Section A.1 Set Notation and Types of Numbers
When we talk about how many or how much of something we have, it often makes sense to use different types of numbers. For example, if we are counting dogs in a shelter, the possibilities are only \(0,1,2,\ldots\text{.}\) (It would be difficult to have \(\frac{1}{2}\) of a dog.) On the other hand if you were weighing a dog in pounds, it doesn't make sense to only allow yourself to work with whole numbers. The dog might weigh something like \(28.35\) pounds. These examples highlight how certain kinds of numbers are appropriate for certain situations. We'll classify various types of numbers in this section.
Subsection A.1.1 Set Notation
What is the mathematical difference between these three âlists?â
To a mathematician, the last one, \((28, 31, 30)\) is an ordered triple. What matters is not merely the three numbers, but also the order in which they come. The ordered triple \((28, 31, 30)\) is not the same as \((30, 31, 28)\text{;}\) they have the same numbers in them, but the order has changed. For some context, February has \(28\) days; then March has \(31\) days; then April has \(30\) days. The order of the three numbers is meaningful in that context.
With curly braces and \(\{28, 31, 30\}\text{,}\) a mathematician sees a collection of numbers and does not particularly care in which order they are written. Such a collection is called a set. All that matters is that these numbers are part of a collection. They've been written in some particular order because that's necessary to write them down. But you might as well have put the three numbers in a bag and shaken up the bag. For some context, maybe your favorite three NBA players have jersey numbers \(30\text{,}\) \(31\text{,}\) and \(28\text{,}\) and you like them all equally well. It doesn't really matter what order you use to list them.
So we can say:
What about just writing \(28, 31, 30\text{?}\) This list of three numbers is ambiguous. Without the curly braces or parentheses, it's unclear to a reader if the order is important. Set notation is the use of curly braces to surround a list/collection of numbers, and we will use set notation frequently in this section.
Checkpoint A.1.2. Set Notation.
Practice using (and not using) set notation.
Evaluate \({4q-9q^{2}-8}\) for \(q=-6\text{.}\)
Subsection A.1.2 Different Number Sets
In the introduction, we mentioned how different sets of numbers are appropriate for different situations. Here are the basic sets of numbers that are used in basic algebra.
- Natural Numbers
-
When we count, we begin: \(1, 2, 3, \dots\) and continue on in that pattern. These numbers are known as natural numbers.
\(\mathbb{N}=\{1,2,3,\dots \}\)
- Whole Numbers
-
If we include zero, then we have the set of whole numbers.
\(\{0,1,2,3,\dots \}\) has no standard symbol, but some options are \(\mathbb{N}_0\text{,}\) \(\mathbb{N}\cup\{0\}\text{,}\) and \(\mathbb{Z}_{\geq0}\text{.}\)
- Integers
-
If we include the negatives of whole numbers, then we have the set of integers.
\(\mathbb{Z}=\{\dots,-3,-2,-1,0,1,2,3,\dots \}\text{.}\)
A \(\mathbb{Z}\) is used because one word in German for ânumbersâ is âZahlen.â
- Rational Numbers
-
A rational number is any number that can be written as a fraction of integers, where the denominator is nonzero. Alternatively, a rational number is any number that can be written with a decimal that terminates or that repeats.
\(\mathbb{Q}=\left\{0,1,-1,2,\frac{1}{2},-\frac{1}{2},-2,3,\frac{1}{3},-\frac{1}{3},-3,\frac{3}{2},\frac{2}{3}\ldots\right\}\)
\(\mathbb{Q}=\left\{0,1,-1,2,0.5,-0.5,-2,3,0.\overline{3},-0.\overline{3},-3,1.5,0.\overline{6}\ldots\right\}\)
A \(\mathbb{Q}\) is used because fractions are quotients of integers.
- Irrational Numbers
-
Any number that cannot be written as a fraction of integers belongs to the set of irrational numbers. Another way to say this is that any number whose decimal places goes on forever without repeating is an irrational number. Some examples include \(\pi\approx3.1415926\ldots\text{,}\) \(\sqrt{15}\approx3.87298\ldots\text{,}\) \(e\approx2.71828\ldots\)
There is no standard symbol for the set of irrational numbers.
- Real Numbers
-
Any number that can be marked somewhere on a number line is a real number. Real numbers might be the only numbers you are familiar with. For a number to not be real, you have to start considering things called complex numbers, which are not our concern right now.
The set of real numbers can be denoted with \(\mathbb{R}\) for short.
Warning A.1.4. Rational Numbers in Other Forms.
Any number that can be written as a ratio of integers is rational, even if it's not written that way at first. For example, these numbers might not look rational to you at first glance: \(-4\text{,}\) \(\sqrt{9}\text{,}\) \(0\pi\text{,}\) and \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\text{.}\) But they are all rational, because they can respectively be written as \(\frac{-4}{1}\text{,}\) \(\frac{3}{1}\text{,}\) \(\frac{0}{1}\text{,}\) and \(\frac{1}{1}\text{.}\)
Example A.1.5. Determine If Numbers Are This Type or That Type.
Determine which numbers from the set \(\left\{-102, -7.25, 0, \frac{\pi}{4}, 2, \frac{10}{3}, \sqrt{19}, \sqrt{25}, 10.\overline{7} \right\}\) are natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers.
All of these numbers are real numbers, because all of these numbers can be positioned on the real number line.
Each real number is either rational or irrational, and not both. \(-102\text{,}\) \(-7.25\text{,}\) \(0\text{,}\) and \(2\) are rational because we can see directly that their decimal expressions terminate. \(10.\overline{7}\) is also rational, because its decimal expression repeats. \(\frac{10}{3}\) is rational because it is a ratio of integers. And last but not least, \(\sqrt{25}\) is rational, because that's the same thing as \(5\text{.}\)
This leaves only \(\frac{\pi}{4}\) and \(\sqrt{19}\) as irrational numbers. Their decimal expressions go on forever without entering a repetitive cycle.
Only \(-102\text{,}\) \(0\text{,}\) \(2\text{,}\) and \(\sqrt{25}\) (which is really \(5\)) are integers.
Of these, only \(0\text{,}\) \(2\text{,}\) and \(\sqrt{25}\) are whole numbers, because whole numbers excludes the negative integers.
Of these, only \(2\) and \(\sqrt{25}\) are natural numbers, because the natural numbers exclude \(0\text{.}\)
Checkpoint A.1.6.
Evaluate \({\left(-2y\right)^{2}}\) for \(y=-2\text{.}\)
Checkpoint A.1.7.
In the introduction, we mentioned that the different types of numbers are appropriate in different situation. Which number set do you think is most appropriate in each of the following situations?
Evaluate \({\left(7r\right)^{2}}\) for \(r=-5\text{.}\)
Subsection A.1.3 Converting Repeating Decimals to Fractions
We have learned that a terminating decimal number is a rational number. It's easy to convert a terminating decimal number into a fraction of integers: you just need to multiply and divide by one of the numbers in the set \(\{10,100,1000,\ldots\}\text{.}\) For example, when we say the number \(0.123\) out loud, we say âone hundred and twenty-three thousandths.â While that's a lot to say, it makes it obvious that this number can be written as a ratio:
Similarly,
demonstrating how any terminating decimal can be written as a fraction.
Repeating decimals can also be written as a fraction. To understand how, use a calculator to find the decimal for, say, \(\frac{73}{99}\) and \(\frac{189}{999}\) You will find that
The pattern is that dividing a number by a number from \(\{9,99,999,\ldots\}\) with the same number of digits will create a repeating decimal that starts as â\(0\text{.}\)â and then repeats the numerator. We can use this observation to reverse engineer some fractions from repeating decimals.
Checkpoint A.1.8.
Evaluate \({\left(-5a\right)^{3}}\) for \(a=2\text{.}\)
Converting a repeating decimal to a fraction is not always quite this straightforward. There are complications if the number takes a few digits before it begins repeating. For your interest, here is one example on how to do that.
Example A.1.9.
Can we convert the repeating decimal \(9.134343434\ldots=9.1\overline{34}\) to a fraction? The trick is to separate its terminating part from its repeating part, like this:
Now note that the terminating part is \(\frac{91}{10}\text{,}\) and the repeating part is almost like our earlier examples, except it has an extra \(0\) right after the decimal. So we have:
With what we learned in the earlier examples and basic fraction arithmetic, we can continue:
Check that this is right by entering \(\frac{9043}{990}\) into a calculator and seeing if it returns the decimal we started with, \(9.134343434\ldots\text{.}\)
Exercises A.1.4 Exercises
Review and Warmup
1.
Evaluate \({\left(5b\right)^{3}}\) for \(b=8\text{.}\)
\(64000\)
2.
Evaluate \({\sqrt{A-7}-3}\) for \(A=32\text{.}\)
\(2\)
3.
Evaluate \({\sqrt{B+1}-8}\) for \(B=3\text{.}\)
\(-6\)
4.
Evaluate \({9\sqrt{m+8}+5}\) for \(m=41\text{.}\)
\(68\)
5.
Evaluate \({9-6\sqrt{p-3}}\) for \(p=19\text{.}\)
\(-15\)
6.
Evaluate \({\left|q+5\right|-5}\) for \(q=-8\text{.}\)
\(-2\)
Set Notation
11.
Evaluate \(\frac{y_2-y_1}{x_2-x_1}\) for \(x_1=1\text{,}\) \(x_2=-4\text{,}\) \(y_1=8\text{,}\) and \(y_2=6\text{.}\)
\({\textstyle\frac{2}{5}}\)
12.
Evaluate \(\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}\) for \(x_1=-5\text{,}\) \(x_2=0\text{,}\) \(y_1=8\text{,}\) and \(y_2=20\text{.}\)
\(13\)
13.
Evaluate \(\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}\) for \(x_1=4\text{,}\) \(x_2=-1\text{,}\) \(y_1=8\text{,}\) and \(y_2=-4\text{.}\)
\(13\)
14.
Evaluate the algebraic expression \(-5 a + b\) for \(a = {\textstyle\frac{8}{7}}\) and \(b = {\textstyle\frac{5}{9}}\text{.}\)
Answer:
\(-{\textstyle\frac{325}{63}}\)
Types of Numbers
15.
Evaluate the algebraic expression \(5 a + b\) for \(a = {\textstyle\frac{9}{7}}\) and \(b = {\textstyle\frac{8}{3}}\text{.}\)
Answer:
\({\textstyle\frac{191}{21}}\)
16.
Evaluate each algebraic expression for the given value(s):
\(\displaystyle\frac{2+4|x-y|}{x+5 y}\text{,}\) for \(x = 10\) and \(y = 11\text{:}\)
\({\textstyle\frac{6}{65}}\)
17.
Evaluate each algebraic expression for the given value(s):
\(\displaystyle\frac{2+3|x-y|}{x+3 y}\text{,}\) for \(x = 11\) and \(y = -2\text{:}\)
\({\textstyle\frac{41}{5}}\)
18.
To convert a temperature measured in degrees Fahrenheit to degrees Celsius, there is a formula:
where \(C\) represents the temperature in degrees Celsius and \(F\) represents the temperature in degrees Fahrenheit.
If a temperature is \(41 {^\circ}\text{F}\text{,}\) what is that temperature measured in Celsius?
\(5\ {\rm degC}\)
We substitute \(F=41\) into the formula, and we have:
\(\displaystyle{ C={{\textstyle\frac{5}{9}}}(F-32)}\)
\(\displaystyle{ C={{\textstyle\frac{5}{9}}}(41-32) }\)
\(\displaystyle{ C={{\textstyle\frac{5}{9}}} \cdot 9 }\)
\(\displaystyle{ C={5\ {\rm degC}}. }\)
So the temperature is \({5\ {\rm degC}}\text{.}\)
19.
To convert a temperature measured in degrees Fahrenheit to degrees Celsius, there is a formula:
where \(C\) represents the temperature in degrees Celsius and \(F\) represents the temperature in degrees Fahrenheit.
If a temperature is \(50 {^\circ}\text{F}\text{,}\) what is that temperature measured in Celsius?
\(10\ {\rm degC}\)
We substitute \(F=50\) into the formula, and we have:
\(\displaystyle{ C={{\textstyle\frac{5}{9}}}(F-32)}\)
\(\displaystyle{ C={{\textstyle\frac{5}{9}}}(50-32) }\)
\(\displaystyle{ C={{\textstyle\frac{5}{9}}} \cdot 18 }\)
\(\displaystyle{ C={10\ {\rm degC}}. }\)
So the temperature is \({10\ {\rm degC}}\text{.}\)
20.
A formula for converting hours into seconds is
where \(H\) is a number of hours, and \(S\) is the corresponding number of seconds.
Use the formula to find the number of seconds that corresponds to six hours.
seconds corresponds to six hours.
\(21600\)
Since we are working with six hours, we have that \(H=6\text{.}\) We can substitute this into the formula:
\(\begin{aligned} S \amp = 3600H\\ \amp = 3600(6)\\ \amp = {21600} \end{aligned}\)
So there are 21600 seconds in six hours.
21.
A formula for converting hours into seconds is
where \(H\) is a number of hours, and \(S\) is the corresponding number of seconds.
Use the formula to find the number of seconds that corresponds to eighteen hours.
seconds corresponds to eighteen hours.
\(64800\)
Since we are working with eighteen hours, we have that \(H=18\text{.}\) We can substitute this into the formula:
\(\begin{aligned} S \amp = 3600H\\ \amp = 3600(18)\\ \amp = {64800} \end{aligned}\)
So there are 64800 seconds in eighteen hours.
22.
The formula
gives the vertical position of an object, at time \(t\text{,}\) thrown with an initial velocity \(v_0\text{,}\) from an initial position \(y_0\) in a place where the acceleration of gravity is \(a\text{.}\) The acceleration of gravity on earth is \({-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}\text{.}\) It is negative, because we consider the upward direction as positive in this situation, and gravity pulls down.
What is the height of a baseball thrown with an initial velocity of \(v_0={87\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{,}\) from an initial position of \(y_0= {76\ {\rm m}}\text{,}\) and at time \(t={9\ {\rm s}}\text{?}\)
Nine seconds after the baseball was thrown, it was high in the air.
\(462.1\ {\rm m}\)
These given values can be substituted into the formula as follows:
\(\displaystyle{\begin{aligned}[t] y \amp = \frac{1}{2}\,a\,t^2 +v_0\,t + y_0 \\ \amp = \frac{1}{2}\left({-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}\right)({9\ {\rm s}})^{2} + \left({87\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\right)({9\ {\rm s}}) +{76\ {\rm m}} \\ \amp = -396.9 \text{ m} + 783 \text{ m} + {76\ {\rm m}} \\ \amp = {462.1\ {\rm m}} \end{aligned} }\)
The baseball was \({462.1\ {\rm m}}\) high in the air.
23.
The formula
gives the vertical position of an object, at time \(t\text{,}\) thrown with an initial velocity \(v_0\text{,}\) from an initial position \(y_0\) in a place where the acceleration of gravity is \(a\text{.}\) The acceleration of gravity on earth is \({-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}\text{.}\) It is negative, because we consider the upward direction as positive in this situation, and gravity pulls down.
What is the height of a baseball thrown with an initial velocity of \(v_0={93\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{,}\) from an initial position of \(y_0= {58\ {\rm m}}\text{,}\) and at time \(t={17\ {\rm s}}\text{?}\)
Seventeen seconds after the baseball was thrown, it was high in the air.
\(222.9\ {\rm m}\)
These given values can be substituted into the formula as follows:
\(\displaystyle{\begin{aligned}[t] y \amp = \frac{1}{2}\,a\,t^2 +v_0\,t + y_0 \\ \amp = \frac{1}{2}\left({-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}\right)({17\ {\rm s}})^{2} + \left({93\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\right)({17\ {\rm s}}) +{58\ {\rm m}} \\ \amp = -1416.1 \text{ m} + 1581 \text{ m} + {58\ {\rm m}} \\ \amp = {222.9\ {\rm m}} \end{aligned} }\)
The baseball was \({222.9\ {\rm m}}\) high in the air.
24.
The percentage of births in the U.S. delivered via C-section can be given by the following formula for the years since 1996:
In this formula \(y\) is a year after 1996 and \(p\) is the percentage of births delivered via C-section for that year.
What percentage of births in the U.S. were delivered via C-section in the year 2013?
of births in the U.S. were delivered via C-section in the year 2013.
\(34.6\%\)
Since we are working with the year 2013, we have that \(y=2013\text{.}\) We can substitute this into the formula:
\(\begin{aligned} p \amp = 0.8(y-1996)+21\\ \amp = 0.8(2013-1996)+21\\ \amp = 0.8(17)+21\\ \amp = {13.6}+21\\ \amp = {34.6} \end{aligned}\)
So in 2013, \({34.6}\)% of births in the U.S were delivered via C-section.
25.
The percentage of births in the U.S. delivered via C-section can be given by the following formula for the years since 1996:
In this formula \(y\) is a year after 1996 and \(p\) is the percentage of births delivered via C-section for that year.
What percentage of births in the U.S. were delivered via C-section in the year 1999?
of births in the U.S. were delivered via C-section in the year 1999.
\(23.4\%\)
Since we are working with the year 1999, we have that \(y=1999\text{.}\) We can substitute this into the formula:
\(\begin{aligned} p \amp = 0.8(y-1996)+21\\ \amp = 0.8(1999-1996)+21\\ \amp = 0.8(3)+21\\ \amp = {2.4}+21\\ \amp = {23.4} \end{aligned}\)
So in 1999, \({23.4}\)% of births in the U.S were delivered via C-section.
26.
Target heart rate for moderate exercise is \(50\%\) to \(70\%\) of maximum heart rate. If we want to represent a certain percent of an individualâs maximum heart rate, weâd use the formula
where \(p\) is the percent, and \(a\) is age in years.
Determine the target heart rate at \(54\%\) level for someone who is \(17\) years old. Round your answer to an integer.
The target heart rate at \(54\%\) level for someone who is \(17\) years old is beats per minute.
\(110\)
We will substitute \(p=0.54\) and \(a=17\) into the formula \({p\!\left(220-a\right)}\text{,}\) and we have:
\(\displaystyle{\begin{aligned} \text{rate} \amp = p(220-a) \\ \amp = 0.54(220-17)\\ \amp = 0.54(203)\\ \amp = 109.62\\ \amp \approx 110 \end{aligned}}\)
The target heart rate at \(54\%\) level for someone who is \(17\) years old is \(110\) beats per minute.
27.
Target heart rate for moderate exercise is \(50\%\) to \(70\%\) of maximum heart rate. If we want to represent a certain percent of an individualâs maximum heart rate, weâd use the formula
where \(p\) is the percent, and \(a\) is age in years.
Determine the target heart rate at \(56\%\) level for someone who is \(57\) years old. Round your answer to an integer.
The target heart rate at \(56\%\) level for someone who is \(57\) years old is beats per minute.
\(91\)
We will substitute \(p=0.56\) and \(a=57\) into the formula \({p\!\left(220-a\right)}\text{,}\) and we have:
\(\displaystyle{\begin{aligned} \text{rate} \amp = p(220-a) \\ \amp = 0.56(220-57)\\ \amp = 0.56(163)\\ \amp = 91.28\\ \amp \approx 91 \end{aligned}}\)
The target heart rate at \(56\%\) level for someone who is \(57\) years old is \(91\) beats per minute.
28.
The diagonal length (\(D\)) of a rectangle with side lengths \(L\) and \(W\) is given by:
Determine the diagonal length of rectangles with \(L={12\ {\rm ft}}\) and \(W={5\ {\rm ft}}\text{.}\)
The diagonal length of rectangles with \(L={12\ {\rm ft}}\) and \(W={5\ {\rm ft}}\) is .
\(13\ {\rm ft}\)
We will substitute \(L=12\) and \(W=5\) into the formula \(D=\sqrt{L^2+W^2}\text{,}\) and we have:
\(\displaystyle{\begin{aligned} D \amp = \sqrt{L^2+W^2} \\ \amp = \sqrt{(12)^2+(5)^2} \\ \amp = \sqrt{144+25} \\ \amp = \sqrt{169} \\ \amp = 13 \end{aligned}}\)
The diagonal length of rectangles with \(L={12\ {\rm ft}}\) and \(W={5\ {\rm ft}}\) is \({13\ {\rm ft}}\text{.}\)
Writing Decimals as Fractions
29.
The diagonal length (\(D\)) of a rectangle with side lengths \(L\) and \(W\) is given by:
Determine the diagonal length of rectangles with \(L={24\ {\rm ft}}\) and \(W={10\ {\rm ft}}\text{.}\)
The diagonal length of rectangles with \(L={24\ {\rm ft}}\) and \(W={10\ {\rm ft}}\) is .
\(26\ {\rm ft}\)
We will substitute \(L=24\) and \(W=10\) into the formula \(D=\sqrt{L^2+W^2}\text{,}\) and we have:
\(\displaystyle{\begin{aligned} D \amp = \sqrt{L^2+W^2} \\ \amp = \sqrt{(24)^2+(10)^2} \\ \amp = \sqrt{576+100} \\ \amp = \sqrt{676} \\ \amp = 26 \end{aligned}}\)
The diagonal length of rectangles with \(L={24\ {\rm ft}}\) and \(W={10\ {\rm ft}}\) is \({26\ {\rm ft}}\text{.}\)
30.
The height inside a camping tent when you are \(d\) feet from the edge of the tent is given by
where \(h\) stands for height in feet.
Determine the height when you are:
-
\({10.2\ {\rm ft}}\) from the edge.
The height inside a camping tent when you are \({10.2\ {\rm ft}}\) from the edge of the tent is .
-
\({3.3\ {\rm ft}}\) from the edge.
The height inside a camping tent when you are \({3.3\ {\rm ft}}\) from the edge of the tent is .
\(3.8\ {\rm ft}\)
\(4.75\ {\rm ft}\)
-
We will substitute \(d=10.2\) into the formula \({-0.5\!\left|d-5.8\right|+6}\text{,}\) and we have: \(\displaystyle{\begin{aligned}[t] h \amp = {-0.5\!\left|d-5.8\right|+6} \\ \amp = -0.5 \lvert 10.2-5.8 \rvert +6 \\ \amp = -0.5 \lvert 4.4 \rvert +6 \\ \amp = -0.5(4.4)+6 \\ \amp = -2.2+6 \\ \amp = 3.8 \end{aligned}}\)
The height inside a camping tent when you are \({10.2\ {\rm ft}}\) from the edge of the tent is \({3.8\ {\rm ft}}\text{.}\)
-
We will substitute \(d=3.3\) into the formula \({-0.5\!\left|d-5.8\right|+6}\text{,}\) and we have: \(\displaystyle{\begin{aligned}[t] h \amp = {-0.5\!\left|d-5.8\right|+6} \\ \amp = -0.5 \lvert 3.3-5.8 \rvert +6 \\ \amp = -0.5 \lvert -2.5 \rvert +6 \\ \amp = -0.5(2.5)+6 \\ \amp = -1.25+6 \\ \amp = 4.75 \end{aligned}}\)
The height inside a camping tent when you are \({3.3\ {\rm ft}}\) from the edge of the tent is \({4.75\ {\rm ft}}\text{.}\)
31.
The height inside a camping tent when you are \(d\) feet from the edge of the tent is given by
where \(h\) stands for height in feet.
Determine the height when you are:
-
\({8.5\ {\rm ft}}\) from the edge.
The height inside a camping tent when you are \({8.5\ {\rm ft}}\) from the edge of the tent is .
-
\({4.8\ {\rm ft}}\) from the edge.
The height inside a camping tent when you are \({4.8\ {\rm ft}}\) from the edge of the tent is .
\(3.43\ {\rm ft}\)
\(4.24\ {\rm ft}\)
-
We will substitute \(d=8.5\) into the formula \({-0.9\!\left|d-6.2\right|+5.5}\text{,}\) and we have: \(\displaystyle{\begin{aligned}[t] h \amp = {-0.9\!\left|d-6.2\right|+5.5} \\ \amp = -0.9 \lvert 8.5-6.2 \rvert +5.5 \\ \amp = -0.9 \lvert 2.3 \rvert +5.5 \\ \amp = -0.9(2.3)+5.5 \\ \amp = -2.07+5.5 \\ \amp = 3.43 \end{aligned}}\)
The height inside a camping tent when you are \({8.5\ {\rm ft}}\) from the edge of the tent is \({3.43\ {\rm ft}}\text{.}\)
-
We will substitute \(d=4.8\) into the formula \({-0.9\!\left|d-6.2\right|+5.5}\text{,}\) and we have: \(\displaystyle{\begin{aligned}[t] h \amp = {-0.9\!\left|d-6.2\right|+5.5} \\ \amp = -0.9 \lvert 4.8-6.2 \rvert +5.5 \\ \amp = -0.9 \lvert -1.4 \rvert +5.5 \\ \amp = -0.9(1.4)+5.5 \\ \amp = -1.26+5.5 \\ \amp = 4.24 \end{aligned}}\)
The height inside a camping tent when you are \({4.8\ {\rm ft}}\) from the edge of the tent is \({4.24\ {\rm ft}}\text{.}\)
32.
List the terms in the expression \(5x-4x+10z\text{.}\)
33.
Which expressions have like terms that you can combine?
\(10x+3y\)
can
cannot
\(4x-8x\)
can
cannot
\(9y-4y\)
can
cannot
\(-6x+17z\)
can
cannot
\(-3x-7x\)
can
cannot
\(5t+8t^2\)
can
cannot
\(\text{cannot}\)
\(\text{can}\)
\(\text{can}\)
\(\text{cannot}\)
\(\text{can}\)
\(\text{cannot}\)
The terms that we can combine have the same variable part, including any exponents.
\(10x+3y\) cannot be combined.
\(\displaystyle 4x-8x=-4x\)
\(\displaystyle 9y-4y=5y\)
\(-6x+17z\) cannot be combined.
\(\displaystyle -3x-7x=-10x\)
\(5t+8t^2\) cannot be combined.
34.
Simplify each expression, if possible, by combining like terms.
\(x+0.25x\)
\(\frac{4}{9}x-\frac{7}{10}y+\frac{2}{3}x\)
\(\frac{5}{6}y-\frac{8}{15}y+\frac{2}{3}x^2\)
\(4x+1.5y-9z\)
\(1.25x\)
\({\textstyle\frac{10}{9}}x-{\textstyle\frac{7}{10}}y\)
\({\textstyle\frac{3}{10}}y+{\textstyle\frac{2}{3}}x^{2}\)
\(4x+1.5y-9z\)
Rewrite this expression as \(1.00x+0.25x\) and simplify to get \(1.25x\text{.}\)
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This expression has two like terms that can be combined: \(\frac{4}{9}x\) and \(\frac{2}{3}x\text{.}\) To combine them, we need to add the fractions \(\frac{4}{9}+\frac{2}{3}\text{.}\) (You may find a review of fraction addition in Section A.2.) Here, we have
\begin{equation*} \begin{aligned} \frac{4}{9}+\frac{2}{3}\amp=\frac{4}{9}+\frac{6}{9}\\ \amp=\frac{10}{9} \end{aligned} \end{equation*}so \(\frac{4}{9}x+\frac{2}{3}x=\frac{10}{9}x\text{.}\) And together with the third term, the answer is \(\frac{10}{9}x-\frac{7}{10}y\text{.}\)
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In this expression we can combine the \(y\) terms, but we need to subtract the fractions \(\frac{5}{6}-\frac{8}{15}\text{.}\) (You may find a review of fraction subtraction in Section A.2.) Here, we have
\begin{equation*} \begin{aligned} \frac{5}{6}-\frac{8}{15}\amp=\frac{25}{30}-\frac{16}{30}\\ \amp=\frac{9}{30}\\ \amp=\frac{3}{10} \end{aligned} \end{equation*}so \(\frac{5}{6}y-\frac{8}{15}y=\frac{3}{10}y\text{.}\) And together with the third term, the answer is \(\frac{3}{10}y+\frac{2}{3}x^2\text{.}\)
This expression cannot be simplified further because there are not any like terms.
Challenge
35.
Add the following.
\(1+(-6)\)
\(10+(-4)\)
\(6+(-6)\)
\(-5\)
\(6\)
\(0\)
Here are two different explanations of how two negative numbers can be added together.
METHOD 1
Use a number line. Letâs find \(1+(-6)\text{.}\)
First, find \(1\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(6\) units. We will reach \({-5}\) on the number line, which is the answer.
So, \(1+(-6)={-5}\text{.}\)
Similarly, \(10+(-4)={6}\text{,}\) and \(6+(-6)={0}\text{.}\)
METHOD 2
A second method asks you to think in terms of money. Letâs find \(1+(-6)\text{.}\)
The first number is \(1\text{.}\) Since itâs positive, itâs like you won \(1\) dollars while gambling at the casino this morning.
The second number is \(-6\text{.}\) Since itâs negative, itâs like you lost \(6\) dollars while gambling at the casino this afternoon.
Since you lost more money than you won, overall you have lost, implying the answer is negative.
Since you won \(1\) dollars and then lost \(6\) dollars, it makes sense that you lost the difference of \(6\) and \(1\) dollars, which is \({5}\) dollars. So the final answer is: \(1+(-6)={-5}\text{.}\)