Subsection Parentheses and Fraction Bars
We can use parentheses to override the multiplicationfirst rule. Compare the two expressions below.
\begin{align*}
\amp\text{The sum of 4 times 6 and 10} \amp\amp \alert{4 \cdot 6} + 10\\
\amp\text{4 times the sum of 6 and 10} \amp\amp 4(\alert{6 + 10})
\end{align*}
In the first expression, we perform the multiplication \(4\times 6\) first, but in the second expression we perform the addition \(6 + 10\) first, because it is enclosed in parentheses.
The location (or absence) of parentheses can drastically alter the meaning of an expression. In the following example, note how the location of the parentheses changes the value of the expression.
Example A.1
\(\begin{aligned}[t]53\cdot \alert{4^2} \amp = 53\cdot 16\\
\amp = 548=43
\end{aligned} \)
\(\begin{aligned}[t]5(\alert{3\cdot 4})^2 \amp = 512^2\\
\amp= 5144 = 139
\end{aligned}\)
\(\begin{aligned}[t](5\alert{3\cdot 4})^2 \amp = (512)^2\\
\amp = (7)^2 = 49
\end{aligned}\)
\(\begin{aligned}[t](\alert{53})\cdot 4^2 \amp = 2\cdot 4^2\\
\amp = 2\cdot 16 = 32
\end{aligned}\)
The order of operations mentions other grouping devices besides parentheses: fraction bars and square root bars. Notice how the placement of the fraction bar affects the expressions in the next example.
Example A.3
\(\begin{aligned}[t]
\frac{1+2}{3\cdot 4} \amp = \frac{3}{12}\\
\amp = \frac{1}{4}
\end{aligned}\)
\(\begin{aligned}[t]
1+\frac{2}{3\cdot 4} \amp = 1+ \frac{2}{12}\\
\amp = 1+\frac{1}{6}=\frac{7}{6}
\end{aligned} \)
\(\begin{aligned}[t]
\frac{1+2}{3}\cdot 4 \amp = \frac{3}{3}\cdot 4\\
\amp = 1\cdot 4=4
\end{aligned}\)
\(\begin{aligned}[t]
1+\frac{2}{3}\cdot 4 \amp = 1+ \frac{8}{3}\\
\amp = \frac{3}{3}+\frac{8}{3}=\frac{11}{3}
\end{aligned} \)
Subsection Radicals
You are already familiar with square roots. Every nonnegative number has two square roots, defined as follows:
\begin{equation*}
\blert{s ~~\text{ is a square root of }~~ n ~~\text{ if }~~s^2 = n}
\end{equation*}
There are several other kinds of roots, one of which is called the cube root, denoted by \(\sqrt[3]{n}\text{.}\) We define the cube root as follows.
Cube Roots
\begin{equation*}
b ~~\text{ is a cube root of }~~~ n ~~~\text{ if }~~~ b ~\text{ cubed equals }~~n.
\end{equation*}
In symbols, we write
\begin{equation*}
\blert{b=\sqrt[3]{n} ~~\text{ if } ~~ b^3=n}
\end{equation*}
In the order of operations, simplifying radicals and powers comes after parentheses but before products and quotients.
Example A.5
Simplify each expression.
 \(3\sqrt[3]{8}\)
 \(2\sqrt[3]{125}\)
 \(\displaystyle{\frac{6\sqrt[3]{27}}{2}}\)
Solution
 \(3\sqrt[3]{8}=3(2)=6\)
 \(2\sqrt[3]{125}=2(5)=7\)
 \(\displaystyle{\frac{6\sqrt[3]{27}}{2}=\frac{6(3)}{2}=\frac{9}{2}}\)
Subsection Scientific Notation
¶Scientists and engineers regularly encounter very large numbers such as
\begin{equation*}
5,980,000,000,000,000,000,000,000
\end{equation*}
(the mass of the Earth in kilograms) and very small numbers such as
\begin{equation*}
0.000 \,000 \,000 \,000 \,000 \,000 \,000 \,001 \,67
\end{equation*}
(the mass of a hydrogen atom in grams). These numbers can be written in a more compact and useful form by using powers of \(10\text{.}\)
In our base \(10\) number system, multiplying a number by a positive power of \(10\) has the effect of moving the decimal place \(k\) places to the right, where \(k\) is the exponent in the power of \(10\text{.}\) For example,
\begin{equation*}
3\alert{.}529 \times 10^\alert{2} = 352\alert{.}9 ~~~\text{ and }~~~ 25 \times 10^4 = 250,000
\end{equation*}
Multiplying by a power of \(10\) with a negative exponent moves the decimal place to the left. For example,
\begin{equation*}
1728 \times 10^{3} = 1.728 ~~~\text{ and }~~~ 4.6 \times 10^{5} = 0.000046
\end{equation*}
Using this property, we can write any number as the product of a number between \(1\) and \(10\) (including \(1\)) and a power of \(10\text{.}\) For example, the mass of the Earth and the mass of a hydrogen atom can be expressed as
\begin{equation*}
5.98 \times 1024 \text{ kilograms}\hphantom{blank} \text{ and }\hphantom{blank}1.67 \times 10^{24} \text{ gram}
\end{equation*}
respectively. A number written in this form is said to be expressed in scientific notation.
To Write a Number in Scientific Notation:
Locate the decimal point so that there is exactly one nonzero digit to its left.

Count the number of places you moved the decimal point: This determines the power of \(10\text{.}\)
If the original number is greater than \(10\text{,}\) the exponent is positive.
If the original number is less than \(1\text{,}\) the exponent is negative.
Example A.6
Write each number in scientific notation.
\(\begin{aligned}[t]
478,000 \amp = 4.78000 \times 10^5 \amp \amp \blert{\text{Move the decimal 5 places.}}\\
\amp = 4.78 \times 10^5
\end{aligned}\)
\(\begin{aligned}[t]
0.00032 \amp= 00003.2 \times 10^{4} \amp \amp \blert{\text{Move the decimal 4 places.}}\\
\amp = 3.2 \times 10^{4}
\end{aligned}\)
Example A.7
The average American eats \(110\) kilograms of meat per year. It takes about \(16\) kilograms of grain to produce \(1\) kilogram of meat, and advanced farming techniques can produce about \(6000\) kilograms of grain on each hectare of arable land. (The hectare is \(10,000\) square meters, or just under \(2 \frac{1}{2}\) acres.) Now, the total land area of the Earth is about \(13\) billion hectares, but only about \(11\%\) of that land is arable. Is it possible for each of the \(7.6\) billion people on Earth to eat as much meat as Americans do?
Solution
First we will compute the amount of meat necessary to feed every person on Earth \(110\) kilograms per year. In 2018 there are \(7.6 \times 10^9\) people on Earth.
\begin{equation*}
(5.5\times 10^9 \text{ people})\times (110 \text{ kg/person}) = 8.36 \times 10^{11} \text{ kg of meat}
\end{equation*}
Next we will compute the amount of grain needed to produce that much meat.
\begin{equation*}
(16 \text{ kg of grain/kg of meat})\times (8.36 \times 10^{11} \text{ kg of meat}) = 1.34 \times 10^{13} \text{ kg of grain}
\end{equation*}
Next we will see how many hectares of land are needed to produce that much grain.
\begin{equation*}
(1.34\times 10^{13} \text{ kg of grain})\div(6000 \text{ kg/hectare}) = 2.23\times 10^{11} \text{ hectares}
\end{equation*}
Finally, we will compute the amount of arable land available for grain production.
\begin{equation*}
0.11\times (13\times 10^9 \text{ hectares}) = 1.43\times 10^9 \text{ hectares}
\end{equation*}
Thus, even if we use every hectare of arable land to produce grain for livestock, we will not have enough to provide every person on Earth with \(110\) kilograms of meat per year.
Subsection Exercises A.1
¶
For Problems 126, simplify each expression according to the order of operations.
1
\(\dfrac{3(68)}{2}  \dfrac{6}{2} \)
2
\(\dfrac{5(35)}{2}  \dfrac{18}{3} \)
3
\(6\left[32(4+1)\right] \)
4
\(5\left[3+4(64)\right] \)
5
\((43)\left[2+3(21)\right] \)
6
\((86)\left[5+7(23)\right] \)
7
\(64\div 8\left[42(3+1)\right] \)
8
\(27\div 3\left[93(42)\right] \)
9
\(5\left[3+(81)\right] \div (25) \)
10
\(3\left[2+(61)\right] \div 9 \)
11
\(\left[3(82)+3\right]\cdot \left[24\div 6\right] \)
12
\(\left[2+3(58)\right] \cdot \left[15\div 3\right] \)
13
\(5^2\)
14
\((15)^2 \)
15
\((3)^4\)
16
\(3^4 \)
17
\(4^3\)
18
\((4)^3 \)
19
\((2)^5\)
20
\(2^5 \)
21
\(\dfrac{4\cdot 2^3}{16}+3\cdot 4^2 \)
22
\(\dfrac{4\cdot 3^2}{6}+(3\cdot 4)^2 \)
23
\(\dfrac{3^25}{62^2}  \dfrac{6^2}{3^2} \)
24
\(\dfrac{3\cdot 2^2}{41}+\dfrac{(3)(2)^3}{6} \)
25
\(\dfrac{(5)^23^2}{46}+\dfrac{(3)^2}{2+1} \)
26
\(\dfrac{7^26^2}{10+3}\dfrac{8^2\cdot (2)}{(4)^2} \)
For Problems 2728, compute each cube root. Round your answers to three decimal places if necessary. Verify your answers by cubing them.
27
 \(\sqrt[3]{512} \)
 \(\sqrt[3]{125} \)
 \(\sqrt[3]{0.064} \)
 \(\sqrt[3]{1.728} \)
28
 \(\sqrt[3]{9} \)
 \(\sqrt[3]{258} \)
 \(\sqrt[3]{0.002} \)
 \(\sqrt[3]{3.1} \)
For Problems 2930, simplify each expression according to the order of operations.
29
 \(\dfrac{43\sqrt[3]{64}}{2} \)
 \(\dfrac{4+\sqrt[3]{216}}{88\sqrt[3]{8}} \)
30
 \(\sqrt[3]{3^3+4^3+5^3} \)
 \(\sqrt[3]{9^3+10^31^3} \)
For Problems 3142, use a calculator to simplify each expression.
31
\(\dfrac{8398}{26\cdot 17} \)
32
\(\dfrac{415.112}{8.58+18.73} \)
33
\(\dfrac{112.78+2599.124}{27.56} \)
34
\(\dfrac{202,4629510}{356} \)
35
\(\sqrt{24\cdot 54} \)
36
\(\sqrt{\dfrac{1216}{19}} \)
37
\(\dfrac{11635}{215242} \)
38
\(\dfrac{842987}{443385} \)
39
\(\sqrt{27^2+36^2} \)
40
\(\sqrt{13^24\cdot 21\cdot 2} \)
41
\(\dfrac{27\sqrt{27^24(4)(35)}}{2\cdot 4} \)
42
\(\dfrac{13+\sqrt{13^24(5)(6)}}{2\cdot 5} \)
For Problems 4350, evaluate the expression for the given values of the variable. Use your calculator where appropriate.
43
\(\dfrac{5(F32)}{9} \text{;}\) \(~~~F=212\)
44
\(\dfrac{a4s}{1r} \text{;}\) \(~~~r=2, ~s=12, ~\) and \(~a=4\)
45
\(P+Prt \text{;}\) \(~~~P=1000, ~ r=0.04, ~\) and \(~t=2\)
46
\(R(1+at) \text{;}\) \(~~~R=2.5, ~a=0.05, ~\) and \(~t=20\)
47
\(\dfrac{1}{2}gt^212t \text{;}\) \(~~~g=32 ~\) and \(~t=\dfrac{3}{4} \)
48
\(\dfrac{Mv^2}{g} \text{;}\) \(~~~M=\dfrac{16}{3}, ~a=\dfrac{3}{2}, ~\) and \(~g=32\)
49
\(\dfrac{32(Vv)^2}{g} \text{;}\) \(~~~V=12.78, ~v=4.26, ~\) and \(~g=32 \)
50
\(\dfrac{32(Vv)^2}{g} \text{;}\) \(~~~V=38.3, ~v=6.7, ~\) and \(~g=9.8\)
For Problems 5152, write each number in scientific notation.
51
\(285\)
\(8,372,000\)
\(0.024\)
\(0.000523\)
52
\(68,742\)
\(481,000,000,000\)
\(0.421\)
\(0.000004\)
For Problems 5354, write each number in standard notation.
53
\(2.4 \times 10^{2} \)
\(6.87 \times 10^{15} \)
\(5.0 \times 10^{3} \)
\(2.02 \times 10^{4} \)
54
\(4.8 \times 10^{3} \)
\(8.31 \times 10^{12} \)
\(8.0 \times 10^{1} \)
\(4.31 \times 10^{5} \)
For Problems 5556, compute with the aid of a calculator. Write your answers in standard notation.
55
\(\dfrac{(2.4\times 10^{8})(6.5\times 10^{32})}{5.2\times 10^{18}} \)
\(\dfrac{(7.5\times 10^{13})(3.6\times 10^{9})}{(1.5\times 10^{15})(1.6\times 10^{11})} \)
56
\(\dfrac{(8.4\times 10^{22})(1.6\times 10^{15})}{3.2\times 10^{11}} \)
\(\dfrac{(9.4\times 10^{24})(7.2\times 10^{18})}{(4.5\times 10^{26})(6.4\times 10^{16})} \)
57
In 2018, the public debt of the United States was over $20,620,000,000,000.
Express this number in scientific notation.
If the population of the United States in 2018 was 327,112,000, what was the per capita debt (the debt per person) in 2018?
58
A lightyear is the number of miles traveled by light in 1 year (365 days). The speed of light is approximately 186,000 miles per second.
Compute the number of miles in 1 lightyear, and express your answer in scientific notation.
The star nearest to the Sun is Proxima Centauri, at a distance of 4.3 lighthears. How long would it take Pioneer 10 (the first space vehicle to achieve escape velocity from the solar system), traveling at 32,114 miles per hour, to reach Proxima Centauri?
59
The diameter of the galactic disk is about \(1.2\times 10^{18} \) kilometers, and our Sun lies about halfway from the center of the galaxy to the edge of the disk. The Sun orbits the galactic center once in 240 million years.
What is the speed of the Sun in its orbit, in kilometers per year?
What is its speed in meters per second?
60
Lake Superior has an area of 31,700 square miles and an average depth of 483 feet.
Find the approximate volume of Lake Superior in cubic feet.
If 1 cubic foot of water is equivalent to 7.48 gallons, how many gallons of water are in Lake Superior?
61
The average distance from the Earth to the Sun is \(1.5\times 10^{11} \) meters. The distance from the Sun to Proxima Centauri, the next closest star, is \(3.99\times 10^{16} \) meters. The most distant star visible to the unaided eye are 2000 times as far away as Proxima Centauri.
How many times farther is Proxima Centauri from the Sun than the Sun is from Earth?
How far from the Sun are the most distant visible stars?
62
The radius of the Earth is \(6.37 \times 10^6 \) meters, and the radius of the Sun is \(6.96\times 10^{8} \) meters. The radii of the other stars range from 1% of the solar radius to 1000 times the solar radius.
What fraction of the solar radius is the Earth's radius?
What is the range of stellar radii, in meters?