch which appears near 2250 cm -1 .
- 1 H NMR of nitriles show protons on the carbon next to the nitrile at around 2-3 ppm.
- 13 C NMR of nitriles show the nitrile carbon in the 115-120 ppm range.
- Mass spectra of nitriles will often not have a visible molecular ion, but instead exhibit a strong M-1 peak.
Skills to Master
- Skill 20.1 Name carboxylic acids and nitriles using IUPAC rules.
- Skill 20.2 Draw the structure of carboxylic acids and nitriles from the IUPAC name.
- Skill 20.3 Describe the geometries and approximate bond angles of carboxylic acids and nitriles.
- Skill 20.4 Explain the acidity of carboxylic acids based on conjugate base stabilization.
- Skill 20.5 Explain physical properties of carboxylic acids.
- Skill 20.6 Describe the structure of carboxylic acids in buffered solutions.
- Skill 20.7 Use the Henderson-Hasselbach equation to relate pH, pK a , and acid/conjugate base concentrations.
- Skill 20.8 Explain inductive effects on acidity.
- Skill 20.9 Explain relative acidity of aromatic acids.
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Skill 20.10 Draw mechanisms for preparing carboxylic acids including:
- Oxidations
- Carboxylation of Grignard reagents
- Hydrolysis of Nitriles
- Skill 20.11 List the four categories of carboxylic acid reactions.
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Skill 20.12 Draw reactions for preparing nitriles including formation from:
- aldehyde or ketone
- 1 o amide
- Skill 20.13 List general reactions of nitriles.
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Skill 20.14 Draw mechanisms for reactions of nitriles including:
- Hydrolysis to acids (acidic and basic mechanisms)
- Reduction to 1 o amines
- Conversion to aldehyde by reaction with DIBALH.
- Conversion to ketones by reaction with organometallic compounds.
- Skill 20.15 Plan a synthesis of a ketone from a nitrile.
- Skill 20.16 Use IR, NMR and MS to identify carboxylic acids and nitriles.


11.2: Climate Change
One of the most pressing concerns in environmental health is the effects of climate change on the planet, and human and animal populations. Climate change refers to the shifts in climate over time in various regions across the globe. Many of these shifts are due to global warming , or the observable increase in average temperature of the earth. The term “global warming” sometimes causes confusion because people think it means all temperatures will be higher throughout the year. Instead, it’s better to think of global warming as the driver of climate change, because the increasing average global temperature is causing shifts to different regional climates across the globe - including more severe storms, droughts, and permanent changes to ecosystems.
How do we know that the earth is warming? Evidence from ice cores collected from Greenland and Antarctica, plus ambient temperature data collected since 1958 demonstrate that although the earth has gone through cycles of warming and ice ages, human activity since the industrial revolution has caused much more warming in the last 150 years than occurred in the ~800,000 years prior. Images from space show ice sheets and glaciers melting and shrinking, dumping fresh water into the ocean. And since the middle of the last century, scientists have tracked a 2 °F increase in the average temperature of the planet, with the most significant increase happening in the last 4 decades (NASA, 2024).
This glacier located along the Icefields Parkway, near Banff National Park, Canada, is melting, showcasing the effect global climate change is having on the environment. The glaciers in this area have been studied since the 1950s, making them some of the longest scientifically observed glaciers in the world. Photo Credit:
This most recent and rapid global warming is considered to be anthropogenic (caused by human activity), not something that would typically happen from natural sources alone. Anthropogenic factors include the release of Greenhouse Gasses (GHG) . These GHGs include carbon dioxide (CO₂), methane (CH₄), and nitrous oxide (N₂O),and are released into the atmosphere from the burning of fossil fuels for energy, as well as from modern agricultural practices. Trees actually remove CO₂ from the atmosphere and store it in their trunks - so forests become “carbon sinks”. Deforestation therefore also contributes to global warming by cutting into carbon sinks - thus decreasing the amount of CO₂ collected (Burke & Weill, 2023). These GHGs left floating in the atmosphere trap heat from the sun and reflect it back down to the earth, causing an increase in the average temperature - much like a greenhouse uses glass walls to trap heat from the sun and keep the interior warmer for plants. Since the 1970s, the release of GHGs has increased much faster than in years prior, with the largest spike happening after the turn of the 21st century (Frumkin, 2016). Part of this is due to global population growth, which certainly contributes to the demand for energy production and thus the burning of fossil fuels to make it. However, the largest contributors to GHG tend to be the wealthiest countries, and per-capita GHG emissions are higher for more economically developed countries than for less economically developed countries. So it’s not just that more people require more energy, it’s also the behaviors and demands of wealthy societies that are driving GHG emissions. Still, as a few countries like Germany and the Netherlands have demonstrated, quality of life can be maintained while reducing GHG emissions through the use of technology and renewable energy sources (Ritchie & Roser, 2020). See figures \(\PageIndex{3}\) and \(\PageIndex{4}\) below.
Increasing the Earth’s average surface temperature has several consequences. When the ocean warms even just a little, it expands. As ice sheets and glaciers on land melt, they also contribute to rising sea levels. (Melting icebergs don’t contribute to increased sea levels, as their volume is already accounted for - like ice cubes in a glass of water). Increases in ocean temperature can cause the “bleaching” of coral reefs, essentially killing them and the marine life that rely on them for food and shelter (NOAA, 2023). On land, the higher temperatures cause more evaporation of water from plants and soil, leading to droughts. Warming air can also hold onto more water vapor, which can also cause more extreme weather - like severe rainstorms, hurricanes, and flooding (Frumkin, 2016). As warm air travels toward the poles, it shifts the cold air from the polar vortex (low pressure air at the north and south poles) which can cause severe winter storms - like the unexpected snowstorms experienced in Texas in 2021 which caused massive failures in the power grid (Burke & Weill, 2023). While some rain is needed for vegetation growth, very wet springs followed by dry summers cause all that vegetation to dry out. In 2021 during the summer, dry timber and windy conditions led to mega wildfires in the Sierra Nevada mountains (Sierra Nevada Conservancy, 2022). Droughts can not only dry out plants, but also make the soil less permeable, causing flooding when substantial rains do fall, and allowing precious rainwater to wash away into the ocean instead of sinking into groundwater aquifers (Fry & Reyes-Velarde, 2019). And if heavy rains follow a wildfire, they can cause mudslides that change terrain and threaten homes and roads (Burke & Weill, 2023). .
Climate change might include changes in temperature, rain or snowfall, drought, or wind that influence the overall climate for that region. Climate is a term used to describe average weather patterns. Southern California for example has a mostly mediterranean (arid or semi-arid) climate around the coastal cities and Los Angeles, gradually shifting to a desert climate in the Mojave desert and inland regions. Los Angeles has pretty mild temperature changes over the seasons, with some rainfall in the winter, and dry, sunny summers. The desert areas get much less rain of course, and have more temperature extremes, getting hotter in the summer and colder in the winter. Weather is made up of the temperature, atmospheric pressure, precipitation, wind, and humidity at any given time and can change rapidly. Certain weather patterns may be predictable for a particular climate, but the two are not the same by definition. Over time, if the climate changes in a specific area, there may be more extreme weather patterns, and there will be changes in the overall ecology of that region - which may positively or negatively influence its ability to support life of the plants and animals who live there (Burke & Weill, 2023).
In our California example, the average temperature in the Los Angeles (LA) region has already increased since the middle of the 20th century, and is expected to be 4-5 °F warmer by the middle of the century. This increase in temperature is causing more of the snowpack to melt from the nearby mountains, and more of the water sources across the state to evaporate. The combination of warm, dry summers and substantial rainstorms in the winter will cause more vegetation to grow in the spring and become dry, wildfire fuel in the summer. Subsequent rains will increase the risk of mudslides and soil erosion. Along the coast, higher sea levels will threaten coastal communities, which, along with mudslides can cause severe danger to lives and homes (California Climate Adaptation Strategy, 2024). Central California has a mediterranean climate similar to LA, which is conducive to growing fruit and nut orchards and crops that require rain in the winter and warm sun in the summer. If the entire climate in central CA becomes more desert-like, it will severely impact the state’s large agricultural economy. For example, almond orchards in CA produce 80% of the world’s almonds - but these trees require many years to mature and need water year round (Chea, 2021).


8.7: References
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Celiac Disease Foundation. (2015a). What can I eat? Retrieved from http :// celiac . org / live – gluten – free / glutenfreediet / food – options /
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Celiac Disease Foundation. (2015c). What is gluten? Retrieved from http :// celiac . org / live – gluten – free / glutenfreediet / what – is – gluten /
Jackson, F. W. (2015). Dysphagia diet: Five levels for difficulty in swallowing diet. Retrieved from http :// gicare . com / diets / dysphagia – diet /
Leahy, W., Fuzy, J., & Grafe, J. (2013). Providing home care : A textbook for home health aides (4th ed.). Albuquerque, NM: Hartman.
Lehman, S. (2014, June 9). Nutrients: What they are and why you need them. Retrieved from http :// nutrition . about . com / od / nutrition 101 / a / nutrients . htm
Mayo Clinic (2014, July 8). Low potassium (hypokalemia). Retrieved from http :// www . mayoclinic . org / symptoms / low – potassium / basics / definition / sym -20050632
National Kidney Foundation. (2015). Potassium and your CKD Diet. Retrieved from https :// www . kidney . org / atoz / content / potassium
Ohio State University Wexner Medical Center. (2013, June 27). Dysphagia: Mechanically altered diet (Level 2). Retrieved from https :// patienteducation . osumc . edu / documents / dys -2. pdf
U.S. Department of Agriculture. (2011, May 25). Cooking meat? Check the new recommended temperatures. Retrieved from http :// blogs . usda . gov /2011/05/25/ cooking – meat – check – the – new – recommended – temperatures /
U.S. Department of Agriculture. (2013a, June 8). Easy lessons in safe food handling. Retrieved from http :// www . fsis . usda . gov / wps / wcm / connect / 18 cece 94 - 747 b - 44 ca - 874 f - 32 d 69 fff 1 f 7 d / Basics _ for _ Safe _ Food _ Handling . pdf ? MOD = AJPERES # page =2
U.S. Department of Agriculture. (2013b, August 8). Cooking safely in the microwave oven. Retrieved from http :// www . fsis . usda . gov / wps / portal / fsis / topics / food – safety – education / get – answers / food – safety – fact – sheets / appliances – and – thermometers / cooking – safely – in – the – microwave / cooking – safely – in – the – microwave – oven
U.S. Department of Agriculture. (2015a). ChooseMyPlate. Retrieved from http :// www . choosemyplate . gov /
U.S. Department of Agriculture. (2015b, March 23). Refrigerator and food safety. Retrieved from http :// www . fsis . usda . gov / wps / portal / fsis / topics / food – safety – education / get – answers / food – safety – fact – sheets / safe – food – handling / refrigeration – and – food – safety / ct _ index
University of Pittsburgh Medical Center. (2015). Tips for increasing protein in your diet. Retrieved from http :// www . upmc . com / patients – visitors / education / nutrition / Pages / tips – for – increasing – protein – in – your – diet . aspx


15.5B: Coccidiomycosis
- Describe the mode of transmission, symptoms and diagnostic test associated with Coccidioidomycosis immitis
Coccidioidomycosis (commonly known as “Valley fever”, as well as “California fever”, “Desert rheumatism”, and “San Joaquin Valley fever”) is a fungal disease caused by Coccidioides immitis or C. posadasii . It is endemic in certain parts of Arizona, California, Nevada, New Mexico, Texas, Utah and northwestern Mexico.
C. immitis resides in the soil in certain parts of the southwestern United States, northern Mexico, and parts of Central and South America. It is dormant during long dry spells, then develops as a mold with long filaments that break off into airborne spores when the rains come. The spores, known as arthroconidia, are swept into the air by disruption of the soil, such as occurs during construction, farming, or an earthquake.
Infection is caused by inhalation of the particles. The disease is not transmitted from person to person. The infection ordinarily resolves leaving the patient with a specific immunity to re-infection. C. immitis is a dimorphic saprophytic organism that grows as a mycelium in the soil and produces a spherule form in the host organism.
The disease is usually mild, with flu-like symptoms and rashes. The Mayo Clinic estimates that half the population in some affected areas have suffered from the disease. On occasion, those particularly susceptible may develop a serious or even fatal illness. Serious complications include severe pneumonia, lung nodules, and disseminated disease, where the fungus spreads throughout the body. The disseminated form of Valley Fever can devastate the body, causing skin ulcers, abscesses, bone lesions, severe joint pain, heart inflammation, urinary tract problems, meningitis, and often death. In order of decreasing risk, people of Filipino, African, Native American, Hispanic, and Asian descent are susceptible to the disseminated form of the disease. Men and pregnant women, and people with weakened immune systems (such as from AIDS), are more susceptible than non-pregnant women.
It has been known to infect humans, cattle, deer, dogs, elk, fish, mules, livestock, apes, kangaroos, wallabies, tigers, bears, badgers, otters and marine mammals.
Symptomatic infection (40% of cases) usually presents as an influenza-like illness with fever, cough, headaches, rash, and myalgia (muscle pain). Some patients fail to recover and develop chronic pulmonary infection or widespread disseminated infection (affecting meninges, soft tissues, joints, and bone). Severe pulmonary disease may develop in HIV-infected persons.
An additional risk is that health care providers who are unfamiliar with it or are unaware that the patient has been exposed to it may misdiagnose it as cancer and subject the patient to unnecessary surgery.
Coccidioidomycosis may be divided into the following types: Primary pulmonary coccidioidomycosis; Disseminated coccidioidomycosis; and Primary cutaneous coccidioidomycosis.
The fungal infection can be demonstrated by microscopic detection of diagnostic cells in body fluids, exudates, sputum and biopsy-tissue. With specific nucleotide primers, C.immitis DNA can be amplified by PCR. It can also be detected in cultures by morphological identification, or by using molecular probes that hybridize with C.immitis RNA. An indirect demonstration of fungal infection can be achieved also by serologic analysis detecting fungal antigen or host antibody produced against the fungus.
There are no published prospective studies that examine optimal antifungal therapy for coccidioidomycosis. Mild cases often do not require treatment. Oral Fluconazole and intravenous Amphotericin B are used in progressive or disseminated disease, or in which patients are immunocompromised. Alternatively, itraconazole or ketoconazole may be used. Posaconazole and voriconazole have also been used. There is currently no practical preventative measures available for people who live or travel through Valley Fever endemic areas. It is recommended to avoid airborne dust or dirt, though this is not a guaranteed manner of prevention. People in certain occupations may be advised to wear face masks.
Key Points
- C. immitis resides in the soil in certain parts of the southwestern United States, northern Mexico, and parts of Central and South America.
- Infection is caused by inhalation of the particles.
- The disease is usually mild, with flu-like symptoms and rashes.
Key Terms
- Coccidioidomycosis : Coccidioidomycosis is a fungal disease caused by Coccidioides immitis or C. posadasii. It is endemic in certain parts of Arizona, California, Nevada, New Mexico, Texas, Utah and northwestern Mexico.
- fungal : Of or pertaining to a fungus or fungi
- inhalation : The substance (medicament) which is inhaled.


2.3: Simple and Compound Interest
Learning outcome
- Calculate simple and compound interest
When a person takes out a loan, most lenders charge interest on the loan. Interest is a fee or change for borrowing money, typically a percent rate charged per year. We can compute simple interest by finding the interest rate percentage of the amount borrowed, then multiply by the number of years interest is earned. Another type of interest calculates interest on both the money initially deposited as well as the interest money earned, and is called compound interest. We’ll start with simple interest.
Simple Interest Equation
\(I=p\cdot{r}\cdot{t}\)
Where:
\(I\) is the interest paid
\(p\) is the principal —the original amount of money borrowed
\(r\) is the interest rate , a per-year rate, written as a decimal
\(t\) is the time of the loan, expressed in years or portions of a year
Example
Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a \($1,000\) T-note with a \(4\%\) annual rate, with a maturity in \(2\) years. How much interest will you earn?
[reveal-answer q=”694437″]Show Solution[/reveal-answer]
[hidden-answer a=”694437″]
Identify the information given in the problem.
Interest (\(I\)): unknown
Principal (\(p\)): \($1000\)
Rate (\(r\)): \(4\%=0.04\)
Time (\(t\)): \(2\) years
Put the information in the simple interest equation.
\(I=1000\cdot0.04\cdot2\)
Multiply.
\(I=80\)
Answer
You would earn \($80\) in interest.
[/hidden-answer]
In the following video, you are shown how to find how much interest is earned on a specified investment amount.
Example
A friend asks to borrow \($240\), offering to repay you \($250\) in 1 month. What annual interest rate is this equivalent to?
[reveal-answer q=”470866″]Show Solution[/reveal-answer]
[hidden-answer a=”470866″]
Identify the information given in the problem. Here your friend is paying back $10 more than he borrowed, so that is the interest paid.
Interest (\(I\)): \($10\)
Principal (\(p\)): \($240\)
Rate (\(r\)): unknown
Time (\(t\)): \(1\) month
Convert the time to years.
\(1\,\,\text{month}=\frac{1}{12}\,\,\text{year}\)
Put the information in the simple interest equation.
\(10=240\cdot{r}\cdot\frac{1}{12}\)
Regroup and simplify.
\(\begin{array}{l}10=r\cdot240\cdot\frac{1}{12}\\\\10=r\cdot\frac{240}{12}\\\\10=r\cdot20\end{array}\)
Divide to undo the multiplication.
\(r=10\div20=0.50\)
Rewrite as a percent.
\(0.50=50\%\)
Answer
This is equivalent to a \(50\%\) annual interest rate.
[/hidden-answer]
The example video that follows shows how to determine the annual simple interest rate.
Applications with simple interest usually involve either investing money or borrowing money. To solve these applications, we continue to use the same strategy for applications that we have used earlier in this chapter. The only difference is that in place of translating to get an equation, we can use the simple interest formula. We will provide examples of how to find interest earned, calculate the rate of interest, and how to find the principal given a rate and the interest earned.
We will start by solving a simple interest application to find the interest.
example
Nathaly deposited \(\text{\$12,500}\) in her bank account where it will earn \(\text{4%}\) interest. How much interest will Nathaly earn in \(5\) years?
Solution
We are asked to find the interest, \(I\).
Organize the given information in a list.
\(I=?\)
\(p = $12,500\)
\(r =4\%\)
\(t =5\) years
| Write the formula. | \(I=Prt\) |
| Substitute the given information. | \(I=\left(12,500\right)\left(0.04\right)\left(5\right)\) |
| Simplify. | \(I=2,500\) |
| Check your answer. Is \(\text{\$2,500}\) a reasonable interest on \(\text{\$12,500}\) over \(5\) years? | |
| At \(4\text{%}\) interest per year, in \(5\) years the interest would be \(20\text{%}\) of the principal. Is \(20\text{%}\) of \(\text{\$12,500}\) equal to \(\text{\$2,500}\)? Yes. | |
| Write a complete sentence that answers the question. | The interest is \(\text{\$2,500}\). |
try it
[ohm_question]146792[/ohm_question]
There may be times when you know the amount of interest earned on a given principal over a certain length of time, but you don’t know the rate. For instance, this might happen when family members lend or borrow money among themselves instead of dealing with a bank. In the next example, we’ll show how to solve for the rate.
example
Loren lent his brother \($3,000\) to help him buy a car. In \(\text{4 years}\) his brother paid him back the \($3,000\) plus \($660\) in interest. What was the rate of interest?
[reveal-answer q=”361018″]Show Answer[/reveal-answer]
[hidden-answer a=”361018″]
Solution
We are asked to find the rate of interest, \(r\).
Organize the given information.
\(I=660\)
\(P=$3,000\)
\(r=?\)
\(t=4\) years
| Write the formula. | \(I=Prt\) |
| Substitute the given information. | \(660=\left(3,000\right)r\left(4\right)\) |
| Multiply. | \(660=\left(12,000\right)r\) |
| Divide. | \(\Large\frac{660}{12,000}\normalsize=\Large\frac{\left(12,000\right)r}{12,000}\) |
| Simplify. | \(0.055=r\) |
| Change to percent form. | \(5.5\%=r\) |
| Check your answer. Is \(5.5\%\) a reasonable interest rate to pay your brother? | |
| \(I=Prt\) | |
| \(660\stackrel{?}{=}\left(3,000\right)\left(0.055\right)\left(4\right)\) | |
| \(660=660\quad\checkmark\) | |
| Write a complete sentence that answers the question. | The rate of interest was \(5.5\%\). |
[/hidden-answer]
try it
[ohm_question]146793[/ohm_question]
In the next video we use the simple interest formula to find the rate of interest given an amount of money borrowed and the amount if interest paid.
There may be times when you take a loan for a large purchase and the amount of the principal is not clear. This might happen, for instance, in making a car purchase when the dealer adds the cost of a warranty to the price of the car. In the next example, we will solve a simple interest application for the principal.
example
Eduardo noticed that his new car loan papers stated that with an interest rate of \(\text{7.5%}\), he would pay \(\text{\$6,596.25}\) in interest over \(5\) years. How much did he borrow to pay for his car?
[reveal-answer q=”754756″]Show Answer[/reveal-answer]
[hidden-answer a=”754756″]
Solution
We are asked to find the principal, \(P\).
Organize the given information.
\(\begin{array}{ccc}\hfill I& =& 6,596.25\hfill \\ \hfill P& =& ?\hfill \\ \hfill r& =& \text{7.5%}\hfill \\ \hfill t& =& \text{5 years}\hfill \end{array}\)
| Write the formula. | \(I=Prt\) |
| Substitute the given information. | \(6,596.25=P\left(0.075\right)\left(5\right)\) |
| Multiply. | \(6,596.25=0.375P\) |
| Divide. | \(\Large\frac{6,596.25}{0.375}\normalsize=\Large\frac{0.375P}{0.375}\) |
| Simplify. | \(17,590=P\) |
| Check your answer. Is \(\text{\$17,590}\) a reasonable amount to borrow to buy a car? | |
| \(I=Prt\) | |
| \(6,596.25\stackrel{?}{=}\left(17,590\right)\left(0.075\right)\left(5\right)\) | |
| \(6,596.25=6,596.25\quad\checkmark\) | |
| Write a complete sentence that answers the question. | The amount borrowed was \($17,590\). |
[/hidden-answer]
TRY IT
[ohm_question]146794[/ohm_question]
In the simple interest formula, the rate of interest is given as an annual rate, the rate for one year. So the units of time must be in years. If the time is given in months, we convert it to years.
example
Caroline got \($900\) as graduation gifts and invested it in a \(10\)-month certificate of deposit that earned \(2.1\%\) interest. How much interest did this investment earn?
[reveal-answer q=”799758″]Show Answer[/reveal-answer]
[hidden-answer a=”799758″]
Solution
We are asked to find the interest, \(I\).
Organize the given information.
\(\begin{array}{ccc}\hfill I& =& ?\hfill \\ \hfill P& =& \text{\$900}\hfill \\ \hfill r& =& \text{2.1%}\hfill \\ \hfill t& =& \text{10 months}\hfill \end{array}\)
| Write the formula. | \(I=Prt\) |
| Substitute the given information, converting 10 months to \(\Large\frac{10}{12}\) of a year. | \(I=$900\left(0.021\right)\left(\Large\frac{10}{12}\normalsize\right)\) |
| Multiply. | \(I=15.75\) |
| Check your answer. Is \(\text{\$15.75}\) a reasonable amount of interest? | |
| If Caroline had invested the \(\text{\$900}\) for a full year at \(2\text{%}\) interest, the amount of interest would have been \(\text{\$18}\). Yes, \(\text{\$15.75}\) is reasonable. | |
| Write a complete sentence that answers the question. | The interest earned was \(\text{\$15.75}\). |
[/hidden-answer]
try it
[ohm_question]146796[/ohm_question]
In the following video we show an example of how to calculate the amount of interest earned on a treasury note.
THE COMPOUND INTEREST FORMULA
Compound interest can be calculated using the formula
\(A(t)=P(1+r)^{t}\)
where
- A ( t ) is the account value,
- t is measured in years,
- P is the starting amount of the account, often called the principal, or more generally present value,
- r is the annual percentage rate (APR) expressed as a decimal, and assuming the interest is only compounded once a year
In our next example we will calculate the value of an account after ten years of interest compounded annually.
Example
If we invest \($3,000\) in an investment account paying \(3\%\) interest compounded annually, how much will the account be worth in \(10\) years?
[reveal-answer q=”689928″]Show Answer[/reveal-answer]
[hidden-answer a=”689928″]
Because we are starting with \($3,000\), \(P= 3000\). Our interest rate is \(3\%\), so \(r=0.03\). We want to know the value of the account in \(10\) years, so we are looking for \(A(10)\), the value when \(t = 10\).
\(\begin{array}{c}A\left(t\right)\hfill & =P\left(1+r\right)^{t}\hfill & \text{Use the compound interest formula}. \\ A\left(10\right)\hfill & =3000\left(1+0.03\right)^{10}\hfill & \text{Substitute using given values}. \\ \text{ }\hfill & \approx 4031.75\hfill & \text{Round to two decimal places}.\end{array}\)
The account will be worth about \($4,031.75\) in \(10\) years.
[/hidden-answer]
Sometimes you will be asked for the final total amount in an account after interest is added.
TRY IT
[ohm_question]2556[/ohm_question]
Now we will look at an example that uses the compound interest formula to solve for the principal.
Example
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to \($40,000\) over \(18\) years. She believes the account will earn 6% compounded annually. To the nearest dollar, how much will Lily need to invest in the account now?
[reveal-answer q=”254680″]Show Answer[/reveal-answer]
[hidden-answer a=”254680″]
The nominal interest rate is \(6\%\), so \(r=0.06\)
We want to find the initial investment, \(P\), needed so that the value of the account will be worth \($40,000\) in \(18\) years. Substitute the given values into the compound interest formula, and solve for \(P\).
\(\begin{array}{c}A\left(t\right)\hfill & =P{\left(1+r\right)}^{t}\hfill & \text{Use the compound interest formula}.\hfill \\ 40,000\hfill & =P{\left(1+0.06\right)}^{18}\hfill & \text{Substitute using given values }A\text{, }r, n\text{, and }t.\hfill \\ 40,000\hfill & =P{\left(1.06\right)}^{18}\hfill & \text{Simplify}.\hfill \\ \frac{40,000}
Callstack:
at (Bookshelves/Accounting/Accounting_for_Managers_(Lumen)/02:_Calculations_and_Solving_Equations/2.03:_Simple_and_Compound_Interest), /content/body/div[5]/div[3]/p[5]/span, line 1, column 1
Lily will need to invest \($14,013.75\) to have \($40,000\) in \(18\) years.
[/hidden-answer]
TRY IT
[ohm_question]157038[/ohm_question]


4.1: Leadership and Management Introduction
Learning Objectives
- Compare and contrast the role of a leader and a manager
- Examine the roles of team members
- Identify the activities managers perform
- Describe the role of the RN as a leader and change agent
- Evaluate the effects of power, empowerment, and motivation in leading and managing a nursing team
- Recognize limitations of self and others and utilize resources
As a nursing student preparing to graduate, you have spent countless hours on developing clinical skills, analyzing disease processes, creating care plans, and cultivating clinical judgment. In comparison, you have likely spent much less time on developing management and leadership skills. Yet, soon after beginning your first job as a registered nurse, you will become involved in numerous situations requiring nursing leadership and management skills. Some of these situations include the following:
- Prioritizing care for a group of assigned clients
- Collaborating with interprofessional team members regarding client care
- Participating in an interdisciplinary team conference
- Acting as a liaison when establishing community resources for a patient being discharged home
- Serving on a unit committee
- Investigating and implementing a new evidence-based best practice
- Mentoring nursing students
Delivering safe, quality client care often requires registered nurses (RN) to manage care provided by the nursing team. Making assignments, delegating tasks, and supervising nursing team members are essential managerial components of an entry-level staff RN role. As previously discussed, nursing team members include RNs, licensed practical/vocational nurses (LPN/VN), and assistive personnel (AP). [1]
Read more about assigning, delegating, and supervising in the “ Delegation and Supervision ” chapter.
An RN is expected to demonstrate leadership and management skills in many facets of the role. Nurses manage care for high-acuity patients as they are admitted, transferred, and discharged; coordinate care among a variety of diverse health professionals; advocate for clients’ needs; and manage limited resources with shrinking budgets. [2]
Read more about collaborating and communicating with the interprofessional team; advocating for clients; and admitting, transferring, and discharging clients in the “ Collaboration Within the Interprofessional Team ” chapter.
An article published in the Online Journal of Issues in Nursing states, “With the growing complexity of healthcare practice environments and pending nurse leader retirements, the development of future nurse leaders is increasingly important.” [3] This chapter will explore leadership and management responsibilities of an RN. Leadership styles are introduced, and change theories are discussed as a means for implementing change in the health care system.
- American Nurses Association & NCSBN. (2019). National guidelines for nursing delegation. https://www.ncsbn.org/NGND-PosPaper_06.pdf ↵
- Cherry, B., & Jacob, S. R. (2017). Nursing leadership and management. In Cherry, B. & Jacob, S. (Eds.), Contemporary nursing: Issues, trends, and management (8th ed.). Elsevier, pp. 294-314. ↵
- Dyess, S. M., Sherman, R. O., Pratt, B. A., & Chiang-Hanisko, L. (2016). Growing nurse leaders: Their perspectives on nursing leadership and today’s practice environment. OJIN: The Online Journal of Issues in Nursing, 21 (1). https://ojin.nursingworld.org/MainMenuCategories/ANAMarketplace/ANAPeriodicals/OJIN/TableofContents/Vol-21-2016/No1-Jan-2016/Articles-Previous-Topics/Growing-Nurse-Leaders.html ↵


10.4: Composition of functions
a function \(A \to C\) created from given functions \(f: A \rightarrow B\) and \(g: B \rightarrow C\) by \(a \mapsto g(f(a))\)
the composition of functions \(f: A \rightarrow B\) and \(g: B \rightarrow C\text{,}\) so that \(g \circ f: A\rightarrow C\) by \((g \circ f)(a) = g(f(a))\)
Consider the functions
\begin{align*} f \colon \mathbb{R} & \to \mathbb{R}_{\geq 0}, & g \colon \mathbb{R}_{\geq 0} & \to \mathbb{R}_{\geq 0},\\ x & \mapsto x^2, & x & \mapsto \sqrt{x}, \end{align*}
Then we have
\begin{align*} g \circ f \colon \mathbb{R} & \to \mathbb{R}_{\geq 0}, \\ x & \mapsto \sqrt{x^2} = \vert x \vert. \end{align*}
The notation for the composition of functions \(f\) and \(g\) involves a reversal of order, so that we write \(g \circ f\text{.}\) This is so that when we use this notation with input-output notation \((g \circ f)(a)\text{,}\) the notation reminds us that \(f\) must first be applied to the input \(a\text{,}\) and then \(g\) is applied to the result \(f(a)\text{.}\)
In general, \(f \circ g \ne g \circ f\text{.}\) Usually, one of the two is not even defined, because domains and codomains of \(f\) and \(g\) will not necessarily match up in both orders. And when both are defined, the two different orders of composition usually have different domains and codomains.
Consider functions
\begin{align*} f \colon \mathbb{N} & \to \mathbb{N}, & g \colon \mathbb{N} & \to \mathbb{N}, \\ n & \mapsto n^2, & n & \mapsto n + 1. \end{align*}
Then, both \(f \circ g: \mathbb{N} \rightarrow \mathbb{N}\) and \(g \circ f: \mathbb{N} \rightarrow \mathbb{N}\) are defined. But they are not equal, as
\begin{align*} (f \circ g)(n) & = (n+1)^2 = n^2+2n+1, & (g \circ f)(n) & = n^2+1. \end{align*}
Consider functions
\begin{align*} \text{sqrt} \colon \mathbb{N} & \to \mathbb{R}, & \text{flr} \colon \mathbb{R} & \to \mathbb{Z}, \\ n & \mapsto \sqrt{n}, & x & \mapsto \lfloor x \rfloor. \end{align*}
(See
Example 10.3.4
for a description of the \(\text{flr}\) function.)
Then, \(\text{flr} \circ \text{sqrt} : \mathbb{N} \rightarrow \mathbb{Z}\) is defined, with
\begin{equation*} (\text{flr} \circ \text{sqrt} )(n) = \lfloor \sqrt{n} \rfloor \text{.} \end{equation*}
But \(\text{sqrt} \circ \text{flr}\) is not defined, as the codomain of \(\text{flr}\) does not match the domain of \(\text{sqrt} \text{.}\) In particular, \(\text{flr}\) will sometimes return a negative output, and we cannot use such an output as an input in \(\text{sqrt} \text{.}\)
Consider functions \(f: A \rightarrow B\) and \(g: B \rightarrow C\text{.}\)
If \(g \circ f\) is injective, are either or both of \(f,g\) necessarily injective?
Answer the same question as above with “injective” replaced by “surjective”.
Demonstrate that if both \(f\) and \(g\) are bijective, then the composition \(g \circ f\) is also bijective.
Of course, we can compose any number of functions.
Let us reconsider the function defined by algorithm in Example 10.1.3 . As the function description involved a multi-step algorithm, we should be able to break the steps involved into their own functions, then recreate the original functions as a composition.
First, define \(\text{abs} : \mathscr{P}( \mathbb{Z}) \rightarrow \mathscr{P}( \mathbb{N})\) by
\begin{equation*} \text{abs}(X) = \{\vert x \vert \; \vert x \in X\} \text{.} \end{equation*}
Next, define \(min : \mathscr{P}( \mathbb{N}) \rightarrow \mathbb{N}\) so that \(\text{min}(X)\) outputs the minimum number in input set \(X\text{,}\) and outputs \(0\) in case \(X = \emptyset\text{.}\)
Finally, define \(f: \mathbb{N} \rightarrow \mathbb{N}\) by \(f(n) = 2 n + 1\text{.}\)
Each of these functions represents one step in the algorithm defining the function in Example 10.1.3 , but to recreate that function we need to compose the functions in the correct order: write \(\varphi = f \circ \text{min} \circ \text{abs}\text{,}\) so that
\begin{equation*} \varphi(X) = 2 \text{min}(\text{abs}(X)) + 1 \end{equation*}
computes the same result for an input set \(X\) as the algorithm described in
Example 10.1.3
.


<IP_ADDRESS>: Rhetorical Criticism Summary
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Summary
By now you have a more complex understanding of the term rhetoric and realize the distinction between the use of the term in the popular press and its meaning within the context of the Communication discipline. At a very basic level, you know that it refers to the process in which humans use symbols to communicate with one another. Moreover, you should know the contributions of classical and well as contemporary theorists to understanding human symbol use. As a way of knowing, rhetorical criticism provides scholars with a unique methodological tool for understanding communication. Through the contributions and paradigm shifts from classical to contemporary scholars, the scope of rhetorical criticism has been widened to include a variety of social events, acts, and artifacts that we encounter in our daily life. This method has a direct impact on our lives by informing us about the communicative options available to us when communicating with others in cultural, professional, personal and political contexts.
Contributions and Affiliations
- Survey of Communication Study. Authored by : Scott T Paynton and Linda K Hahn. Provided by : Humboldt State University. Located at : en.wikibooks.org/wiki/Survey_of_Communication_Study/Preface. License : CC BY-SA: Attribution-ShareAlike


15: Green Revolution
In the last few chapters, we have seen Americans spread across the continent. Beginning with the original thirteen colonies, the history of the United States often seems to be a straightforward story of expansion westward. After a Revolution motivated partly by the colonists’ resistance to the Crown’s restrictive Proclamation, citizens of the new nation took full advantage of opportunities provided by the Northwest Ordinance, the Louisiana Purchase, the Mexican-American War, and the Oregon Treaty, to seek their fortunes on the frontier. Expansion was achieved partly by hardworking farmers, ranchers, and miners on the countryside, and partly by urban wage-workers and businesses in new western cities. These pioneers were aided by capital from eastern and overseas financial markets, by improving transportation networks, and by eastern city workers and capitalists who processed and consumed the products of fields, forests, and mines.
But not all the significant movement in American History was east to west. The growth of eastern cities depended on resources acquired on the frontier, so the flow of raw materials like farm products, timber, and minerals (which we will examine in Chapter Twelve) from west to east was vitally important to the nation’s growth. The ideal of Manifest Destiny was as much about securing access to these resources as it was about planting the stars and stripes from sea to shining sea. And sometimes, important changes actually originated in the west, and expanded eastward.
The expansion and growth of both the West and the East depended on the ability of farmers to continue producing food for the nation’s growing population and for export. The biggest single factor in the continuing farm bonanza was soil fertility. In the 1840s, German chemist Justus von Liebig discovered that the chemicals nitrogen, phosphorus, and potassium were essential to plant growth. Farm soils naturally lose fertility unless fields are given time to regenerate between crop plantings or amendments are added. Liebig announced that instead of leaving fields fallow for years between crop plantings, farmers could amend the soil with the chemicals plants needed. We will examine each of these chemicals, beginning with the element often considered the most important soil amendment, nitrogen. During the crucial period when America’s commercial farmers became increasingly responsible for feeding the growing nation, nitrogen came from the west.
Green Manure, Guano, Caliche
When early settlers cut and burned forests to build farms, the thin soils of the forest floors received an initial productivity boost from leaf mulch and from the ashes of burned trees. This fertility was exhausted in a few years unless the farmer rotated crops and pasture or added manure to the soil. The prairie soils later settlers found farther west were deeper and more fertile than eastern forest soils. Tallgrass prairies grew from the Canadian border to Texas, and as the name suggests, some of their grasses grew six to nine feet tall. When Laura Ingalls’s parents told her not to stray too far onto the prairie in the Little House books, it was because they were afraid of losing her in the tall grass.
In addition to being tall, the prairies were deep. After millennia of growth and decay, prairie vegetation had left a layer of humus up to ten feet deep. The roots of perennial grasses took advantage of this depth, drawing water and nutrients from well below the surface. When settlers arrived with their new John Deere cast-steel plows, they only managed to turn over the top foot or so of this organic layer. But even so, the prairie offered some of the richest soils American settlers had ever seen. Western farmers wrote their families and friends back east boasting of the land and the great yields they got farming it.
In the long run, even the most fertile soils will be depleted after years of growing annual crops that don’t return any nutrients. Smaller farms were able to spread manure from livestock, but as their cultivated fields grew from a few acres to a few hundred, most farmers found there was just not enough manure to spread. Commercial farmers, who specialized in growing wheat or corn for the market, began planting their fields with the cash crops every year rather than rotating grain planting with pasture planting for grazing animals that would return at least some manure to the land. A new source of fertility was needed to prevent the agricultural balloon from deflating. The solution was alfalfa, a crop farmers called Chili Clover.
It has long been known that in addition to spreading manure, planting legumes is an effective method of returning nitrogen to the soil. Although nitrogen is the most common element in Earth’s atmosphere, atmospheric nitrogen is very stable and unreactive. Plants cannot use nitrogen until it has been converted to nitrogen oxides, nitrates, or ammonia. Legumes have evolved a symbiotic relationship with Rhizobium bacteria which live in nodules on the plants’ roots and break the bonds of atmospheric nitrogen. When legumes are grown on a field and then plowed under, the nitrogen accumulated in the plants enters the soil and becomes available for the next crop. One of the most effective legume crops used as a green manure is alfalfa.
Alfalfa is native to Eurasia, where Siberian horse tribes began feeding it to their animals over three thousand years ago. Cortés and Pizarro brought alfalfa to Mexico and Peru in the early 1500s and the aggressive species quickly spread to across Central and South America. When English colonists planted alfalfa, which they called lucerne, the plant did poorly in the cool, acidic soils of the Atlantic colonies. After this early failure the plant was forgotten by American farmers. But a Chilean miner introduced alfalfa to California during the Gold Rush. The climates of Chile and California are nearly identical, and the Chili Clover thrived in western pastures, renewing soils and providing ten times the nutrition of regular grasses for the growing dairy herds of California. Impressed by alfalfa’s success in California in the early 1850s, the U.S. Patent Office mailed out thousands of packets of alfalfa seed to farmers across the country. Alfalfa plantings spread eastward from California as the plant was quickly adopted by farmers in Utah, Colorado, Kansas, and Nebraska. Today, the Chilean import is so common in American fields, almost no one remembers it was not a native species.
But some farmers lacked the patience to rotate cropping with pasture, or were unwilling to accept the idea that a significant portion of their farms would be fallow each year. They preferred to amend their soils rather than waiting for fertility to regenerate naturally. But as mentioned, few farmers had access to enough manure to supplement all their soil. The first commercial fertilizers were made from guano, the droppings of seabirds living on islands off the western shores of South America. Guano comes from the Quechua Indian word Wanu , which means any excrement used as a soil additive in farming. The advantage guano had over locally-available manure was that it was dry, light, and highly concentrated. Natives of the Andes have mined guano on the coast and islands for at least 1,500 years, according to archeologists. Spanish colonial records noted that Inca rulers considered protecting the cormorants that were the main source of guano so important, that disturbing the birds’ nesting areas a capital offense. Guano was carried from the coast up into the Andes on the backs of llamas, for use on the terraced farms surrounding highland cities like Machu Picchu.
Although surrounded by ocean, the islands off the western coast of South America are arid. Like the deserts they face on the mainland, some experience no annual rainfall at all. Seabirds such as cormorants and pelicans have lived on these islands by the millions, for thousands of years. Over that time, they’ve left literal mountains of droppings, which due to the lack of rain have simply piled up. The guano contains 8 to 16 percent nitrogen, 8 to 12 percent phosphorus, and 2 to 3 percent potassium, which makes it an excellent fertilizer without any mixing. It simply needs to be chopped off the mountain, ground up, and spread on fields.
Prussian explorer Alexander von Humboldt visited the islands around 1802, and publicized guano’s value as a fertilizer throughout Europe. Seeing a lucrative business opportunity, Europeans and Americans fell on the area in a guano rush, and by the middle of the century several nations had enlisted the work of Chinese peasants in a Pacific labor system that has been compared with the slavery of the Atlantic world. Although the Chinese workers were technically free, many had been tricked into labor contracts promising work in California. Once they reached the guano islands and realized they had been duped, there was no way off. Over a hundred thousand Chinese workers were imported to the islands in the second half of the nineteenth century.
Guano was so profitable that the U.S. Congress passed a Guano Islands Act in 1856. The law provided an incentive for American sailors to find and claim undefended islands for America by giving the discoverer exclusive rights to the guano recovered. Islands claimed under the Guano Islands Act include parts of the Hawaiian chain, Midway Atoll, part of American Samoa, and several islands still disputed with Colombia. The guano islands off the western coast of South America were so valuable that two wars were fought over them. Chile and Peru fought Spain in the Chincha Islands War, 1864-66, and defeated the Spanish Empire. Then, Spain’s claim had been successfully set aside, Chile took many of the guano islands from Peru, along with the nitrate fields of the Atacama Desert, in the War of the Pacific, 1879-83.
After about 1870, guano was overtaken as a source of nitrogen by nitrate-rich desert soils called Caliche. These soils were discovered in the Atacama desert, a region regarded as the driest place on earth that lay partly in Chile, partly in Peru, and partly in Bolivia. When all three nations rushed to extract and process the Caliche, Chile challenged its northern rivals for the nitrate fields. Chile’s victory in the War of the Pacific extended its border northwards to encompass the Atacama desert, including all the coastal territory that had belonged to Bolivia. Many ethnic Bolivians living around the port city of Arica still dream of throwing off the Chilean yoke and winning their country access to the Pacific again.
Defeating its northern neighbors in the War of the Pacific made Chile the undisputed power on the west coast of the Americas and generated an economic boom. The nitrate Chile monopolized was valuable both as a fertilizer and as a key ingredient in explosives and munitions. But mining and processing nitrate from Chile’s desert soil required much more capital than digging guano. Chile attracted British investors, and soon joint ventures began shipping a million tons of nitrate per year out of the South American desert. Production grew steadily until 1914, when World War I created new incentives for Britain’s enemies to find an alternative to Caliche nitrate.
Fritz Haber was the German chemist who in 1911 developed the high-pressure method of extracting nitrogen from the atmosphere that is used today to produce nearly all the nitrogen used in industry and agriculture. Carl Bosch, working for German chemical company BASF, scaled up Haber’s laboratory experiment to industrial production. In 1914, BASF produced 20 tons a day for the war effort. After the war, production of synthetic nitrate fertilizer quickly cut prices by more than half, and annual Chilean caliche shipments dropped to less than a third of their 2.5 million ton pre-war volume. Haber and Bosch were each awarded Nobel Prizes for their work in chemistry.
The Haber-Bosch process requires not only extreme high pressures to extract atmospheric nitrogen, but a great deal of energy. Approximately five percent of the world’s production of natural gas is required to produce about 500 million tons of ammonia each year. But the results have been spectacular. The application of concentrated nitrogen to farm fields increased production even over the yields that had been achieved using guano and nitrate, causing a global explosion of crop yields known as the Green Revolution. By 2010, the Haber-Bosch process produced 133 million tons of ammonia annually, about three quarters of which was applied as fertilizer. Corn yields per acre increased five-fold, and other staple crops such as rice saw increases of three or four times. Historian of technology Vaclav Smil estimates that if crop yields had stayed where they were in 1900, by the year 2000 farm fields would have needed to cover at least half the land on the ice-free continents in order to feed the world’s population, rather than the fifteen percent they now occupy. Another way of putting Smil’s point is that nearly half the people alive today would probably starve, without the Haber-Bosch process.
Although he won a Nobel Prize and worldwide admiration for his work leading to nitrogen fertilizers, Fritz Haber’s legacy is clouded with tragedy. In addition to the nitrate Germany required for munitions, Fritz Haber’s other contributions to the German war effort were the chlorine-based chemical warfare agents used against allied troops in World War I and a cyanide-based pesticide called Zyklon A that was a direct predecessor of the poison used against prisoners in the concentration camps during World War II. Haber’s wife, Clara Immerwahr, was also a scientist and was the first woman ever to receive a Ph.D. from the University of Breslau. Immerwahr killed herself with her husband’s gun when she discovered he had supervised the first successful use of chlorine gas at the battle of Ypres in 1915. Their son Hermann emigrated to the U.S. and committed suicide in 1946 after discovering that his father’s invention had been used to kill millions of their fellow German Jews. Fritz Haber did not live to see Zyklon B used in the concentration camps. He died while moving to Palestine in 1934.
The Haber-Bosch process is an effective way to synthesize nitrogen as long as energy costs remain low, but it is not the only way to make nitrogen fertilizer. As mentioned earlier, nitrogen fixating soil bacteria and symbiotic microbes living on legumes fix nitrogen from the atmosphere without any additional energy inputs. And there are other chemical processes that can be applied on industrial scales. Some of these industrial processes are much less energy-intensive than the Haber-Bosch. For example, a hydro-electric plant in Norway used its surplus energy to make ammonia for fertilizer from 1911 to 1971, using a process that had actually been developed before Fritz Haber made his discovery. But in spite of these alternatives, almost all the nitrogen used in commercial fertilizer is currently produced through the Haber-Bosch process. As a result, fertilizer prices tend to follow natural gas prices, and ammonia-based fertilizers will probably continue to be made using the high-energy process until energy costs rise substantially.
Phosphorus and Potassium
The second element Liebig discovered was crucial to plant growth is phosphorus. Unlike nitrogen, which makes up seventy-eight percent of Earth’s atmosphere, phosphorus is a mineral. Traditional agriculture recycled phosphorus by returning manure to fields as fertilizer. During the green revolution, most of the phosphorus applied by the world’s farmers has been superphosphate, a concentrate manufactured from a raw material called phosphate rock. Phosphate rock is a sedimentary mixture of minerals deposited in earlier geological eras on the bottoms of ancient oceans. Guano was also rich in concentrated phosphate, and after the Atacama Desert became the main source of nitrates, much of the guano remaining on the Pacific islands was used to produce phosphate fertilizer. The deposits currently used to make fertilizer contain about thirty percent phosphorus.
Although mineral phosphorus is plentiful in the Earth’s crust, known deposits of sedimentary phosphate rock are finite. Estimates of the concentrated phosphorus remaining vary, and depend on assumptions about yet to be developed technologies for discovering and recovering the mineral. Some scientists believe phosphorus production may peak around 2030, and then decline as exploitable reserves are exhausted. Like nitrogen, inexpensive and plentiful phosphorus has boosted crop yields greatly. Eighty percent of the phosphate rock mined each year goes to produce fertilizer. Phosphorus shortages would create serious threats to world food supply, which could impact global security. But since phosphorus is still available in animal and human waste, a shift back toward applying manure to agricultural fields would greatly reduce demand for superphosphate. Recognizing the growing value of the chemicals contained in our waste, some urban sewage treatment operations have begun producing concentrated phosphorus fertilizers from city sewer sludge.
Potassium, the third vital plant nutrient, is actually named after the potash that early American colonists made from trees they burned to clear their fields. Ashes were soaked in water, and then the water was evaporated in cast iron pots to produce lye (potassium hydroxide) and potash (a mixture of potassium chloride and potassium carbonate). Wood ashes yielded about ten percent potash by weight, and by clearing a few acres of trees each year as they expanded their cultivation settlers could also produce a product they could sell for cash. As previously mentioned, the first patent issued by the United States in 1790 was for a “new apparatus and process” for making potash.
Like phosphorus, commercial potassium fertilizers are manufactured from sedimentary deposits mined at the sites of ancient inland seas. Current world production is dominated by two cartels, but known reserves total more than three hundred times annual production, so the element is not considered strategic.
Hazards
One of the problems associated with using concentrated soluble nitrogen, phosphorus, and potassium (NPK) in agriculture is that uptake by crops is less than 50 percent. In the old days of ten-foot deep prairie soils, this would not have been a problem. Organic matter binds elemental nutrients, and then decays slowly and releases the nutrient chemicals into the soil gradually over many seasons. The deep root systems of perennial grasses stored nutrients in biomass. And the chemistry of organic soils and the microscopic insect, bacterial, and fungal life-forms that thrived in those soils held onto the rest. However, in the shallow-plowed fields of modern farms, the root systems of annual crops are not deep enough to catch and hold nitrogen, phosphorus, and potassium as it runs off with the water. And much of the soil life that once thrived in farm fields and pastures has been eliminated by pesticides, herbicides, and fungicides.
In spite of the additional food produced to support growing world population, the green revolution has been criticized by environmentalists for ignoring the problems created by fertilizer runoff. Excess agricultural fertilizer contributes nearly half of the nutrients that are currently poisoning American streams and rivers. Fertilizers entering the watershed encourage algal blooms. When these prolific mats of single-celled plants die, the decaying algae traps dissolved oxygen, creating hypoxic dead zones. The dead zone in the Gulf of Mexico has shrunk somewhat from its 2002 peak of 8,500 square miles. But excess fertilizer continues to flow down the Mississippi into the ocean.
Video \(\PageIndex{1}\): A YouTube element has been excluded from this version of the text. You can view it online here: https://youtu.be/rK_mEHqx7rw . Video from NOAA, Public Domain
The green revolution has also been criticized by less developed nations for increasing poor people’s dependence on technological solutions created in the faraway laboratories and factories in the United States and other industrialized nations. In many cases, development loans from organizations like the World Bank are tied to spending on products from the donor nations. Critics argue this benefits the donors more than the recipients and puts poor farmers at risk when they become dependent on new technologies and then fail to earn enough with their crops to pay rising prices for seeds, fertilizers, and pesticides. Farmers in developing nations have been encouraged with generous loans to invest in expensive capital equipment they cannot afford when the loans disappear. Since the 1990s, more than a quarter million farmers in India have killed themselves because they cannot escape the cycle of debt created by their green-revolution involvement in high-tech agriculture.
Dust Bowl
In America, the early green revolution that began with the use of guano, potash, and nitrate led to bonanza farming and the extension of agriculture onto marginal lands. The Russian Revolution and World War I at the beginning of the twentieth century reduced European farm production and drove up the price of American grain in international markets. Marginal high plains grazing lands like those of western Kansas and Nebraska, Oklahoma and northwestern Texas were put under the plow. Cropland in this region doubled between 1900 and 1920, and then tripled again between 1925 and 1930.
Plowing fields for annual corn and wheat planting is such a common farming practice, it seems normal. On the high plains, perennial grasses had evolved to find water deep in the soil and hold onto it. Plowing exposed the soil to the sun and wind, and cut the roots that trapped moisture and bound the soil together. Worse, generations of American farmers who had been taught to depend on plentiful commercial fertilizers, had forgotten the value of humus. Since Justus von Liebig had focused their attention on soil chemistry in the 1840s, most soil scientists had promoted the idea that chemical nutrients were the important thing, and that soil itself was just an inert medium. Although the prairies settlers discovered when they moved westward had been rich in living organic material, the farms built on those original soils relied on imported chemicals rather than on nutrients produced by biological activity in the soil. Many farmers even burned off the stems and stalks left after each harvest. Then the soil blew away.
The western edge of the prairie was actually a completely different ecosystem from the eastern edge by the Great Lakes, but the change in climate was very gradual. Although many farmers moving west to convert the high plains rangeland to cultivation failed to notice it, rainfall was scarcer and the wind blew harder. Those who did notice were mollified by the assurances of experts that “rain followed the plow.” But of course that was just boosterism and wishful thinking: rain did not follow the plow. Between 1933 and 1935, drought struck the area. Actually, dry conditions returned after a few years of unusual wetness that had been taken by hopeful farmers and optimistic boosters as the region’s permanent climate. Over half a million people were left homeless when their topsoil blew away.
In a single storm, beginning on November 11, 1933, topsoil from Oklahoma was blown all the way to Chicago, where over 12 million pounds of it fell on the city like snow. Like alfalfa, guano, and nitrate in the nineteenth century, America’s topsoil was traveled from west to east. On Black Sunday, April 14, 1935, dust storms were reported from the Canadian border to Texas. Newspaper reporters throughout the affected area wrote that they could not see five feet through the blowing dust. The agricultural disaster that became known as the Dust Bowl caused an exodus from the high plains region that should never have been put under the plow. But the disaster was not just just agricultural. Of 116,000 refugee families surveyed on their way into California, only four out of ten were farmer families. A full third of the heads of families who fled Oklahoma, Kansas, Nebraska, and Texas were white collar professionals. When the farms blew away the whole region was wiped out.
Alarmingly, less than a century later the lessons of history have been forgotten and much of the Dust Bowl region is again under the plow. The wheat-fields of Kansas, Texas, Oklahoma, and Colorado produced 411 million bushels of grain in 1933, just before the drought began. The Dust Bowl disaster cut production by three quarters, but by 2012, the region was once again producing 700 million bushels of wheat. This time, farmers have taken advantage of irrigation.
The Ogallala is one of the world’s largest aquifers, stretching from South Dakota to West Texas. Farmers have been drawing water from it for over fifty years. But the aquifer, which was charged with water over centuries before American farmers arrived, is not an unlimited resource. Sections of West Texas and Kansas have already run out of water, and irrigation is still accelerating. In just seven years between 2001 and 2008, farmers used about a third of the water that was taken from the aquifer during the entire twentieth century. The rate of depletion far exceeds the rate of recharge, even in the occasional rainy years. If the aquifer is emptied and irrigation becomes impossible on the high plains, American grain production will be reduced by 700 million bushels per year. Farmers will have no choice but to replant perennial prairie grasses and convert the plains back to rangeland, if they hope to prevent the soil blowing away again. The question is, will American agribusiness change willingly and thoughtfully before a crisis, or out of desperate reaction after the crisis strikes?
Further Reading
- Edward Dallam Melillo, Strangers on Familiar Soil: Rediscovering the Chile-California Connection. 2015.
- Vaclav Smil, Enriching the Earth: Fritz Haber, Carl Bosch, and the Transformation of World Food Production. 2004.
- Donald Worster, Dust Bowl: The Southern Plains in the 1930s. 1979.


5.5: Introduction to Contemporary Visual Aids
What you’ll learn to do: Discuss the use of video in business messages
Every day, hundreds of hours of video are uploaded to YouTube, and a millions of hours of video are watched. Impressive, to be sure, and while it doesn’t mean that your audience is out on YouTube searching for business communications, it does mean that they’ve grown accustomed to, or have even established a preference for, consuming information delivered on video.
Luckily, videos are amazingly easy and inexpensive to make these days. A communicator can create and deliver a video using nothing more than a phone; or, with minimal investment, purchasing a few microphones and a camera with a decent lens can create a more polished, higher-quality product.
If it’s true that facts bore and stories sell, then perhaps your message is better delivered in the format of a video. In this section, we’ll talk about the kind of message that works best for a video and how to put together your first blockbuster communication hit.


1.3: Add Whole Numbers (Part 1)
- Use addition notation
- Model addition of whole numbers
- Add whole numbers without models
- Translate word phrases to math notation
- Add whole numbers in applications
Before you get started, take this readiness quiz.
- What is the number modeled by the base-\(10\) blocks? If you missed this problem, review Example 1.1.2 .
Figure \(\PageIndex{1}\)
- Write the number three hundred forty-two thousand six using digits? If you missed this problem, review Example 1.1.6 .
Use Addition Notation
A college student has a part-time job. Last week he worked \(3\) hours on Monday and \(4\) hours on Friday. To find the total number of hours he worked last week, he added \(3\) and \(4\).
The operation of addition combines numbers to get a sum . The notation we use to find the sum of \(3\) and \(4\) is:
\[3 + 4 \nonumber \]
We read this as three plus four and the result is the sum of three and four. The numbers \(3\) and \(4\) are called the addends. A math statement that includes numbers and operations is called an expression.
To describe addition, we can use symbols and words.
| Operation | Notation | Expression | Read as | Result |
|---|---|---|---|---|
| Addition | + | 3+4 | three plus four | the sum of 3 and 4 |
Translate from math notation to words:
- \(7 + 1\)
- \(12 + 14\)
Solution
- The expression consists of a plus symbol connecting the addends \(7\) and \(1\). We read this as seven plus one . The result is the sum of seven and one .
- The expression consists of a plus symbol connecting the addends \(12\) and \(14\). We read this as twelve plus fourteen. The result is the sum of twelve and fourteen.
Translate from math notation to words:
- \(8 + 4\)
- \(18 + 11\)
- Answer a
-
eight plus four; the sum of eight and four
- Answer b
-
eighteen plus eleven; the sum of eighteen and eleven
Translate from math notation to words:
- \(21 + 16\)
- \(100 + 200\)
- Answer a
-
twenty-one plus sixteen; the sum of twenty-one and sixteen
- Answer b
-
one hundred plus two hundred; the sum of one hundred and two hundred
Model Addition of Whole Numbers
Addition is really just counting. We will model addition with base-\(10\) blocks. Remember, a block represents \(1\) and a rod represents \(10\). Let’s start by modeling the addition expression we just considered, \(3 + 4\).
Each addend is less than 10, so we can use ones blocks.
|
We start by modeling the first number with 3 blocks. |
|
|
Then we model the second number with 4 blocks. |
|
|
Count the total number of blocks. |
There are \(7\) blocks in all. We use an equal sign (\(=\)) to show the sum. A math sentence that shows that two expressions are equal is called an equation. We have shown that \(3 + 4 = 7\).
Model the addition \(2 + 6\).
Solution
\(2 + 6\) means the sum of \(2\) and \(6\)
Each addend is less than \(10\), so we can use ones blocks.
| Model the first number with 2 blocks. | |
| Model the second number with 6 blocks. | |
| Count the total number of blocks | |
| There are 8 blocks in all, so 2 + 6 = 8. |
Model: \(3 + 6\).
- Answer
-
Model: \(5 + 1\).
- Answer
-
When the result is \(10\) or more ones blocks, we will exchange the \(10\) blocks for one rod.
Model the addition \(5 + 8\).
Solution
\(5 + 8\) means the sum of \(5\) and \(8\).
| Each addend is less than 10, se we can use ones blocks. | |
| Model the first number with 5 blocks. | |
| Model the second number with 8 blocks. | |
| Count the result. There are more than 10 blocks so we exchange 10 ones blocks for 1 tens rod. | |
| Now we have 1 ten and 3 ones, which is 13. | $$5 + 8 = 13$$ |
Notice that we can describe the models as ones blocks and tens rods, or we can simply say ones and tens. From now on, we will use the shorter version but keep in mind that they mean the same thing.
Model the addition: \(5 + 7\)
- Answer
-
Model the addition: \(6 + 8\).
- Answer
-
Model the addition: \(17 + 26\).
Solution
\(17 + 26\) means the sum of \(17\) and \(26\).
| Model the 17. | 1 ten and 7 ones | |
| Model the 26. | 2 tens and 6 ones | |
| Combine. | 3 tens and 13 ones | |
| Exchange 10 ones for 1 ten. |
4 tens and 3 ones 40 + 3 = 43 |
|
| We have shown that 17 + 26 = 43 |
Model each addition: \(15 + 27\).
- Answer
-
Model each addition: \(16 + 29\).
- Answer
-
Add Whole Numbers Without Models
Now that we have used models to add numbers, we can move on to adding without models. Before we do that, make sure you know all the one digit addition facts. You will need to use these number facts when you add larger numbers.
Imagine filling in Table \(\PageIndex{1}\) by adding each row number along the left side to each column number across the top. Make sure that you get each sum shown. If you have trouble, model it. It is important that you memorize any number facts you do not already know so that you can quickly and reliably use the number facts when you add larger numbers.
| + | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 2 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 3 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 4 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| 5 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| 6 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 7 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| 8 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| 9 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
Did you notice what happens when you add zero to a number? The sum of any number and zero is the number itself. We call this the Identity Property of Addition. Zero is called the additive identity.
The sum of any number \(a\) and \(0\) is the number.
\[a + 0 = a\]
\[0 + a = a\]
Find each sum:
- \(0 + 11\)
- \(42 + 0\)
Solution
|
0 + 11 = 11 |
|
42 + 0 = 42 |
Find each sum:
- \(0 + 19\)
- \(39 + 0\)
- Answer a
-
\(0+19=19\)
- Answer b
-
\(39+0=39\)
Find each sum:
- \(0 + 24\)
- \(57 + 0\)
- Answer a
-
\(0+24=24\)
- Answer b
-
\(57+0=57\)
Look at the pairs of sums.
| 2 + 3 = 5 | 3 + 2 = 5 |
| 4 + 7 = 11 | 7 + 4 = 11 |
| 8 + 9 = 17 | 9 + 8 = 17 |
Notice that when the order of the addends is reversed, the sum does not change. This property is called the Commutative Property of Addition, which states that changing the order of the addends does not change their sum.
Changing the order of the addends a and b does not change their sum.
\[a + b = b + a\]
Add:
- \(8 + 7\)
- \(7 + 8\)
Solution
- \(\begin{align*} 8+7 & \\ 15 & \end{align*}\)
- \(\begin{align*} 7 + 8 & \\ 15 & \end{align*}\)
Add: \(9 + 7\) and \(7 + 9\).
- Answer
-
\(9+7=16; 7+9=16\)
Add: \(8 + 6\) and \(6 + 8\).
- Answer
-
\(8+6=14; 6+8=14\)
Add: \(28 + 61\).
Solution
To add numbers with more than one digit, it is often easier to write the numbers vertically in columns.
| Write the numbers so the ones and tens digits line up vertically. | |
| Then add the digits in each place value. Add the ones: 8 + 1 = 9. | |
| Add the tens: 2 + 6 = 8. | 89 |
Add: \(32 + 54\).
- Answer
-
\(32+54=86\)
Add: \(25 + 74\).
- Answer
-
\(25+74=99\)
In the previous example, the sum of the ones and the sum of the tens were both less than \(10\). But what happens if the sum is \(10\) or more? Let’s use our base-\(10\) model to find out. Figure \(\PageIndex{2}\) shows the addition of \(17\) and \(26\) again.
Figure \(\PageIndex{2}\)
When we add the ones, \(7 + 6\), we get \(13\) ones. Because we have more than \(10\) ones, we can exchange \(10\) of the ones for \(1\) ten. Now we have \(4\) tens and \(3\) ones. Without using the model, we show this as a small red \(1\) above the digits in the tens place.
When the sum in a place value column is greater than \(9\), we carry over to the next column to the left. Carrying is the same as regrouping by exchanging. For example, \(10\) ones for \(1\) ten or \(10\) tens for \(1\) hundred.
Step 1. Write the numbers so each place value lines up vertically.
Step 2. Add the digits in each place value. Work from right to left starting with the ones place. If a sum in a place value is more than \(9\), carry to the next place value.
Step 3. Continue adding each place value from right to left, adding each place value and carrying if needed.
Add: \(43 + 69\).
Solution
| Write the numbers so the digits line up vertically. | |
| Add the digits in each place. Add the ones: 3 + 9 = 12. | |
| Write the 2 in the ones place in the sum. Add the 1 ten to the tens place. | |
| Now add the tens: 1 + 4 + 6 = 11. Write the 11 in the sum. |
Add: \(35 + 98\).
- Answer
-
\(35+98=133\)
Add: \(72 + 89\).
- Answer
-
\(72+89=161\)
Add: \(324 + 586\).
Solution
| Write the numbers so the digits line up vertically. | |
| Add the digits in each place value. Add the ones: 4 + 6 = 10. Write the 0 in the ones place in the sum and carry the 1 ten to the tens place. | |
| Add the tens: 1 + 2 + 8 = 11. Write the 1 in the tens place in the sum and carry the 1 hundred to the hundreds. | |
| Add the hundreds: 1 + 3 + 5 = 9. Write the 9 in the hundreds place. |
Add: \(456 + 376\).
- Answer
-
\(456+376=832\)
Add: \(269 + 578\).
- Answer
-
\(269+578=847\)
Add: \(1,683 + 479\).
Solution
| Write the numbers so the digits line up vertically. | |
| Add the digits in each place value | |
| Add the ones: 3 + 9 = 12. Write the 2 in the ones place of the sum and carry the 1 ten to the tens place. | |
| Add the tens: 1 + 7 + 8 = 16. Write the 6 in the tens place and carry the 1 hundred to the hundreds place. | |
| Add the hundreds: 1 + 6 + 4 = 11. Write the 1 in the hundreds place and carry the 1 thousand to the thousands place | |
| Add the thousands 1 + 1 = 2. Write the 2 in the thousands place of the sum. |
When the addends have different numbers of digits, be careful to line up the corresponding place values starting with the ones and moving toward the left.
Add: \(4,597 + 685\).
- Answer
-
\(4,597+685=5,282\)
Add: \(5,837 + 695\).
- Answer
-
\(5,837+695=6,532\)
Add: \(21,357 + 861 + 8,596\).
Solution
| Write the numbers so the place values line up vertically. | |
| Add the digits in each place value. | |
| Add the ones: 7 + 1 + 6 = 14. Write the 4 in the ones place of the sum and carry the 1 to the tens place. | |
| Add the tens: 1 + 5 + 6 + 9 = 21. Write the 1 in the tens place and carry the 2 to the hundreds place. | |
| Add the hundreds: 2 + 3 + 8 + 5 = 18. Write the 8 in the hundreds place and carry the 1 to the thousands place. | |
| Add the thousands 1 + 1 + 8 = 10. Write the 0 in the thousands place and carry the 1 to the ten thousands place. | |
| Add the ten-thousands 1 + 2 = 3. Write the 3 in the ten thousands place in the sum. |
This example had three addends. We can add any number of addends using the same process as long as we are careful to line up the place values correctly.
Add: \(46,195 + 397 + 6,281\).
- Answer
-
\(46,195 + 397 + 6,281=52,873\)
Add: \(53,762 + 196 + 7,458\).
- Answer
-
\(53,762 + 196 + 7,458=61,416\)
Contributors and Attributions
- Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.