Adelante/CompleteCheck-LargeBatch · Datasets at Hugging Face

Q stringlengths 4 3.96k	A stringlengths 1 3k	Result stringclasses 4 values
4.32 is what percent of 72 ? Missing factor statement.	\n\n\\begin{matrix} \\text{(percentage)} & \\underset{ \\downarrow }{ = } & \\text{(percent)} & & \\vdots & \\text{(base)} & {\\[ \\text{(product)} = \\text{(factor)} \\cdot \\text{(factor)}\\] } \\\\ & \\downarrow & & \\downarrow & & \\downarrow & \\end{matrix}\n\n\\( \\begin{array}{lllll} {4.32} & = & Q & \\cdot & {72} \\end{array} \\)\n\n\\( Q = {4.32} \\div {72}\\; \\) Divide.\n\n\\[ Q = {0.06}\\;\\text{Convert to a percent.} \\]\n\n\\[ Q = 6\\% \\]\n\nThus,4.32 is \\( 6\\% \\) of 72 .\n	Yes
On a 160 question exam, a student got 125 correct answers. What percent is this? Round the result to two decimal places.	\[ \begin{matrix} \underset{ \downarrow }{125} & \text{ is } & \underbrace{\text{ what percent }} & \text{ of } & \underset{ \downarrow }{160}? & \text{ Missing factor statement. } & \\ \text{ (percentage) } & = & \text{ (percent) } & \downarrow & \downarrow & \text{ (base) } & \text{ (product) } = \text{ (factor) } \cdot \text{ (factor)] } \\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \text{ [(product) } = \text{ (factor) } \cdot \text{ (factor)] } & \\ {125} & = & Q & \cdot & {160} & & \end{matrix} \] \n\n\( Q = {125} \div {160}\; \) Divide.\n\n\( Q = {0.78125} \) Round to two decimal places.\n\n\[ Q = {.78} \]\n\nThus, this student received a \( {78}\% \) on the exam.	Yes
A bottle contains 80 milliliters of hydrochloric acid (HCl) and 30 milliliters of water. What percent of \( \mathrm{{HCl}} \) does the bottle contain? Round the result to two decimal places.	We need to determine the percent. The total amount of liquid in the bottle is\n\n80 milliliters +30 milliliters \( = {110} \) milliliters.\n\n\[ \begin{matrix} {80} & \\downarrow & \\left( \\text{ percentage }\\right) & = & \\text{ (percent) } & \\downarrow & \\text{ (base) } & \\left\\lbrack {\\text{ (product) } = \\text{ (factor) } \\cdot \\text{ (factor) }}\\right\\rbrack \\ \\downarrow & \\downarrow & \\downarrow & \\downarrow & \\downarrow & \\ {80} & = & Q & \\cdot & {110} & \\end{matrix} \]\n\n\( Q = {80} \\div {110}\\; \) Divide.\n\n\( Q = {0.727272}\\ldots \\; \) Round to two decimal places.\n\n\( Q \\approx {73}\\% \\; \) The symbol \	No
Five years ago a woman had an annual income of $19,200. She presently earns $42,000 annually. By what percent has her salary increased? Round the result to two decimal places.	We need to determine the percent.\n\n\\( Q = {42},{000} \\div {19},{200}\\; \\) Divide.\n\n\\( Q = {2.1875}\\; \\) Round to two decimal places.\n\n\\( Q = \\;{2.19}\\; \\) Convert to a percent.\n\n\\( Q = {219}\\% \\; \\) Convert to a percent.\n\nThus, this woman’s annual salary has increased \\( {219}\\% \\) .	No
15 is 30% of what number?	<table><thead><tr><th>15 \( \downarrow \)</th><th>is \( \downarrow \)</th><th>30% \( \downarrow \)</th><th>of \( \downarrow \)</th><th>what number?</th><th>Missing factor statement.</th></tr></thead><tr><td>(percentage) \( \downarrow \)</td><td>1</td><td>(percent) \( \downarrow \)</td><td>\( \downarrow \)</td><td>(base) \( \downarrow \)</td><td>\( \left\lbrack {\left( \text{percentage}\right) = \left( \text{factor}\right) \cdot \left( \text{factor}\right) }\right\rbrack \)</td></tr><tr><td>15</td><td>\( = \)</td><td>30%</td><td>.</td><td>\( B \)</td><td>Convert to decimals.</td></tr><tr><td>15</td><td>\( = \)</td><td>.30</td><td>.</td><td>\( B \)</td><td>\( \lbrack \left( \text{missing factor}\right) = \left( \text{product}\right) \div ( \) known fa</td></tr></table>\n\n\[ B = {15} \div {.30} \]\n\n\[ B = {50} \]\n\nThus, 15 is \( {30}\% \) of 50.	Yes
56,43 is 33% of what number? Missing factor statement.	\n\[ \n{56.43} = {33}\% \; \cdot \;B\;\text{Convert to decimals.} \n\]\n\[ \n{56.43} = {.33}\;B\;\text{Divide.} \n\]\n\[ \nB = {56.43} \div {.33} \n\]\n\[ \nB = \;{171} \n\]\n\nThus, 56.43 is 33% of 171.	Yes
Fifteen milliliters of water represents \( 2\% \) of a hydrochloric acid (HCl) solution. How many milliliters of solution are there?	We need to determine the total supply. The word supply indicates base.\n\n\[ \n\begin{array}{llllll} {15} & \text{ is } & 2\% & \text{ of } & \underbrace{\text{ what number? }} & \text{ Missing factor statement. } \end{array} \]\n\n\[ \n\begin{matrix} \text{ (percentage) } & = & \text{ (percent) } \\ \downarrow & \downarrow & \downarrow \end{matrix}\;\text{ (base) } \]\n\n\[ \n\begin{array}{lllll} {15} & = & {.02} & \cdot & \text{ Divide. } \end{array} \]\n\n\[ \nB = {15} \div {.02} \]\n\n\[ \nB = {750} \]\n\nThus, there are 750 milliliters of solution in the bottle.	Yes
A clothing item is priced at $20.40. This marked price includes a 15% discount. What is the original price?	We need to determine the original price. We can think of the original price as the starting place. Starting place indicates base. We need to determine the base. The new price, $20.40, represents 100% - 15% = 85% of the original price.\n\n<table><tr><td>20.40</td><td>is \( \downarrow \)</td><td>85% \( \downarrow \)</td><td>of \( \downarrow \)</td><td>what number?</td><td>Missing factor statement.</td></tr><tr><td>\( \begin{matrix} \text{(percentage)} \\ \downarrow \end{matrix} \)</td><td>1</td><td>(percent) \( \downarrow \)</td><td>\( \downarrow \)</td><td>(base) \( \downarrow \)</td><td></td></tr><tr><td>20.40</td><td>\( = \)</td><td>85%</td><td>.</td><td>\( B \)</td><td>Convert to decimals.</td></tr><tr><td>20.40</td><td>\( = \)</td><td>.85</td><td>.</td><td>\( B \)</td><td>\( \lbrack \left( \text{missing factor}\right) = \left( \text{product}\right) \div \left( \text{known factor}\right) \)</td></tr></table>\n\n\[ B = {20.40} \div {.85}\text{Divide.} \]\n\n\[ B = \;{24} \]\n\nThus, the original price of the item is $24.00.	Yes
Estimate the sum: \( 2,{357} + 6,{106} \) .	The sum can be estimated by \( 2,{400} + 6,{100} = 8,{500} \) . (It is quick and easy to add 24 and 61 .) Thus, \( 2,{357} + 6,{106} \) is about 8,400 . In fact, \( 2,{357} + 6,{106} = 8,{463} \) .	Yes
Estimate the difference: \( 5,{203} - 3,{015} \) .	Notice that 5,203 is near \( \underset{\text{two nonzero }}{\underbrace{5,{200},}} \) and that 3,015 is near \( \underset{\text{one nonzero }}{\underbrace{3,{000}.}} \) \n\nThe difference can be estimated by \( 5,{200} - 3,{000} = 2,{200} \) . \n\nThus, \( 5,{203} - 3,{015} \) is about 2,200 . In fact, \( 5,{203} - 3,{015} = 2,{188} \) .	Yes
Estimate the product: \( {87} \cdot 4,{316} \) .	Notice that 87 is close to \( \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{90},}} \) and that 4,316 is close to \( \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{4,{000}.}} \)\n\nThe product can be estimated by \( {90} \cdot 4,{000} = {360},{000} \).\n\nThus, \( {87} \cdot 4,{316} \) is about 360,000 . In fact, \( {87} \cdot 4,{316} = {375},{492} \).	Yes
Estimate the quotient: \( {153} \div {17} \) .	Notice that 153 is close to \( \underset{\begin{matrix} \text{ two nonzero } \\ \text{ digits } \end{matrix}}{\underbrace{150}} \) and that 17 is close to \( \underset{\text{two nonzero }}{\underbrace{\text{ 15. }}} \)\n\nThe quotient can be estimated by \( {150} \div {15} = {10} \).\n\nThus, \( {153} \div {17} \) is about 10 . In fact, \( {153} \div {17} = 9 \) .	Yes
Estimate the quotient: \( {742},{000} \div 2,{400} \) .	Notice that 742,000 is close to \( \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{700},{000}}} \), and that 2,400 is close to \( \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{2,{000}.}} \)\n\nThe quotient can be estimated by \( {700},{000} \div 2,{000} = {350} \).\n\nThus, \( {742},{000} \div 2,{400} \) is about 350 . In fact, \( {742},{000} \div 2,{400} = {309.1}\overline{6} \) .	Yes
Estimate the product: (31.28) (14.2).	Notice that 31.28 is close to \( \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{30},}} \) and that 14.2 is close to \( \underset{\text{two nonzero }}{\underbrace{{15}.}} \)\n\nThe product can be estimated by \( {30} \cdot {15} = {450} \) . ( \( 3 \cdot {15} = {45} \), then affix one zero.)\n\nThus,(31.28)(14.2) is about 450 . In fact,(31.28)(14.2) \( = {444.176} \) .	Yes
Estimate \( {21}\% \) of 5.42.	Notice that \( {21}\% = {.21} \) as a decimal, and that .21 is close to one nonzero digit\n\nNotice also that 5.42 is close to\n\nThen, \( {21}\% \) of 5.42 can be estimated by \( \left( {.2}\right) \left( 5\right) = 1 \) .\n\nThus, \( {21}\% \) of 5.42 is about 1 . In fact, \( {21}\% \) of 5.42 is 1.1382 .	No
\( {27} + {48} + {31} + {52} \) .	27 and 31 cluster near 30 . Their sum is about \( 2 \cdot {30} = {60} \) .\n\n48 and 52 cluster near 50 . Their sum is about \( 2 \cdot {50} = {100} \) .\n\nThus, \( {27} + {48} + {31} + {52} \) is about \( \begin{aligned} \left( {2 \cdot {30}}\right) + \left( {2 \cdot {50}}\right) & = {60} + {100} \\ & = {160} \end{aligned} \)\n\nIn fact, \( {27} + {48} + {31} + {52} = {158} \) .	Yes
\( {88} + {21} + {19} + {91} \)	88 and 91 cluster near 90 . Their sum is about \( 2 \cdot {90} = {180} \) . 21 and 19 cluster near 20 . Their sum is about \( 2 \cdot {20} = {40} \) . Thus, \( {88} + {21} + {19} + {91} \) is about \( \begin{aligned} \left( {2 \cdot {90}}\right) + \left( {2 \cdot {20}}\right) & = {180} + {40} \\ & = {220} \end{aligned} \) In fact, \( {88} + {21} + {19} + {91} = {219} \)	Yes
\( {17} + {21} + {48} + {18} \)	\( {17},{21} \), and 18 cluster near 20 . Their sum is about \( 3 \cdot {20} = {60} \) . 48 is about 50 . Thus, \( {17} + {21} + {48} + {18} \) is about \( \begin{aligned} \left( {3 \cdot {20}}\right) + {50} & = {60} + {50} \\ & = {110} \end{aligned} \). In fact, \( {17} + {21} + {48} + {18} = {104} \).	Yes
\( {61} + {48} + {49} + {57} + {52} \)	61 and 57 cluster near 60 . Their sum is about \( 2 \cdot {60} = {120} \) . \n\n\( {48},{49} \), and 52 cluster near 50 . Their sum is about \( 3 \cdot {50} = {150} \) . \n\nThus, \( {61} + {48} + {49} + {57} + {52} \) is about \( \begin{aligned} \left( {2 \cdot {60}}\right) + \left( {3 \cdot {50}}\right) & = {120} + {150} \\ & = {270} \end{aligned} \) \n\nIn fact, \( {61} + {48} + {49} + {57} + {52} = {267} \)	Yes
\( {706} + {321} + {293} + {684} \) .	706 and 684 cluster near 700 . Their sum is about \( 2 \cdot {700} = 1,{400} \) .\n\n321 and 293 cluster near 300 . Their sum is about \( 2 \cdot {300} = {600} \) .\n\nThus, \( {706} + {321} + {293} + {684} \) is about \( \begin{aligned} \left( {2 \cdot {700}}\right) + \left( {2 \cdot {300}}\right) & = 1,{400} + {600} \\ & = 2,{000} \end{aligned} \)\n\nIn fact, \( {706} + {321} + {293} + {684} = 2,{004} \) .	Yes
\[ 4\left( {6 + 2}\right) = 4 \cdot 6 + 4 \cdot 2 \]	\[ = {24} + 8 \] \[ = {32} \] Using the order of operations, we get \[ 4\left( {6 + 2}\right) = 4 \cdot 8 \] \[ = \;{32} \]	Yes
\[ 8\left( {9 + 6}\right) = 8 \cdot 9 + 8 \cdot 6 \]	\[ = {72} + {48} \] \[ = {120} \] Using the order of operations, we get \[ 8\left( {9 + 6}\right) = 8 \cdot {15} \] \[ = \;{120} \]	Yes
\[ 4\left( {9 - 5}\right) = 4 \cdot 9 - 4 \cdot 5 \]	\[ = {36} - {20} \] \[ = {16} \]	Yes
\( {25} \cdot {23} \)	Notice that \( {23} = {20} + 3 \) . We now write\n\n\[ \n{25} \cdot {23} = \underset{25}{\underbrace{{55}\left( {{20} + 3}\right) }} \n\]\n\n\[ \n= {25} \cdot {20} + {25} \cdot 3 \n\]\n\n\[ \n= {500} + {75} \n\]\n\n\[ \n= {575} \n\]\n\nThus, \( {25} \cdot {23} = {575} \)	Yes
\( {15} \cdot {37} \)	Notice that \( {37} = {30} + 7 \) . We now write\n\n\[ \n{15} \cdot {37} = {15}\left( {{30} + 7}\right) \n\]\n\n\[ \n= {15} \cdot {30} + {15} \cdot 7 \n\]\n\n\[ \n= {450} + {105} \n\]\n\n\[ \n= {555} \n\]\n\nThus, \( {15} \cdot {37} = {555} \)	Yes
\( {15} \cdot {86} \)	Notice that \( {86} = {80} + 6 \) . We now write\n\n\[ \n{15} \cdot {86} = \underset{15}{\underbrace{{15}\left( {{80} + 6}\right) }} \n\]\n\n\[ \n= {15} \cdot {80} + {15} \cdot 6 \n\]\n\n\[ \n= 1,{200} + {90} \n\]\n\n\[ \n= 1,{290} \n\]\n\nWe could have proceeded by writing 86 as \( {90} - 4 \) .\n\n\[ \n{15} \cdot {86} = {15}\left( {{90} - 4}\right) \n\]\n\n\[ \n= {15} \cdot {90} - {15} \cdot 4 \n\]\n\n\[ \n= 1,{350} - {60} \n\]\n\n\[ \n= 1,{290} \n\]	Yes
Estimate \( \frac{3}{5} + \frac{5}{12} \) .	Notice that \( \frac{3}{5} \) is about \( \frac{1}{2} \), and that \( \frac{5}{12} \) is about \( \frac{1}{2} \). Thus, \( \frac{3}{5} + \frac{5}{12} \) is about \( \frac{1}{2} + \frac{1}{2} = 1 \). In fact, \( \frac{3}{5} + \frac{5}{12} = \frac{61}{60} \), a little more than 1.	Yes
Estimate \( 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} \) .	Adding the whole number parts, we get 20 . Notice that \( \frac{3}{8} \) is close to \( \frac{1}{4},\frac{9}{10} \) is close to 1, and \( \frac{1}{5} \) is close to \( \frac{1}{4} \) . Then \( \frac{3}{8} + \frac{9}{10} + \frac{1}{5} \) is close to \( \frac{1}{4} + 1 + \frac{1}{4} = 1\frac{1}{2} \) . Thus, \( 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} \) is close to \( {20} + 1\frac{1}{2} = {21}\frac{1}{2} \) . In fact, \( 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} = {21}\frac{19}{40} \), a little less than \( {21}\frac{1}{2} \) .	Yes
Convert 11 yards to feet.	Looking in the unit conversion table under length, we see that \( 1\mathrm{{yd}} = 3\mathrm{{ft}} \) . There are two corresponding unit fractions, \( \frac{1\mathrm{{yd}}}{3\mathrm{{ft}}} \) and \( \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} \) . Which one should we use? Look to see which unit we wish to convert to. Choose the unit fraction with this unit in the numerator. We will choose \( \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} \) since\n\nthis unit fraction has feet in the numerator. Now, multiply 11 yd by the unit fraction. Notice that since the unit fraction has the value of 1 , multiplying by it does not change the value of 11 yd.\n\n\[ \n{11}\mathrm{{yd}} = \frac{{11}\mathrm{{yd}}}{1} \cdot \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} \]\n\n\( = \frac{{11}\overset{1}{\overbrace{)\mathrm{{yd}}}}}{1} \cdot \frac{3\mathrm{{ft}}}{1\overline{\overline{)\mathrm{{yd}}}}}\; \) (Units can be added, subtracted, multiplied, and divided, just as numbers can.)\n\n\[ \n= \;\frac{{11} \cdot 3\mathrm{{ft}}}{1} \]\n\n\[ \n= \;{33}\mathrm{{ft}} \]\n\nThus, \( {11}\mathrm{{yd}} = {33}\mathrm{{ft}} \).	Yes
Convert \( {36}\mathrm{{fl}} \) oz to pints.	\n\n\( {36}\mathrm{{fl}}\mathrm{{oz}} = \frac{{36}\mathrm{{fl}}\mathrm{{oz}}}{1} \cdot \frac{1\mathrm{{pt}}}{{16}\mathrm{{fl}}\mathrm{{oz}}}\; \) Divide out common units.\n\n\[ \n= \frac{{36}\overline{)\mathrm{{fl}}\text{ oz }}}{1} \cdot \frac{1\mathrm{{pt}}}{{16}\overline{)\mathrm{{fl}}\text{ oz }}} \n\] \n\n\[ \n= \;\frac{{36} \cdot 1\mathrm{{pt}}}{16} \n\] \n\n\( = \;\frac{{36}\mathrm{{pt}}}{16}\; \) Reduce.\n\n\( = \;\frac{9}{4}\mathrm{{pt}}\; \) Convert to decimals: \( \frac{9}{4} = {2.25} \). \n\nThus, \( {36}\mathrm{{fl}} \) oz \( = {2.25}\mathrm{{pt}} \) .	Yes
Convert 3 kilograms to grams.	(a) \( 3\mathrm{\;{kg}} \) can be written as \( {3.0}\mathrm{\;{kg}} \) . Then, ![7d43c799-07cb-4976-9fbc-3035178b5209_529_0.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_529_0.jpg)\n\nThus, \( 3\mathrm{\;{kg}} = 3,{000}\mathrm{\;g} \) .\n\n(b) We can also use unit fractions to make this conversion. Since we are converting to grams, and\n\n\( 1,{000}\mathrm{\;g} = 1\mathrm{\;{kg}} \), we choose the unit fraction \( \frac{1,{000}\mathrm{\;g}}{1\mathrm{l}\mathrm{g}} \) since grams is in the numerator.\n\n\[ 3\mathrm{\;{kg}} = 3\mathrm{\;{kg}} \cdot \frac{1,{000}\mathrm{\;g}}{1\mathrm{\;{kg}}} \]\n\n\[ = 3\overline{)\mathrm{{kg}}} \cdot \frac{1,{000}\mathrm{g}}{1\overline{)\mathrm{{kg}}}} \]\n\n\[ = \;3 \cdot 1,{000}\mathrm{\;g} \]\n\n\[ \text{=}\;\text{3,000}\;\text{g} \]	Yes
Convert 67.2 hectoliters to milliliters.	\( {67.2}\mathrm{\;{hL}} = \underset{1\text{ 2 }3\text{ 4 }5}{{67}\underbrace{20000}},\mathrm{\;{mL}} \)\n\nThus, \( {67.2}\mathrm{\;{hL}} = 6,{720},{000}\mathrm{\;{mL}} \).	Yes
Convert 100.07 centimeters to meters.	Thus, \( {100.07}\mathrm{\;{cm}} = {1.0007}\mathrm{\;m} \).	Yes
Convert 0.16 milligrams to grams.	\( {0.16}\mathrm{{mg}} = 0,\underset{3}{000}{16}\mathrm{\;g} \)\n\nThus, \( {0.16}\mathrm{{mg}} = {0.00016}\mathrm{\;g} \).	Yes
Simplify 19 in.	Since 12 in. \( = 1\mathrm{{ft}} \), and \( {19} = {12} + 7 \) ,\n\n19 in. \( = {12} \) in. +7 in.\n\n\( = \;1\mathrm{{ft}} + 7\mathrm{{in}} \) .\n\n\( = \;1\mathrm{{ft}}7\mathrm{{in}} \) .	Yes
Simplify 2 hr 75 min.	Since \( {60}\mathrm{\;{min}} = 1\mathrm{{hr}} \), and \( {75} = {60} + {15} \) ,\n\n\( 2\mathrm{{hr}}{75}\mathrm{\;{min}} = 2\mathrm{{hr}} + {60}\mathrm{\;{min}} + {15}\mathrm{\;{min}} \)\n\n\[ \n= \;2\mathrm{{hr}} + 1\mathrm{{hr}} + {15}\mathrm{\;{min}} \n\]\n\n\[ \n= \;3\mathrm{{hr}} + {15}\mathrm{\;{min}} \n\]\n\n\[ \n= \;3\mathrm{{hr}}{15}\mathrm{\;{min}} \n\]	Yes
Simplify \( {43}\mathrm{{fl}} \) oz.	Since \( 8\mathrm{{fl}} \) oz \( = 1\mathrm{c}\left( {1\mathrm{{cup}}}\right) \), and \( {43} \div 8 = 5\mathrm{R}3 \) ,\n\n\( {43}\mathrm{{fl}}\mathrm{{oz}} = {40}\mathrm{{fl}}\mathrm{{oz}} + 3\mathrm{{fl}}\mathrm{{oz}} \)\n\n\( = 5 \cdot 8\mathrm{{fl}}\mathrm{{oz}} + 3\mathrm{{fl}}\mathrm{{oz}} \)\n\n\[ = \;5 \cdot 1\mathrm{c} + 3\mathrm{{fl}}\text{oz} \]\n\n\[ = \;5\mathrm{c} + 3\mathrm{{fl}}\text{oz} \]\n\nBut, \( 2\mathrm{c} = 1\mathrm{{pt}} \) and \( 5 \div 2 = 2\mathrm{R}1 \) . So,\n\n\[ 5\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} = 2 \cdot 2\mathrm{c} + 1\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} \]\n\n\[ = 2 \cdot 1\mathrm{{pt}} + 1\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} \]	Yes
Add \( 6\mathrm{{ft}}8 \) in. to \( 2\mathrm{{ft}}9 \) in.	\n\( 6\mathrm{{ft}}8\mathrm{{in}} \) .\n\n\( + 2\mathrm{{ft}}9\mathrm{{in}} \) .\n\n8 ft 17 in. Simplify this denominate number.\n\nSince 12 in. \( = 1\mathrm{{ft}} \) ,\n\n\[ 8\mathrm{{ft}} + {12}\mathrm{{in}}. + 5\mathrm{{in}}. = 8\mathrm{{ft}} + 1\mathrm{{ft}} + 5\mathrm{{in}}. \]\n\n\[ = \;9\mathrm{{ft}} + 5\mathrm{{in}}\text{.} \]\n\n\[ = \;9\mathrm{{ft}}5\mathrm{{in}}\text{.} \]\n	Yes
Subtract 5 da 3 hr from 8 da 11 hr.	\( 8\mathrm{{da}}{11}\mathrm{{hr}} \)\n\n\( - 5\mathrm{{da}}\;3\mathrm{{hr}} \)\n\n\( 3\mathrm{{da}}\;8\mathrm{{hr}} \)	Yes
Subtract 3 lb 14 oz from 5 lb 3 oz.	We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz from the pounds.\n\n5 lb 3 oz \( = \;5\mathrm{{lb}} + 3\mathrm{{oz}} \)\n\n\[ = \;4\mathrm{{lb}} + 1\mathrm{{lb}} + 3\mathrm{{oz}} \]\n\n\( = {4lb} + {16oz} + {3oz}\; \) (Since \( {1lb} = {16oz} \) .)\n\n\( = \;4\mathrm{{lb}} + {19}\mathrm{{oz}} \)\n\n\( = \;4\mathrm{{lb}}{19}\mathrm{{oz}} \)\n\n\( 4\mathrm{{lb}}{19}\mathrm{{oz}} \)\n\n\( - 3\mathrm{{lb}}{14}\mathrm{{oz}} \)\n\n\( 1\mathrm{{lb}}\;5\mathrm{{oz}} \)	Yes
Subtract 4 da 9 hr \( {21}\mathrm{\;{min}} \) from 7 da 10 min.	Borrow 1 do from the 7 da.\n\n\( - 4\mathrm{{da}}9\mathrm{{hr}}{21}\mathrm{\;{min}} \)\n\noud 24 m 10 mm\n\n-4 da 9 hr \( {21}\mathrm{\;{min}} \)\n\n6 da \( {23}\mathrm{{hr}}{70}\mathrm{\;{min}} \)\n\n-4 da 9 hr \( {21}\mathrm{\;{min}} \)\n\n2 da 14 hr \( {49}\mathrm{\;{min}} \)	No
\[ 6 \cdot \left( {2\mathrm{{ft}}4\mathrm{{in}}.}\right) = 6 \cdot 2\mathrm{{ft}} + 6 \cdot 4\mathrm{{in}}. \]	\[ = \;{12}\mathrm{{ft}} + {24}\text{in.} \] Since \( 3\mathrm{{ft}} = 1\mathrm{{yd}} \) and 12 in. \( = 1\mathrm{{ft}} \), \[ {12}\mathrm{{ft}} + {24} \) in. \( = 4\mathrm{{yd}} + 2\mathrm{{ft}} \] \[ = \;4\mathrm{{yd}}2\mathrm{{ft}} \]	Yes
\[ 8 \cdot \left( {5\mathrm{{hr}}{21}\mathrm{\;{min}}{55}\mathrm{{sec}}}\right) = 8 \cdot 5\mathrm{{hr}} + 8 \cdot {21}\mathrm{\;{min}} + 8 \cdot {55}\mathrm{{sec}} \]	\[ = \;{40}\mathrm{{hr}} + {168}\mathrm{\;{min}} + {440}\mathrm{{sec}} \]\n\[ = {40}\mathrm{{hr}} + {168}\mathrm{\;{min}} + 7\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{40}\mathrm{{hr}} + {175}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{40}\mathrm{{hr}} + 2\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{42}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{24}\mathrm{{hr}} + {18}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = 1\mathrm{{da}} + {18}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;1\mathrm{{da}}{18}\mathrm{{hr}}{55}\mathrm{\;{min}}{20}\mathrm{{sec}} \]	Yes
\( \left( {{12}\min {40}\sec }\right) \div 4 \)	Thus \( \left( {{12}\mathrm{\;{min}}{40}\mathrm{{sec}}}\right) \div 4 = 3\mathrm{\;{min}}{10}\mathrm{{sec}} \)	Yes
\( \left( {5\mathrm{{yd}}2\mathrm{{ft}}9\mathrm{{in}}\text{.}}\right) \div 3 \)	\( 1\mathrm{{yd}}2\mathrm{{ft}}{11}\mathrm{{in}} \) . 3) 5 yd 2 ft 9 in. / \( 3\mathrm{{yd}} \) 2 yd 2 ft \( \rightarrow 8\mathrm{{ft}} \) \( 6\mathrm{{ft}} \) 2 ft 9 in. \( \rightarrow {33} \) in. \n\n\( \frac{{33}\text{ in. }}{0} \)\n\nConvert to feet: \( 2\mathrm{{yd}}2\mathrm{{ft}} = 8\mathrm{{ft}} \) .\n\nConvert to inches: \( 2\mathrm{{ft}}9\mathrm{{in}} \) . \( = {33}\mathrm{{in}} \) .\n\nThus \( \left( {5\mathrm{{yd}}2\mathrm{{ft}}9\mathrm{{in}}.}\right) \div 3 = 1\mathrm{{yd}}2\mathrm{{ft}}{11}\mathrm{{in}} \) .	No
Find the approximate circumference of the circle.	Use the formula \( C = {\pi d} \). \( C \approx \left( {3.14}\right) \left( {6.2}\right) \). \( C \approx {19.648}\mathrm{\;{mm}} \). This result is approximate since \( \pi \) has been approximated by 3.14.	Yes
Find the approximate circumference of a circle with radius 18 inches.	Since we’re given that the radius, \( r \), is \( {18} \) in., we’ll use the formula \( C = {2\pi r} \) .\n\n\[ C \approx \left( 2\right) \left( {3.14}\right) \left( {{18}\text{ in. }}\right) \]\n\n\( C \approx {113.04} \) in.	Yes
Find the approximate perimeter of the figure.	We notice that we have two semicircles (half circles).\n\nThe larger radius is \( {6.2}\mathrm{\\;{cm}} \).\n\nThe smaller radius is \( {6.2}\mathrm{\\;{cm}} - {2.0}\mathrm{\\;{cm}} = {4.2}\mathrm{\\;{cm}} \).\n\nThe width of the bottom part of the rectangle is \( {2.0}\mathrm{\\;{cm}} \). Perimeter \( = \) \( {2.0}\mathrm{\\;{cm}} \) \( {5.1}\mathrm{\\;{cm}} \) \( {2.0}\mathrm{\\;{cm}} \) \( {5.1}\mathrm{\\;{cm}} \)\n\n\( \left( {0.5}\right) \cdot \left( 2\right) \cdot \left( {3.14}\right) \cdot \left( {{6.2}\mathrm{\\;{cm}}}\right) \) Circumference of outer semicircle.\n\n\( + \underline{\left( {0.5}\right) \cdot \left( 2\right) \cdot \left( {3.14}\right) \cdot \left( {{4.2}\mathrm{\\;{cm}}}\right) } \) Circumference of inner semicircle \( {6.2}\mathrm{\\;{cm}} - {2.0}\mathrm{\\;{cm}} = {4.2}\mathrm{\\;{cm}} \) The 0.5 appears because we want the perimeter of only half a circle.	Yes
Find the area of the triangle.	\[ \n{A}_{T} = \frac{1}{2} \cdot b \cdot h \n\] \n\[ \n= \;\frac{1}{2} \cdot {20} \cdot 6\;\text{sq ft} \n\] \n\[ \n= {10} \cdot 6\text{ sq ft } \n\] \n\[ \n= \;{60}\mathrm{{sq}}\mathrm{{ft}} \n\] \n\[ \n= \;{60}{\mathrm{{ft}}}^{2} \n\] \nThe area of this triangle is \( {60}\mathrm{{sq}}\mathrm{{ft}} \), which is often written as \( {60}{\mathrm{{ft}}}^{2} \) .	Yes
Find the area of the rectangle.	Let's first convert \( 4\mathrm{{ft}}2\mathrm{{in}} \) . to inches. Since we wish to convert to inches, we’ll use the unit fraction \( \frac{{12}\text{ in. }}{1\mathrm{{ft}}} \) since it has inches in the numerator. Then,\n\n\[ 4\mathrm{{ft}} = \frac{4\mathrm{{ft}}}{1} \cdot \frac{{12}\mathrm{{in}}.}{1\mathrm{{ft}}} \]\n\n\[ = \frac{4\overline{)\mathrm{{ft}}}}{1} \cdot \frac{{12}\mathrm{{in}}.}{1\overline{)\mathrm{{ft}}}} \]\n\n\[ = \;{48}\text{ in. } \]\n\nThus, \( 4\mathrm{{ft}}2\mathrm{{in}}. = {48}\mathrm{{in}} \cdot + 2\mathrm{{in}}. = {50}\mathrm{{in}} \) .\n\n\[ {A}_{R} = \;l \cdot w \]\n\n\[ = {50}\text{in.} \cdot 8\text{in.} \]\n\n\[ = \;{400}\mathrm{{sq}}\text{in.} \]\n\nThe area of this rectangle is \( {400}\mathrm{{sq}} \) in.	Yes
Find the area of the parallelogram.	\[ \n{A}_{P} = \;b \cdot h \]\n\[ \n= {10.3}\mathrm{\;{cm}} \cdot {6.2}\mathrm{\;{cm}} \]\n\[ \n= \;{63.86}\mathrm{{sq}}\mathrm{{cm}} \]\n\nThe area of this parallelogram is \( {63.86}\mathrm{{sq}}\mathrm{{cm}} \) .	Yes
Find the area of the trapezoid.	\[ \n{A}_{\text{Trap }} = \;\frac{1}{2} \cdot \left( {{b}_{1} + {b}_{2}}\right) \cdot h \]\n\[ \n= \frac{1}{2} \cdot \left( {{14.5}\mathrm{\;{mm}},+,{20.4}\mathrm{\;{mm}}}\right) \cdot \left( {{4.1}\mathrm{\;{mm}}}\right) \]\n\[ \n= \;\frac{1}{2} \cdot \left( {{34.9}\mathrm{\;{mm}}}\right) \cdot \left( {{4.1}\mathrm{\;{mm}}}\right) \]\n\[ \n= \;\frac{1}{2} \cdot \left( {{143.09}\mathrm{{sq}}\mathrm{{mm}}}\right) \]\n\[ \n= \;{71.545}\mathrm{{sq}}\mathrm{{mm}} \]\n\nThe area of this trapezoid is \( {71.545}\mathrm{{sq}}\mathrm{{mm}} \) .	Yes
Find the approximate area of the circle.	\[ \n{A}_{c} = \;\pi \cdot {r}^{2} \n\] \n\[ \n\approx \;\left( {3.14}\right) \cdot {\left( {16.8}\mathrm{{ft}}\right) }^{2} \n\] \n\[ \n\approx \left( {3.14}\right) \cdot \left( {{282.24}\mathrm{{sq}}\mathrm{{ft}}}\right) \n\] \n\[ \n\approx \;{888.23}\mathrm{{sq}}\mathrm{{ft}} \n\] \n\nThe area of this circle is approximately \( {886.23}\mathrm{{sq}} \) ft.	Yes
Find the volume of the rectangular solid.	\[ {V}_{R} = \;l \cdot w \cdot h \] \[ = 9\text{in.} \cdot {10}\text{in.} \cdot 3\text{in.} \] \[ = \;{270}\mathrm{{cu}}\text{in.} \] \[ = \;{270}\text{in.}{}^{3} \] The volume of this rectangular solid is \( {270}\mathrm{{cu}} \) in.	Yes
Find the approximate volume of the sphere.	\[ \n{V}_{S} = \frac{4}{3} \cdot \pi \cdot {r}^{3} \]\n\[ \n\approx \;\left( \frac{4}{3}\right) \cdot \left( {3.14}\right) \cdot {\left( 6\mathrm{\;{cm}}\right) }^{3} \]\n\[ \n\approx \;\left( \frac{4}{3}\right) \cdot \left( {3.14}\right) \cdot \left( {{216}\mathrm{{cu}}\mathrm{{cm}}}\right) \]\n\[ \n\approx \;{904.32}\mathrm{{cu}}\mathrm{{cm}} \]\n\nThe approximate volume of this sphere is \( {904.32}\mathrm{{cu}}\mathrm{{cm}} \), which is often written as \( {904.32}{\mathrm{\;{cm}}}^{3} \) .	Yes
Is every whole number a natural number?	No. The number 0 is a whole number but it is not a natural number.	Yes
Is there an integer that is not a natural number?	Yes. Some examples are 0, -1, -2, -3, and -4.	Yes
Is there an integer that is a whole number?	Yes. In fact, every whole number is an integer.	Yes
What integers can replace \( x \) so that the following statement is true?\n\n\( - 3 \leq x < 2 \) ![7d43c799-07cb-4976-9fbc-3035178b5209_589_0.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_589_0.jpg)	The integers are \( - 3, - 2, - 1,0,1 \) .	Yes
Write each expression in words. Exercise 10.3.1	Solution on p. 621.	No
If \( a = - 4 \), then \( - a = - \left( {-4}\right) = 4 \) . Also, \( - \left( {-a}\right) = a = - 4 \) .	If \( a = - 4 \), then \( - a = - \left( {-4}\right) = 4 \) . Also, \( - \left( {-a}\right) = a = - 4 \) .	Yes
\( - \left\| {-3}\right\| = - 3 \)	The quantity on the left side of the equal sign is read as \	No
The number enclosed within the absolute value bars is a nonnegative number, so the upper part of the definition applies. This part says that the absolute value of 8 is 8 itself.	\[ \left\| 8\right\| = 8 \]	Yes
\( \left\| {-3}\right\| \)	The number enclosed within absolute value bars is a negative number, so the lower part of the definition applies. This part says that the absolute value of -3 is the opposite of -3 , which is\n\n- (-3). By the definition of absolute value and the double-negative property,\n\n\[ \left\| {-3}\right\| = - \left( {-3}\right) = 3 \]	Yes
\( \left( {-4}\right) + \left( {-9}\right) \)	\( \begin{array}{l} \left\| {-4}\right\| = 4 \\ \left\| {-9}\right\| = 9 \end{array}\} \) Add these absolute values. \n\n\( 4 + 9 = {13} \)\n\nThe common sign is \	No
\( 7 + \left( {-2}\right) \)	Subtract absolute values: \( 7 - 2 = 5 \). Attach the proper sign: \	No
\( 3 + \left( {-{11}}\right) \)	Subtract absolute values: \( {11} - 3 = 8 \) .\n\nAttach the proper sign: \	No
The morning temperature on a winter's day in Lake Tahoe was -12 degrees. The afternoon temperature was 25 degrees warmer. What was the afternoon temperature?	We need to find \( - {12} + {25} \) . ![7d43c799-07cb-4976-9fbc-3035178b5209_603_2.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_603_2.jpg)\n\nSubtract absolute values: \( {25} - {12} = {16} \) .\n\nAttach the proper sign: \	No
The high temperature today in Lake Tahoe was \( {26}^{ \circ }\mathrm{F} \) . The low temperature tonight is expected to be \( - {7}^{ \circ }\mathrm{F} \) . How many degrees is the temperature expected to drop?	We need to find the difference between 26 and -7 . \[ {26} - \left( {-7}\right) = {26} + 7 = {33} \] Thus, the expected temperature drop is \( {33}^{ \circ }\mathrm{F} \) .	Yes
\[ - 6 - \left( {-5}\right) - {10} = - 6 + 5 + \left( {-{10}}\right) \]	\[ = \left( {-6 + 5}\right) + \left( {-{10}}\right) \]\n\[ = \; - 1 + \left( {-{10}}\right) \]\n\[ = \; - {11} \]	Yes
\( 3,{187} - 8,{719} \)	Thus, \( 3,{187} - 8,{719} = - 5,{532} \).	Yes
\( - {156} - \left( {-{211}}\right) \)	Method A:\n\n<table><thead><tr><th colspan=\	No
\( \left( {-8}\right) \left( {-6}\right) \)	\( \left. \begin{array}{lll} \left\| {-8}\right\| & = & 8 \\ \left\| {-6}\right\| & = & 6 \end{array}\right\} \) Multiply these absolute values.\n\n\( 8 \cdot 6 = {48} \)\n\nSince the numbers have the same sign, the product is positive.\n\nThus, \( \left( {-8}\right) \left( {-6}\right) = + {48} \), or \( \left( {-8}\right) \left( {-6}\right) = {48} \).	Yes
Example 10.38\n\n\( 6\left( {-3}\right) \)	\n\n\( \left. \begin{aligned} \left\| 6\right\| & = 6 \\ \left\| {-3}\right\| & = 3 \end{aligned}\right\} \) Multiply these absolute values.\n\n\( 6 \cdot 3 = {18} \)\n\nSince the numbers have opposite signs, the product is negative.\n\nThus, \( 6\left( {-3}\right) = - {18} \) .	Yes
Example 10.39 \( \frac{-{10}}{2} \)	\( \begin{aligned} \left\| {-{10}}\right\| & = {10} \\ \left\| 2\right\| & = 2 \end{aligned} \) Divide these absolute values.\n\n\( \frac{10}{2} = 5 \)\n\nSince the numbers have opposite signs, the quotient is negative.\n\nThus \( \frac{-{10}}{2} = - 5 \) .	Yes
\( \frac{-{35}}{-7} \)	\( \left. \begin{matrix} \left\| {-{35}}\right\| & = & {35} \\ \left\| {-7}\right\| & = & 7 \end{matrix}\right\} \) Divide these absolute values.\n\n\( \frac{35}{7} = 5 \)\n\nSince the numbers have the same signs, the quotient is positive.\n\nThus, \( \frac{-{35}}{-7} = 5 \) .	Yes
\( \frac{18}{-9} \)	\( \left. \begin{aligned} \left\| {18}\right\| & = {18} \\ \left\| {-9}\right\| & = 9 \end{aligned}\right\} \) Divide these absolute values. \n\n\( \frac{18}{9} = 2 \)\n\nSince the numbers have opposite signs, the quotient is negative.\n\nThus, \( \frac{18}{-9} = 2 \) .	No
Find the value of \( \frac{-6\left( {4 - 7}\right) - 2\left( {8 - 9}\right) }{-\left( {4 + 1}\right) + 1} \) .	Using the order of operations and what we know about signed numbers, we get,\n\n\[ \frac{-6\left( {4 - 7}\right) - 2\left( {8 - 9}\right) }{-\left( {4 + 1}\right) + 1} = \frac{-6\left( {-3}\right) - 2\left( {-1}\right) }{-\left( 5\right) + 1} \]\n\n\[ = \;\frac{{18} + 2}{-5 + 1} \]\n\n\[ = \;\frac{20}{ - } \]\n\n\[ = \; - 5 \]	Yes
\( \left( {-{186}}\right) \cdot \left( {-{43}}\right) \)	Since this product involves a (negative) (negative), we know the result should be a positive number. We'll illustrate this on the calculator.\n\n<table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>186</td><td>186</td></tr><tr><td>Press</td><td>\( + / - \)</td><td>-186</td></tr><tr><td>Press</td><td>\( \times \)</td><td>-186</td></tr><tr><td>Type</td><td>43</td><td>43</td></tr><tr><td>Press</td><td><img src=\	Yes
\( \frac{158.64}{-{54.3}} \) . Round to one decimal place.	<table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>158.64</td><td>158.64</td></tr><tr><td>Press</td><td>\( \div \)</td><td>158.64</td></tr><tr><td>Type</td><td>54.3</td><td>54.3</td></tr><tr><td>Press</td><td>\( + / - \)</td><td>-54.3</td></tr><tr><td>Press</td><td>\( = \)</td><td>-2.921546961</td></tr></table>\n\nTable 10.6\n\nRounding to one decimal place we get -2.9.	Yes
\( a + 7 - b - m \)	Associating the sign with the individual quantities, we see that this expression consists of the four terms \( a,7, - b, - m \).	No
Specify the terms in the expression \( x + 7 \). Exercise 11.2.1	Solution on p. 682.	No
\( \frac{5a}{b} + \frac{8b}{12} \), if \( a = 6 \) and \( b = - 3 \) .	Replace a with 6 and \( b \) with -3 .\n\n\[ \frac{5a}{b} + \frac{8b}{12} = \frac{5\left( 6\right) }{-3} + \frac{8\left( {-3}\right) }{12} \]\n\n\[ = \;\frac{30}{ - } + \frac{ - }{24} \]\n\n\[ = - {10} + \left( {-2}\right) \]\n\n\[ = \; - {12} \]\n\nThus, when \( a = 6 \) and \( b = - 3,\frac{5a}{b} + \frac{8b}{12} = - {12} \) .	No
\( 3{x}^{2} - {2x} + 1 \), if \( x = 4 \)	Replace \( x \) with 4 .\n\n\[ 3{x}^{2} - {2x} + 1 = 3{\left( 4\right) }^{2} - 2\left( 4\right) + 1 \]\n\n\[ = 3 \cdot {16} - 2\left( 4\right) + 1 \]\n\n\[ = \;{48} - 8 + 1 \]\n\n\[ = \;{41} \]\n\nThus, when \( x = 4,3{x}^{2} - {2x} + 1 = {41} \).	Yes
\( - {x}^{2} - 4 \), if \( x = 3 \)	Replace \( x \) with 3 . \n\n\( - {x}^{2} - 4 = - {}^{3} - 4 \) Be careful to square only the 3 . The exponent 2 is connected only to 3, not -3\n\n\[ \n= - 9 - 4 \n\] \n\n\[ \n= \; - {13} \n\]	Yes
\( {\left( -x\right) }^{2} - 4 \), if \( x = 3 \).	Replace \( x \) with 3 . \n\n\( {\left( -x\right) }^{2} - 4 = {\left( -3\right) }^{2} - 4 \) The exponent is connected to -3, not 3 as in problem 5 above. \n\n\[ \n= \;9 - 4 \n\] \n\n\[ \n= \; - 5 \n\] \n\nThe exponent is connected to -3 , not 3 as in the problem above.	No
\( - {3a} + {2b} - {5a} + a + {6b} \) . The like terms are	\n\[ \underset{\begin{matrix} {-3 - 5 + 1 = - 7} \\ {-{7a}} \end{matrix}}{\underbrace{-{3a},\; - {5a},\;a}}\underset{\begin{matrix} {2 + 6 = 8} \\ {8b} \end{matrix}}{\underbrace{{2b},\;{6b}}} \] \nThus, \( - {3a} + {2b} - {5a} + a + {6b} = - {7a} + {8b} \)	Yes
Example 11.15\n\n\( r - {2s} + {7s} + {3r} - {4r} - {5s} \) . The like terms are	\[ \underset{1 + 3 - 4 = 0}{\underbrace{r,{3r}, - {4r}}}\;\underset{-2 + 7 - 5 = 0}{\underbrace{-{2s},{7s}, - {5s}}} \]\n\n\[ \frac{0r}{{0r} + {0s}} = 0 \]\n\nThus, \( r - {2s} + {7s} + {3r} - {4r} - {5s} = 0 \) .	Yes
Verify that 3 is a solution to \( x + 7 = {10} \) .	When \( x = 3 \) , \[ x + 7 = {10} \] becomes \( 3 + 7 = {10} \) \( \begin{matrix} {10} = {10} & \text{ which is a true statement, verifying that } \\ 3\text{ is a solution to }x + 7 = {10} & \end{matrix} \)	Yes
Verify that -6 is a solution to \( {5y} + 8 = - {22} \)	When \( y = - 6 \) ,\n\n\[ \n{5y} + 8 = - {22} \n\]\n\nbecomes \( 5\left( {-6}\right) + 8 = - {22} \)\n\n\[ \n- {30} + 8 = - {22} \n\]\n\n99. - 99. which is a true statement, verifying that\n\n-6 is a solution to \( {5y} + 8 = - {22} \)	Yes
Verify that -2 is a solution to \( {3m} - 2 = - {4m} - {16} \) .	When \( m = - 2 \) ,\n\n\[ \n{3m} - 2 = - {4m} - {16} \n\] \n\nbecomes \( 3\left( {-2}\right) - 2 = - 4\left( {-2}\right) - {16} \)\n\n\[ \n- 6 - 2 = \;8 - {16} \n\] \n\nwhich is a true statement, verifying that -2 is a solution to \( {3m} - 2 = - {4m} - {16} \)	Yes
\( m - 8 = 5 \) is associated with \( m \) by subtraction. Undo the association by adding 8 to both sides.	\[ m - 8 + 8 = 5 + 8 \] \[ m + 0 = 13 \] \[ m = 13 \] Check: When \( m = 13 \), becomes \( m - 8 = 5 \) \( 13 - 8 = 5 \) \( 5 = 5 \) a true statement. The solution to \( m - 8 = 5 \) is \( m = 13 \).	Yes
Example 11.22 \( - 3 - 5 = y - 2 + 8 \)	Before we use the addition/subtraction property, we should simplify as much as possible.\n\n\[ \n- 3 - 5 = y - 2 + 8 \n\]\n\n\[ \n- 8 = y + 6 \n\]\n\n6 is associated with \( y \) by addition. Undo the association by subtracting 6 from both sides.\n\n\[ \n- 8 - 6 = y + 6 - 6 \n\]\n\n\[ \n- {14} = y + 0 \n\]\n\n\[ \n- {14} = y \n\]\n\nThis is equivalent to \( y = - {14} \).\n\nCheck: When \( y = - {14} \), \n\n\( - 3 - 5 = y - 2 + 8 \)\n\nbecomes\n\n\( - 3 - 5 \geq - {14} - 2 + 8 \)\n\n\[ \n- 8 \leq - 8\text{.} \n\]\n\na true statement.\n\nThe solution to \( - 3 - 5 = y - 2 + 8 \) is \( y = - {14} \).	Yes
- {5a} + 1 + {6a} = - 2	Begin by simplifying the left side of the equation.\n\n\[\n\underset{-5 + 6 = 1}{\underbrace{-{5a} + 1 + {6a}}} = - 2\n\]\n\n\( a + 1 = - {21} \) is associated with \( a \) by addition. Undo the association by subtracting 1 from both sides.\n\n\[\n a + 1 - 1 = - 2 - 1\n\]\n\n\[\n a + 0 = - 3\n\]\n\n\[\n a = - 3\n\]\n\nCheck: When \( a = - 3 \) ,\n\n\[\n- {5a} + 1 + {6a} = - 2\n\]\n\nbecomes\n\n\( - 5\left( {-3}\right) + 1 + 6\left( {-3}\right) \geq - 2 \)\n\n\[\n- 2 \neq - 2\text{,}\n\]\n\na true statement.\n\nThe solution to \( - {5a} + 1 + {6a} = - 2 \) is \( a = - 3 \) .	No
\( {7k} - 4 = {6k} + 1 \)	\n\( {7k} - 4 = {6k} + 1 \) Since \( {6k} \) represents \( + {6k} \), subtract \( {6k} \) from each side.\n\n\[ \n\underset{7 - 6 = 1}{\underbrace{{7k} - 4 - {6k}}} = \underset{6 - 6 = 0}{\underbrace{{6k} + 1 - {6k}}} \n\]\n\n\( k - 4 = {14} \) is associated with \( k \) by subtraction. Undo the association by adding 4 to both sides.\n\n\[ \nk - 4 + 4 = 1 + 4 \n\]\n\n\[ \nk = 5 \n\]\n\nCheck: When \( k = 5 \) ,\n\n\[ \n{7k} - 4 = {6k} + 1 \n\]\n\nbecomes\n\n\( 7 \cdot 5 - 4 \geqq 6 \cdot 5 + 1 \)\n\n\( {35} - 4 \geqq {30} + 1 \)\n\n\( {31} \pm {31} \).\n\na true statement.\n\nThe solution to \( {7k} - 4 = {6k} + 1 \) is \( k = 5 \).	Yes
\( - 8 + x = 5 \)	\n\( - 8 + x = 5 \) . -8 is associated with \( x \) by addition. Undo the by subtracting -8 from both sides. Subtracting \( - 8 \) we get \( - \left( {-8}\right) = + 8 \) . We actually add 8 to both sides.\n\n\( - 8 + x + 8 = 5 + 8 \)\n\n\( x = {13} \)\n\nCheck: When \( x = {13} \)\n\n\( - 8 + x = 5 \)\n\nbecomes\n\n\( - 8 + {13} \geqq 5 \)\n\n\[ 5 \leq 5 \]\n\na true statement.\n\nThe solution to \( - 8 + x = 5 \) is \( x = {13} \).	Yes
\( {6y} = {54} \)	6 is associated with \( \mathrm{y} \) by multiplication. Undo the association by dividing both sides by 6\n\n\[ \frac{6y}{6} = \frac{54}{6} \]\n\n\[ \frac{\overset{―}{)6}y}{\overset{―}{)6}} = \frac{\frac{9}{\overset{―}{){54}}}}{\overset{―}{)6}} \]\n\n\[ y = 9 \]\n\nCheck: When \( y = 9 \)\n\n\[ {6y} = {54} \]\n\nbecomes\n\n\( 6 \cdot 9 \geqq {54} \)\n\n\( {54} \leq {54} \)\n\na true statement.\n\nThe solution to \( {6y} = {54} \) is \( y = 9 \).	Yes
\[ \frac{x}{-2} = {27} \]	-2 is associated with \( x \) by division. Undo the association by multiplying both sides by -2 .\n\n\[ \left( {-2}\right) \frac{x}{-2} = \left( {-2}\right) {27} \]\n\n\[ \left( \overset{―}{) - 2}\right) \frac{x}{\overset{―}{) - 2}} = \left( { - 2}\right) \;{27} \]\n\n\[ x = - {54} \]\n\nCheck: When \( x = - {54} \) ,\n\n\[ \frac{x}{-2} = {27} \]\n\nbecomes\n\n\[ \frac{-{54}}{-2} \leq {27} \]\n\n\( {27} \leq {27} \)\n\na true statement.\n\nThe solution to \( \frac{x}{-2} = {27} \) is \( x = - {54} \)	Yes
\( \frac{3a}{7} = 6 \)	7 is associated with \( a \) by division. Undo the association by multiplying both sides by 7 .\n\n\( 7 \cdot \frac{3a}{7} = 7 \cdot 6 \)\n\nDivide out the 7's.\n\n\[ \overset{―}{)7} \cdot \frac{3a}{\overset{―}{)7}} = {42} \]\n\n\[ {3a} = {42} \]\n\n3 is associated with \( a \) by multiplication. Undo the association by dividing both sides by 3 .\n\n\[ \frac{3a}{3} = \frac{42}{3} \]\n\n\[ \frac{\overset{―}{)3}a}{\overset{―}{)3}} = {14} \]\n\n\[ a = {14} \]\n\nCheck: When \( a = {14} \), \n\n\[ \frac{3a}{7} = 6 \]\n\nbecomes\n\n\( \frac{3 \cdot {14}}{7} \geqq 6 \)\n\n\[ \frac{42}{7} = 6 \]\n\n\[ 6 \preccurlyeq 6 \]\n\na true statement.\n\nThe solution to \( \frac{3a}{7} = 6 \) is \( a = {14} \).	Yes
- {8x} = {24}	-8 is associated with \( x \) by multiplication. Undo the association by dividing both sides by -8 .\n\n\[ \frac{-{8x}}{-8} = \frac{24}{-8} \]\n\n\[ \frac{-{8x}}{-8} = \frac{24}{-8} \]\n\n\[ x = - 3 \]\n\nCheck: When \( x = - 3 \) ,\n\n\( - {8x} = {24} \)\n\nbecomes\n\n\( - 8\left( {-3}\right) \geqq {24} \)\n\n\( {24} \leq {24} \),\n\na true statement.	Yes

4.32 is what percent of 72 ? Missing factor statement.

\n\n\\begin{matrix} \\text{(percentage)} & \\underset{ \\downarrow }{ = } & \\text{(percent)} & & \\vdots & \\text{(base)} & {\\[ \\text{(product)} = \\text{(factor)} \\cdot \\text{(factor)}\\] } \\\\ & \\downarrow & & \\downarrow & & \\downarrow & \\end{matrix}\n\n\$ \\begin{array}{lllll} {4.32} & = & Q & \\cdot & {72} \\end{array} \$\n\n\$ Q = {4.32} \\div {72}\\; \$ Divide.\n\n\\[ Q = {0.06}\\;\\text{Convert to a percent.} \\]\n\n\\[ Q = 6\\% \\]\n\nThus,4.32 is \$ 6\\% \$ of 72 .\n

Yes

On a 160 question exam, a student got 125 correct answers. What percent is this? Round the result to two decimal places.

\[ \begin{matrix} \underset{ \downarrow }{125} & \text{ is } & \underbrace{\text{ what percent }} & \text{ of } & \underset{ \downarrow }{160}? & \text{ Missing factor statement. } & \\ \text{ (percentage) } & = & \text{ (percent) } & \downarrow & \downarrow & \text{ (base) } & \text{ (product) } = \text{ (factor) } \cdot \text{ (factor)] } \\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \text{ [(product) } = \text{ (factor) } \cdot \text{ (factor)] } & \\ {125} & = & Q & \cdot & {160} & & \end{matrix} \] \n\n$ Q = {125} \div {160}\; $ Divide.\n\n$ Q = {0.78125} $ Round to two decimal places.\n\n\[ Q = {.78} \]\n\nThus, this student received a $ {78}\% $ on the exam.

Yes

A bottle contains 80 milliliters of hydrochloric acid (HCl) and 30 milliliters of water. What percent of $ \mathrm{{HCl}} $ does the bottle contain? Round the result to two decimal places.

We need to determine the percent. The total amount of liquid in the bottle is\n\n80 milliliters +30 milliliters $ = {110} $ milliliters.\n\n\[ \begin{matrix} {80} & \\downarrow & \\left( \\text{ percentage }\\right) & = & \\text{ (percent) } & \\downarrow & \\text{ (base) } & \\left\\lbrack {\\text{ (product) } = \\text{ (factor) } \\cdot \\text{ (factor) }}\\right\\rbrack \\ \\downarrow & \\downarrow & \\downarrow & \\downarrow & \\downarrow & \\ {80} & = & Q & \\cdot & {110} & \\end{matrix} \]\n\n$ Q = {80} \\div {110}\\; $ Divide.\n\n$ Q = {0.727272}\\ldots \\; $ Round to two decimal places.\n\n$ Q \\approx {73}\\% \\; $ The symbol \

No

Five years ago a woman had an annual income of $19,200. She presently earns $42,000 annually. By what percent has her salary increased? Round the result to two decimal places.

We need to determine the percent.\n\n\$ Q = {42},{000} \\div {19},{200}\\; \$ Divide.\n\n\$ Q = {2.1875}\\; \$ Round to two decimal places.\n\n\$ Q = \\;{2.19}\\; \$ Convert to a percent.\n\n\$ Q = {219}\\% \\; \$ Convert to a percent.\n\nThus, this woman’s annual salary has increased \$ {219}\\% \$ .

No

15 is 30% of what number?

<table><thead><tr><th>15 $ \downarrow $</th><th>is $ \downarrow $</th><th>30% $ \downarrow $</th><th>of $ \downarrow $</th><th>what number?</th><th>Missing factor statement.</th></tr></thead><tr><td>(percentage) $ \downarrow $</td><td>1</td><td>(percent) $ \downarrow $</td><td>$ \downarrow $</td><td>(base) $ \downarrow $</td><td>$ \left\lbrack {\left( \text{percentage}\right) = \left( \text{factor}\right) \cdot \left( \text{factor}\right) }\right\rbrack $</td></tr><tr><td>15</td><td>$ = $</td><td>30%</td><td>.</td><td>$ B $</td><td>Convert to decimals.</td></tr><tr><td>15</td><td>$ = $</td><td>.30</td><td>.</td><td>$ B $</td><td>$ \lbrack \left( \text{missing factor}\right) = \left( \text{product}\right) \div ( $ known fa</td></tr></table>\n\n\[ B = {15} \div {.30} \]\n\n\[ B = {50} \]\n\nThus, 15 is $ {30}\% $ of 50.

Yes

56,43 is 33% of what number? Missing factor statement.

\n\[ \n{56.43} = {33}\% \; \cdot \;B\;\text{Convert to decimals.} \n\]\n\[ \n{56.43} = {.33}\;B\;\text{Divide.} \n\]\n\[ \nB = {56.43} \div {.33} \n\]\n\[ \nB = \;{171} \n\]\n\nThus, 56.43 is 33% of 171.

Yes

Fifteen milliliters of water represents $ 2\% $ of a hydrochloric acid (HCl) solution. How many milliliters of solution are there?

We need to determine the total supply. The word supply indicates base.\n\n\[ \n\begin{array}{llllll} {15} & \text{ is } & 2\% & \text{ of } & \underbrace{\text{ what number? }} & \text{ Missing factor statement. } \end{array} \]\n\n\[ \n\begin{matrix} \text{ (percentage) } & = & \text{ (percent) } \\ \downarrow & \downarrow & \downarrow \end{matrix}\;\text{ (base) } \]\n\n\[ \n\begin{array}{lllll} {15} & = & {.02} & \cdot & \text{ Divide. } \end{array} \]\n\n\[ \nB = {15} \div {.02} \]\n\n\[ \nB = {750} \]\n\nThus, there are 750 milliliters of solution in the bottle.

Yes

A clothing item is priced at $20.40. This marked price includes a 15% discount. What is the original price?

We need to determine the original price. We can think of the original price as the starting place. Starting place indicates base. We need to determine the base. The new price, $20.40, represents 100% - 15% = 85% of the original price.\n\n<table><tr><td>20.40</td><td>is $ \downarrow $</td><td>85% $ \downarrow $</td><td>of $ \downarrow $</td><td>what number?</td><td>Missing factor statement.</td></tr><tr><td>$ \begin{matrix} \text{(percentage)} \\ \downarrow \end{matrix} $</td><td>1</td><td>(percent) $ \downarrow $</td><td>$ \downarrow $</td><td>(base) $ \downarrow $</td><td></td></tr><tr><td>20.40</td><td>$ = $</td><td>85%</td><td>.</td><td>$ B $</td><td>Convert to decimals.</td></tr><tr><td>20.40</td><td>$ = $</td><td>.85</td><td>.</td><td>$ B $</td><td>$ \lbrack \left( \text{missing factor}\right) = \left( \text{product}\right) \div \left( \text{known factor}\right) $</td></tr></table>\n\n\[ B = {20.40} \div {.85}\text{Divide.} \]\n\n\[ B = \;{24} \]\n\nThus, the original price of the item is $24.00.

Yes

Estimate the sum: $ 2,{357} + 6,{106} $ .

The sum can be estimated by $ 2,{400} + 6,{100} = 8,{500} $ . (It is quick and easy to add 24 and 61 .) Thus, $ 2,{357} + 6,{106} $ is about 8,400 . In fact, $ 2,{357} + 6,{106} = 8,{463} $ .

Yes

Estimate the difference: $ 5,{203} - 3,{015} $ .

Notice that 5,203 is near $ \underset{\text{two nonzero }}{\underbrace{5,{200},}} $ and that 3,015 is near $ \underset{\text{one nonzero }}{\underbrace{3,{000}.}} $ \n\nThe difference can be estimated by $ 5,{200} - 3,{000} = 2,{200} $ . \n\nThus, $ 5,{203} - 3,{015} $ is about 2,200 . In fact, $ 5,{203} - 3,{015} = 2,{188} $ .

Yes

Estimate the product: $ {87} \cdot 4,{316} $ .

Notice that 87 is close to $ \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{90},}} $ and that 4,316 is close to $ \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{4,{000}.}} $\n\nThe product can be estimated by $ {90} \cdot 4,{000} = {360},{000} $.\n\nThus, $ {87} \cdot 4,{316} $ is about 360,000 . In fact, $ {87} \cdot 4,{316} = {375},{492} $.

Yes

Estimate the quotient: $ {153} \div {17} $ .

Notice that 153 is close to $ \underset{\begin{matrix} \text{ two nonzero } \\ \text{ digits } \end{matrix}}{\underbrace{150}} $ and that 17 is close to $ \underset{\text{two nonzero }}{\underbrace{\text{ 15. }}} $\n\nThe quotient can be estimated by $ {150} \div {15} = {10} $.\n\nThus, $ {153} \div {17} $ is about 10 . In fact, $ {153} \div {17} = 9 $ .

Yes

Estimate the quotient: $ {742},{000} \div 2,{400} $ .

Notice that 742,000 is close to $ \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{700},{000}}} $, and that 2,400 is close to $ \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{2,{000}.}} $\n\nThe quotient can be estimated by $ {700},{000} \div 2,{000} = {350} $.\n\nThus, $ {742},{000} \div 2,{400} $ is about 350 . In fact, $ {742},{000} \div 2,{400} = {309.1}\overline{6} $ .

Yes

Estimate the product: (31.28) (14.2).

Notice that 31.28 is close to $ \underset{\begin{matrix} \text{ one nonzero } \\ \text{ digit } \end{matrix}}{\underbrace{{30},}} $ and that 14.2 is close to $ \underset{\text{two nonzero }}{\underbrace{{15}.}} $\n\nThe product can be estimated by $ {30} \cdot {15} = {450} $ . ( $ 3 \cdot {15} = {45} $, then affix one zero.)\n\nThus,(31.28)(14.2) is about 450 . In fact,(31.28)(14.2) $ = {444.176} $ .

Yes

Estimate $ {21}\% $ of 5.42.

Notice that $ {21}\% = {.21} $ as a decimal, and that .21 is close to one nonzero digit\n\nNotice also that 5.42 is close to\n\nThen, $ {21}\% $ of 5.42 can be estimated by $ \left( {.2}\right) \left( 5\right) = 1 $ .\n\nThus, $ {21}\% $ of 5.42 is about 1 . In fact, $ {21}\% $ of 5.42 is 1.1382 .

No

$ {27} + {48} + {31} + {52} $ .

27 and 31 cluster near 30 . Their sum is about $ 2 \cdot {30} = {60} $ .\n\n48 and 52 cluster near 50 . Their sum is about $ 2 \cdot {50} = {100} $ .\n\nThus, $ {27} + {48} + {31} + {52} $ is about $ \begin{aligned} \left( {2 \cdot {30}}\right) + \left( {2 \cdot {50}}\right) & = {60} + {100} \\ & = {160} \end{aligned} $\n\nIn fact, $ {27} + {48} + {31} + {52} = {158} $ .

Yes

$ {88} + {21} + {19} + {91} $

88 and 91 cluster near 90 . Their sum is about $ 2 \cdot {90} = {180} $ . 21 and 19 cluster near 20 . Their sum is about $ 2 \cdot {20} = {40} $ . Thus, $ {88} + {21} + {19} + {91} $ is about $ \begin{aligned} \left( {2 \cdot {90}}\right) + \left( {2 \cdot {20}}\right) & = {180} + {40} \\ & = {220} \end{aligned} $ In fact, $ {88} + {21} + {19} + {91} = {219} $

Yes

$ {17} + {21} + {48} + {18} $

$ {17},{21} $, and 18 cluster near 20 . Their sum is about $ 3 \cdot {20} = {60} $ . 48 is about 50 . Thus, $ {17} + {21} + {48} + {18} $ is about $ \begin{aligned} \left( {3 \cdot {20}}\right) + {50} & = {60} + {50} \\ & = {110} \end{aligned} $. In fact, $ {17} + {21} + {48} + {18} = {104} $.

Yes

$ {61} + {48} + {49} + {57} + {52} $

61 and 57 cluster near 60 . Their sum is about $ 2 \cdot {60} = {120} $ . \n\n$ {48},{49} $, and 52 cluster near 50 . Their sum is about $ 3 \cdot {50} = {150} $ . \n\nThus, $ {61} + {48} + {49} + {57} + {52} $ is about $ \begin{aligned} \left( {2 \cdot {60}}\right) + \left( {3 \cdot {50}}\right) & = {120} + {150} \\ & = {270} \end{aligned} $ \n\nIn fact, $ {61} + {48} + {49} + {57} + {52} = {267} $

Yes

$ {706} + {321} + {293} + {684} $ .

706 and 684 cluster near 700 . Their sum is about $ 2 \cdot {700} = 1,{400} $ .\n\n321 and 293 cluster near 300 . Their sum is about $ 2 \cdot {300} = {600} $ .\n\nThus, $ {706} + {321} + {293} + {684} $ is about $ \begin{aligned} \left( {2 \cdot {700}}\right) + \left( {2 \cdot {300}}\right) & = 1,{400} + {600} \\ & = 2,{000} \end{aligned} $\n\nIn fact, $ {706} + {321} + {293} + {684} = 2,{004} $ .

Yes

\[ 4\left( {6 + 2}\right) = 4 \cdot 6 + 4 \cdot 2 \]

\[ = {24} + 8 \] \[ = {32} \] Using the order of operations, we get \[ 4\left( {6 + 2}\right) = 4 \cdot 8 \] \[ = \;{32} \]

Yes

\[ 8\left( {9 + 6}\right) = 8 \cdot 9 + 8 \cdot 6 \]

\[ = {72} + {48} \] \[ = {120} \] Using the order of operations, we get \[ 8\left( {9 + 6}\right) = 8 \cdot {15} \] \[ = \;{120} \]

Yes

\[ 4\left( {9 - 5}\right) = 4 \cdot 9 - 4 \cdot 5 \]

\[ = {36} - {20} \] \[ = {16} \]

Yes

$ {25} \cdot {23} $

Notice that $ {23} = {20} + 3 $ . We now write\n\n\[ \n{25} \cdot {23} = \underset{25}{\underbrace{{55}\left( {{20} + 3}\right) }} \n\]\n\n\[ \n= {25} \cdot {20} + {25} \cdot 3 \n\]\n\n\[ \n= {500} + {75} \n\]\n\n\[ \n= {575} \n\]\n\nThus, $ {25} \cdot {23} = {575} $

Yes

$ {15} \cdot {37} $

Notice that $ {37} = {30} + 7 $ . We now write\n\n\[ \n{15} \cdot {37} = {15}\left( {{30} + 7}\right) \n\]\n\n\[ \n= {15} \cdot {30} + {15} \cdot 7 \n\]\n\n\[ \n= {450} + {105} \n\]\n\n\[ \n= {555} \n\]\n\nThus, $ {15} \cdot {37} = {555} $

Yes

$ {15} \cdot {86} $

Notice that $ {86} = {80} + 6 $ . We now write\n\n\[ \n{15} \cdot {86} = \underset{15}{\underbrace{{15}\left( {{80} + 6}\right) }} \n\]\n\n\[ \n= {15} \cdot {80} + {15} \cdot 6 \n\]\n\n\[ \n= 1,{200} + {90} \n\]\n\n\[ \n= 1,{290} \n\]\n\nWe could have proceeded by writing 86 as $ {90} - 4 $ .\n\n\[ \n{15} \cdot {86} = {15}\left( {{90} - 4}\right) \n\]\n\n\[ \n= {15} \cdot {90} - {15} \cdot 4 \n\]\n\n\[ \n= 1,{350} - {60} \n\]\n\n\[ \n= 1,{290} \n\]

Yes

Estimate $ \frac{3}{5} + \frac{5}{12} $ .

Notice that $ \frac{3}{5} $ is about $ \frac{1}{2} $, and that $ \frac{5}{12} $ is about $ \frac{1}{2} $. Thus, $ \frac{3}{5} + \frac{5}{12} $ is about $ \frac{1}{2} + \frac{1}{2} = 1 $. In fact, $ \frac{3}{5} + \frac{5}{12} = \frac{61}{60} $, a little more than 1.

Yes

Estimate $ 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} $ .

Adding the whole number parts, we get 20 . Notice that $ \frac{3}{8} $ is close to $ \frac{1}{4},\frac{9}{10} $ is close to 1, and $ \frac{1}{5} $ is close to $ \frac{1}{4} $ . Then $ \frac{3}{8} + \frac{9}{10} + \frac{1}{5} $ is close to $ \frac{1}{4} + 1 + \frac{1}{4} = 1\frac{1}{2} $ . Thus, $ 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} $ is close to $ {20} + 1\frac{1}{2} = {21}\frac{1}{2} $ . In fact, $ 5\frac{3}{8} + 4\frac{9}{10} + {11}\frac{1}{5} = {21}\frac{19}{40} $, a little less than $ {21}\frac{1}{2} $ .

Yes

Convert 11 yards to feet.

Looking in the unit conversion table under length, we see that $ 1\mathrm{{yd}} = 3\mathrm{{ft}} $ . There are two corresponding unit fractions, $ \frac{1\mathrm{{yd}}}{3\mathrm{{ft}}} $ and $ \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} $ . Which one should we use? Look to see which unit we wish to convert to. Choose the unit fraction with this unit in the numerator. We will choose $ \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} $ since\n\nthis unit fraction has feet in the numerator. Now, multiply 11 yd by the unit fraction. Notice that since the unit fraction has the value of 1 , multiplying by it does not change the value of 11 yd.\n\n\[ \n{11}\mathrm{{yd}} = \frac{{11}\mathrm{{yd}}}{1} \cdot \frac{3\mathrm{{ft}}}{1\mathrm{{yd}}} \]\n\n$ = \frac{{11}\overset{1}{\overbrace{)\mathrm{{yd}}}}}{1} \cdot \frac{3\mathrm{{ft}}}{1\overline{\overline{)\mathrm{{yd}}}}}\; $ (Units can be added, subtracted, multiplied, and divided, just as numbers can.)\n\n\[ \n= \;\frac{{11} \cdot 3\mathrm{{ft}}}{1} \]\n\n\[ \n= \;{33}\mathrm{{ft}} \]\n\nThus, $ {11}\mathrm{{yd}} = {33}\mathrm{{ft}} $.

Yes

Convert $ {36}\mathrm{{fl}} $ oz to pints.

\n\n$ {36}\mathrm{{fl}}\mathrm{{oz}} = \frac{{36}\mathrm{{fl}}\mathrm{{oz}}}{1} \cdot \frac{1\mathrm{{pt}}}{{16}\mathrm{{fl}}\mathrm{{oz}}}\; $ Divide out common units.\n\n\[ \n= \frac{{36}\overline{)\mathrm{{fl}}\text{ oz }}}{1} \cdot \frac{1\mathrm{{pt}}}{{16}\overline{)\mathrm{{fl}}\text{ oz }}} \n\] \n\n\[ \n= \;\frac{{36} \cdot 1\mathrm{{pt}}}{16} \n\] \n\n$ = \;\frac{{36}\mathrm{{pt}}}{16}\; $ Reduce.\n\n$ = \;\frac{9}{4}\mathrm{{pt}}\; $ Convert to decimals: $ \frac{9}{4} = {2.25} $. \n\nThus, $ {36}\mathrm{{fl}} $ oz $ = {2.25}\mathrm{{pt}} $ .

Yes

Convert 3 kilograms to grams.

(a) $ 3\mathrm{\;{kg}} $ can be written as $ {3.0}\mathrm{\;{kg}} $ . Then, ![7d43c799-07cb-4976-9fbc-3035178b5209_529_0.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_529_0.jpg)\n\nThus, $ 3\mathrm{\;{kg}} = 3,{000}\mathrm{\;g} $ .\n\n(b) We can also use unit fractions to make this conversion. Since we are converting to grams, and\n\n$ 1,{000}\mathrm{\;g} = 1\mathrm{\;{kg}} $, we choose the unit fraction $ \frac{1,{000}\mathrm{\;g}}{1\mathrm{l}\mathrm{g}} $ since grams is in the numerator.\n\n\[ 3\mathrm{\;{kg}} = 3\mathrm{\;{kg}} \cdot \frac{1,{000}\mathrm{\;g}}{1\mathrm{\;{kg}}} \]\n\n\[ = 3\overline{)\mathrm{{kg}}} \cdot \frac{1,{000}\mathrm{g}}{1\overline{)\mathrm{{kg}}}} \]\n\n\[ = \;3 \cdot 1,{000}\mathrm{\;g} \]\n\n\[ \text{=}\;\text{3,000}\;\text{g} \]

Yes

Convert 67.2 hectoliters to milliliters.

$ {67.2}\mathrm{\;{hL}} = \underset{1\text{ 2 }3\text{ 4 }5}{{67}\underbrace{20000}},\mathrm{\;{mL}} $\n\nThus, $ {67.2}\mathrm{\;{hL}} = 6,{720},{000}\mathrm{\;{mL}} $.

Yes

Convert 100.07 centimeters to meters.

Thus, $ {100.07}\mathrm{\;{cm}} = {1.0007}\mathrm{\;m} $.

Yes

Convert 0.16 milligrams to grams.

$ {0.16}\mathrm{{mg}} = 0,\underset{3}{000}{16}\mathrm{\;g} $\n\nThus, $ {0.16}\mathrm{{mg}} = {0.00016}\mathrm{\;g} $.

Yes

Simplify 19 in.

Since 12 in. $ = 1\mathrm{{ft}} $, and $ {19} = {12} + 7 $ ,\n\n19 in. $ = {12} $ in. +7 in.\n\n$ = \;1\mathrm{{ft}} + 7\mathrm{{in}} $ .\n\n$ = \;1\mathrm{{ft}}7\mathrm{{in}} $ .

Yes

Simplify 2 hr 75 min.

Since $ {60}\mathrm{\;{min}} = 1\mathrm{{hr}} $, and $ {75} = {60} + {15} $ ,\n\n$ 2\mathrm{{hr}}{75}\mathrm{\;{min}} = 2\mathrm{{hr}} + {60}\mathrm{\;{min}} + {15}\mathrm{\;{min}} $\n\n\[ \n= \;2\mathrm{{hr}} + 1\mathrm{{hr}} + {15}\mathrm{\;{min}} \n\]\n\n\[ \n= \;3\mathrm{{hr}} + {15}\mathrm{\;{min}} \n\]\n\n\[ \n= \;3\mathrm{{hr}}{15}\mathrm{\;{min}} \n\]

Yes

Simplify $ {43}\mathrm{{fl}} $ oz.

Since $ 8\mathrm{{fl}} $ oz $ = 1\mathrm{c}\left( {1\mathrm{{cup}}}\right) $, and $ {43} \div 8 = 5\mathrm{R}3 $ ,\n\n$ {43}\mathrm{{fl}}\mathrm{{oz}} = {40}\mathrm{{fl}}\mathrm{{oz}} + 3\mathrm{{fl}}\mathrm{{oz}} $\n\n$ = 5 \cdot 8\mathrm{{fl}}\mathrm{{oz}} + 3\mathrm{{fl}}\mathrm{{oz}} $\n\n\[ = \;5 \cdot 1\mathrm{c} + 3\mathrm{{fl}}\text{oz} \]\n\n\[ = \;5\mathrm{c} + 3\mathrm{{fl}}\text{oz} \]\n\nBut, $ 2\mathrm{c} = 1\mathrm{{pt}} $ and $ 5 \div 2 = 2\mathrm{R}1 $ . So,\n\n\[ 5\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} = 2 \cdot 2\mathrm{c} + 1\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} \]\n\n\[ = 2 \cdot 1\mathrm{{pt}} + 1\mathrm{c} + 3\mathrm{{fl}}\mathrm{{oz}} \]

Yes

Add $ 6\mathrm{{ft}}8 $ in. to $ 2\mathrm{{ft}}9 $ in.

\n$ 6\mathrm{{ft}}8\mathrm{{in}} $ .\n\n$ + 2\mathrm{{ft}}9\mathrm{{in}} $ .\n\n8 ft 17 in. Simplify this denominate number.\n\nSince 12 in. $ = 1\mathrm{{ft}} $ ,\n\n\[ 8\mathrm{{ft}} + {12}\mathrm{{in}}. + 5\mathrm{{in}}. = 8\mathrm{{ft}} + 1\mathrm{{ft}} + 5\mathrm{{in}}. \]\n\n\[ = \;9\mathrm{{ft}} + 5\mathrm{{in}}\text{.} \]\n\n\[ = \;9\mathrm{{ft}}5\mathrm{{in}}\text{.} \]\n

Yes

Subtract 5 da 3 hr from 8 da 11 hr.

$ 8\mathrm{{da}}{11}\mathrm{{hr}} $\n\n$ - 5\mathrm{{da}}\;3\mathrm{{hr}} $\n\n$ 3\mathrm{{da}}\;8\mathrm{{hr}} $

Yes

Subtract 3 lb 14 oz from 5 lb 3 oz.

We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz from the pounds.\n\n5 lb 3 oz $ = \;5\mathrm{{lb}} + 3\mathrm{{oz}} $\n\n\[ = \;4\mathrm{{lb}} + 1\mathrm{{lb}} + 3\mathrm{{oz}} \]\n\n$ = {4lb} + {16oz} + {3oz}\; $ (Since $ {1lb} = {16oz} $ .)\n\n$ = \;4\mathrm{{lb}} + {19}\mathrm{{oz}} $\n\n$ = \;4\mathrm{{lb}}{19}\mathrm{{oz}} $\n\n$ 4\mathrm{{lb}}{19}\mathrm{{oz}} $\n\n$ - 3\mathrm{{lb}}{14}\mathrm{{oz}} $\n\n$ 1\mathrm{{lb}}\;5\mathrm{{oz}} $

Yes

Subtract 4 da 9 hr $ {21}\mathrm{\;{min}} $ from 7 da 10 min.

Borrow 1 do from the 7 da.\n\n$ - 4\mathrm{{da}}9\mathrm{{hr}}{21}\mathrm{\;{min}} $\n\noud 24 m 10 mm\n\n-4 da 9 hr $ {21}\mathrm{\;{min}} $\n\n6 da $ {23}\mathrm{{hr}}{70}\mathrm{\;{min}} $\n\n-4 da 9 hr $ {21}\mathrm{\;{min}} $\n\n2 da 14 hr $ {49}\mathrm{\;{min}} $

No

\[ 6 \cdot \left( {2\mathrm{{ft}}4\mathrm{{in}}.}\right) = 6 \cdot 2\mathrm{{ft}} + 6 \cdot 4\mathrm{{in}}. \]

\[ = \;{12}\mathrm{{ft}} + {24}\text{in.} \] Since $ 3\mathrm{{ft}} = 1\mathrm{{yd}} $ and 12 in. $ = 1\mathrm{{ft}} $, \[ {12}\mathrm{{ft}} + {24} \) in. \( = 4\mathrm{{yd}} + 2\mathrm{{ft}} \] \[ = \;4\mathrm{{yd}}2\mathrm{{ft}} \]

Yes

\[ 8 \cdot \left( {5\mathrm{{hr}}{21}\mathrm{\;{min}}{55}\mathrm{{sec}}}\right) = 8 \cdot 5\mathrm{{hr}} + 8 \cdot {21}\mathrm{\;{min}} + 8 \cdot {55}\mathrm{{sec}} \]

\[ = \;{40}\mathrm{{hr}} + {168}\mathrm{\;{min}} + {440}\mathrm{{sec}} \]\n\[ = {40}\mathrm{{hr}} + {168}\mathrm{\;{min}} + 7\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{40}\mathrm{{hr}} + {175}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{40}\mathrm{{hr}} + 2\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{42}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;{24}\mathrm{{hr}} + {18}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = 1\mathrm{{da}} + {18}\mathrm{{hr}} + {55}\mathrm{\;{min}} + {20}\mathrm{{sec}} \]\n\[ = \;1\mathrm{{da}}{18}\mathrm{{hr}}{55}\mathrm{\;{min}}{20}\mathrm{{sec}} \]

Yes

$ \left( {{12}\min {40}\sec }\right) \div 4 $

Thus $ \left( {{12}\mathrm{\;{min}}{40}\mathrm{{sec}}}\right) \div 4 = 3\mathrm{\;{min}}{10}\mathrm{{sec}} $

Yes

$ \left( {5\mathrm{{yd}}2\mathrm{{ft}}9\mathrm{{in}}\text{.}}\right) \div 3 $

$ 1\mathrm{{yd}}2\mathrm{{ft}}{11}\mathrm{{in}} $ . 3) 5 yd 2 ft 9 in. / $ 3\mathrm{{yd}} $ 2 yd 2 ft $ \rightarrow 8\mathrm{{ft}} $ $ 6\mathrm{{ft}} $ 2 ft 9 in. $ \rightarrow {33} $ in. \n\n$ \frac{{33}\text{ in. }}{0} $\n\nConvert to feet: $ 2\mathrm{{yd}}2\mathrm{{ft}} = 8\mathrm{{ft}} $ .\n\nConvert to inches: $ 2\mathrm{{ft}}9\mathrm{{in}} $ . $ = {33}\mathrm{{in}} $ .\n\nThus $ \left( {5\mathrm{{yd}}2\mathrm{{ft}}9\mathrm{{in}}.}\right) \div 3 = 1\mathrm{{yd}}2\mathrm{{ft}}{11}\mathrm{{in}} $ .

No

Find the approximate circumference of the circle.

Use the formula $ C = {\pi d} $. $ C \approx \left( {3.14}\right) \left( {6.2}\right) $. $ C \approx {19.648}\mathrm{\;{mm}} $. This result is approximate since $ \pi $ has been approximated by 3.14.

Yes

Find the approximate circumference of a circle with radius 18 inches.

Since we’re given that the radius, $ r $, is $ {18} $ in., we’ll use the formula $ C = {2\pi r} $ .\n\n\[ C \approx \left( 2\right) \left( {3.14}\right) \left( {{18}\text{ in. }}\right) \]\n\n$ C \approx {113.04} $ in.

Yes

Find the approximate perimeter of the figure.

We notice that we have two semicircles (half circles).\n\nThe larger radius is $ {6.2}\mathrm{\\;{cm}} $.\n\nThe smaller radius is $ {6.2}\mathrm{\\;{cm}} - {2.0}\mathrm{\\;{cm}} = {4.2}\mathrm{\\;{cm}} $.\n\nThe width of the bottom part of the rectangle is $ {2.0}\mathrm{\\;{cm}} $. Perimeter $ = $ $ {2.0}\mathrm{\\;{cm}} $ $ {5.1}\mathrm{\\;{cm}} $ $ {2.0}\mathrm{\\;{cm}} $ $ {5.1}\mathrm{\\;{cm}} $\n\n$ \left( {0.5}\right) \cdot \left( 2\right) \cdot \left( {3.14}\right) \cdot \left( {{6.2}\mathrm{\\;{cm}}}\right) $ Circumference of outer semicircle.\n\n$ + \underline{\left( {0.5}\right) \cdot \left( 2\right) \cdot \left( {3.14}\right) \cdot \left( {{4.2}\mathrm{\\;{cm}}}\right) } $ Circumference of inner semicircle $ {6.2}\mathrm{\\;{cm}} - {2.0}\mathrm{\\;{cm}} = {4.2}\mathrm{\\;{cm}} $ The 0.5 appears because we want the perimeter of only half a circle.

Yes

Find the area of the triangle.

\[ \n{A}_{T} = \frac{1}{2} \cdot b \cdot h \n\] \n\[ \n= \;\frac{1}{2} \cdot {20} \cdot 6\;\text{sq ft} \n\] \n\[ \n= {10} \cdot 6\text{ sq ft } \n\] \n\[ \n= \;{60}\mathrm{{sq}}\mathrm{{ft}} \n\] \n\[ \n= \;{60}{\mathrm{{ft}}}^{2} \n\] \nThe area of this triangle is $ {60}\mathrm{{sq}}\mathrm{{ft}} $, which is often written as $ {60}{\mathrm{{ft}}}^{2} $ .

Yes

Find the area of the rectangle.

Let's first convert $ 4\mathrm{{ft}}2\mathrm{{in}} $ . to inches. Since we wish to convert to inches, we’ll use the unit fraction $ \frac{{12}\text{ in. }}{1\mathrm{{ft}}} $ since it has inches in the numerator. Then,\n\n\[ 4\mathrm{{ft}} = \frac{4\mathrm{{ft}}}{1} \cdot \frac{{12}\mathrm{{in}}.}{1\mathrm{{ft}}} \]\n\n\[ = \frac{4\overline{)\mathrm{{ft}}}}{1} \cdot \frac{{12}\mathrm{{in}}.}{1\overline{)\mathrm{{ft}}}} \]\n\n\[ = \;{48}\text{ in. } \]\n\nThus, $ 4\mathrm{{ft}}2\mathrm{{in}}. = {48}\mathrm{{in}} \cdot + 2\mathrm{{in}}. = {50}\mathrm{{in}} $ .\n\n\[ {A}_{R} = \;l \cdot w \]\n\n\[ = {50}\text{in.} \cdot 8\text{in.} \]\n\n\[ = \;{400}\mathrm{{sq}}\text{in.} \]\n\nThe area of this rectangle is $ {400}\mathrm{{sq}} $ in.

Yes

Find the area of the parallelogram.

\[ \n{A}_{P} = \;b \cdot h \]\n\[ \n= {10.3}\mathrm{\;{cm}} \cdot {6.2}\mathrm{\;{cm}} \]\n\[ \n= \;{63.86}\mathrm{{sq}}\mathrm{{cm}} \]\n\nThe area of this parallelogram is $ {63.86}\mathrm{{sq}}\mathrm{{cm}} $ .

Yes

Find the area of the trapezoid.

\[ \n{A}_{\text{Trap }} = \;\frac{1}{2} \cdot \left( {{b}_{1} + {b}_{2}}\right) \cdot h \]\n\[ \n= \frac{1}{2} \cdot \left( {{14.5}\mathrm{\;{mm}},+,{20.4}\mathrm{\;{mm}}}\right) \cdot \left( {{4.1}\mathrm{\;{mm}}}\right) \]\n\[ \n= \;\frac{1}{2} \cdot \left( {{34.9}\mathrm{\;{mm}}}\right) \cdot \left( {{4.1}\mathrm{\;{mm}}}\right) \]\n\[ \n= \;\frac{1}{2} \cdot \left( {{143.09}\mathrm{{sq}}\mathrm{{mm}}}\right) \]\n\[ \n= \;{71.545}\mathrm{{sq}}\mathrm{{mm}} \]\n\nThe area of this trapezoid is $ {71.545}\mathrm{{sq}}\mathrm{{mm}} $ .

Yes

Find the approximate area of the circle.

\[ \n{A}_{c} = \;\pi \cdot {r}^{2} \n\] \n\[ \n\approx \;\left( {3.14}\right) \cdot {\left( {16.8}\mathrm{{ft}}\right) }^{2} \n\] \n\[ \n\approx \left( {3.14}\right) \cdot \left( {{282.24}\mathrm{{sq}}\mathrm{{ft}}}\right) \n\] \n\[ \n\approx \;{888.23}\mathrm{{sq}}\mathrm{{ft}} \n\] \n\nThe area of this circle is approximately $ {886.23}\mathrm{{sq}} $ ft.

Yes

Find the volume of the rectangular solid.

\[ {V}_{R} = \;l \cdot w \cdot h \] \[ = 9\text{in.} \cdot {10}\text{in.} \cdot 3\text{in.} \] \[ = \;{270}\mathrm{{cu}}\text{in.} \] \[ = \;{270}\text{in.}{}^{3} \] The volume of this rectangular solid is $ {270}\mathrm{{cu}} $ in.

Yes

Find the approximate volume of the sphere.

\[ \n{V}_{S} = \frac{4}{3} \cdot \pi \cdot {r}^{3} \]\n\[ \n\approx \;\left( \frac{4}{3}\right) \cdot \left( {3.14}\right) \cdot {\left( 6\mathrm{\;{cm}}\right) }^{3} \]\n\[ \n\approx \;\left( \frac{4}{3}\right) \cdot \left( {3.14}\right) \cdot \left( {{216}\mathrm{{cu}}\mathrm{{cm}}}\right) \]\n\[ \n\approx \;{904.32}\mathrm{{cu}}\mathrm{{cm}} \]\n\nThe approximate volume of this sphere is $ {904.32}\mathrm{{cu}}\mathrm{{cm}} $, which is often written as $ {904.32}{\mathrm{\;{cm}}}^{3} $ .

Yes

Is every whole number a natural number?

No. The number 0 is a whole number but it is not a natural number.

Yes

Is there an integer that is not a natural number?

Yes. Some examples are 0, -1, -2, -3, and -4.

Yes

Is there an integer that is a whole number?

Yes. In fact, every whole number is an integer.

Yes

What integers can replace $ x $ so that the following statement is true?\n\n$ - 3 \leq x < 2 $ ![7d43c799-07cb-4976-9fbc-3035178b5209_589_0.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_589_0.jpg)

The integers are $ - 3, - 2, - 1,0,1 $ .

Yes

Write each expression in words. Exercise 10.3.1

Solution on p. 621.

No

If $ a = - 4 $, then $ - a = - \left( {-4}\right) = 4 $ . Also, $ - \left( {-a}\right) = a = - 4 $ .

Yes

$ - \left| {-3}\right| = - 3 $

The quantity on the left side of the equal sign is read as \

No

The number enclosed within the absolute value bars is a nonnegative number, so the upper part of the definition applies. This part says that the absolute value of 8 is 8 itself.

\[ \left| 8\right| = 8 \]

Yes

$ \left| {-3}\right| $

The number enclosed within absolute value bars is a negative number, so the lower part of the definition applies. This part says that the absolute value of -3 is the opposite of -3 , which is\n\n- (-3). By the definition of absolute value and the double-negative property,\n\n\[ \left| {-3}\right| = - \left( {-3}\right) = 3 \]

Yes

$ \left( {-4}\right) + \left( {-9}\right) $

$ \begin{array}{l} \left| {-4}\right| = 4 \\ \left| {-9}\right| = 9 \end{array}\} $ Add these absolute values. \n\n$ 4 + 9 = {13} $\n\nThe common sign is \

No

$ 7 + \left( {-2}\right) $

Subtract absolute values: $ 7 - 2 = 5 $. Attach the proper sign: \

No

$ 3 + \left( {-{11}}\right) $

Subtract absolute values: $ {11} - 3 = 8 $ .\n\nAttach the proper sign: \

No

The morning temperature on a winter's day in Lake Tahoe was -12 degrees. The afternoon temperature was 25 degrees warmer. What was the afternoon temperature?

We need to find $ - {12} + {25} $ . ![7d43c799-07cb-4976-9fbc-3035178b5209_603_2.jpg](images/7d43c799-07cb-4976-9fbc-3035178b5209_603_2.jpg)\n\nSubtract absolute values: $ {25} - {12} = {16} $ .\n\nAttach the proper sign: \

No

The high temperature today in Lake Tahoe was $ {26}^{ \circ }\mathrm{F} $ . The low temperature tonight is expected to be $ - {7}^{ \circ }\mathrm{F} $ . How many degrees is the temperature expected to drop?

We need to find the difference between 26 and -7 . \[ {26} - \left( {-7}\right) = {26} + 7 = {33} \] Thus, the expected temperature drop is $ {33}^{ \circ }\mathrm{F} $ .

Yes

\[ - 6 - \left( {-5}\right) - {10} = - 6 + 5 + \left( {-{10}}\right) \]

\[ = \left( {-6 + 5}\right) + \left( {-{10}}\right) \]\n\[ = \; - 1 + \left( {-{10}}\right) \]\n\[ = \; - {11} \]

Yes

$ 3,{187} - 8,{719} $

Thus, $ 3,{187} - 8,{719} = - 5,{532} $.

Yes

$ - {156} - \left( {-{211}}\right) $

Method A:\n\n<table><thead><tr><th colspan=\

No

$ \left( {-8}\right) \left( {-6}\right) $

$ \left. \begin{array}{lll} \left| {-8}\right| & = & 8 \\ \left| {-6}\right| & = & 6 \end{array}\right\} $ Multiply these absolute values.\n\n$ 8 \cdot 6 = {48} $\n\nSince the numbers have the same sign, the product is positive.\n\nThus, $ \left( {-8}\right) \left( {-6}\right) = + {48} $, or $ \left( {-8}\right) \left( {-6}\right) = {48} $.

Yes

Example 10.38\n\n$ 6\left( {-3}\right) $

\n\n$ \left. \begin{aligned} \left| 6\right| & = 6 \\ \left| {-3}\right| & = 3 \end{aligned}\right\} $ Multiply these absolute values.\n\n$ 6 \cdot 3 = {18} $\n\nSince the numbers have opposite signs, the product is negative.\n\nThus, $ 6\left( {-3}\right) = - {18} $ .

Yes

Example 10.39 $ \frac{-{10}}{2} $

$ \begin{aligned} \left| {-{10}}\right| & = {10} \\ \left| 2\right| & = 2 \end{aligned} $ Divide these absolute values.\n\n$ \frac{10}{2} = 5 $\n\nSince the numbers have opposite signs, the quotient is negative.\n\nThus $ \frac{-{10}}{2} = - 5 $ .

Yes

$ \frac{-{35}}{-7} $

$ \left. \begin{matrix} \left| {-{35}}\right| & = & {35} \\ \left| {-7}\right| & = & 7 \end{matrix}\right\} $ Divide these absolute values.\n\n$ \frac{35}{7} = 5 $\n\nSince the numbers have the same signs, the quotient is positive.\n\nThus, $ \frac{-{35}}{-7} = 5 $ .

Yes

$ \frac{18}{-9} $

$ \left. \begin{aligned} \left| {18}\right| & = {18} \\ \left| {-9}\right| & = 9 \end{aligned}\right\} $ Divide these absolute values. \n\n$ \frac{18}{9} = 2 $\n\nSince the numbers have opposite signs, the quotient is negative.\n\nThus, $ \frac{18}{-9} = 2 $ .

No

Find the value of $ \frac{-6\left( {4 - 7}\right) - 2\left( {8 - 9}\right) }{-\left( {4 + 1}\right) + 1} $ .

Using the order of operations and what we know about signed numbers, we get,\n\n\[ \frac{-6\left( {4 - 7}\right) - 2\left( {8 - 9}\right) }{-\left( {4 + 1}\right) + 1} = \frac{-6\left( {-3}\right) - 2\left( {-1}\right) }{-\left( 5\right) + 1} \]\n\n\[ = \;\frac{{18} + 2}{-5 + 1} \]\n\n\[ = \;\frac{20}{ - } \]\n\n\[ = \; - 5 \]

Yes

$ \left( {-{186}}\right) \cdot \left( {-{43}}\right) $

Since this product involves a (negative) (negative), we know the result should be a positive number. We'll illustrate this on the calculator.\n\n<table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>186</td><td>186</td></tr><tr><td>Press</td><td>$ + / - $</td><td>-186</td></tr><tr><td>Press</td><td>$ \times $</td><td>-186</td></tr><tr><td>Type</td><td>43</td><td>43</td></tr><tr><td>Press</td><td><img src=\

Yes

$ \frac{158.64}{-{54.3}} $ . Round to one decimal place.

<table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>158.64</td><td>158.64</td></tr><tr><td>Press</td><td>$ \div $</td><td>158.64</td></tr><tr><td>Type</td><td>54.3</td><td>54.3</td></tr><tr><td>Press</td><td>$ + / - $</td><td>-54.3</td></tr><tr><td>Press</td><td>$ = $</td><td>-2.921546961</td></tr></table>\n\nTable 10.6\n\nRounding to one decimal place we get -2.9.

Yes

$ a + 7 - b - m $

Associating the sign with the individual quantities, we see that this expression consists of the four terms $ a,7, - b, - m $.

No

Specify the terms in the expression $ x + 7 $. Exercise 11.2.1

Solution on p. 682.

No

$ \frac{5a}{b} + \frac{8b}{12} $, if $ a = 6 $ and $ b = - 3 $ .

Replace a with 6 and $ b $ with -3 .\n\n\[ \frac{5a}{b} + \frac{8b}{12} = \frac{5\left( 6\right) }{-3} + \frac{8\left( {-3}\right) }{12} \]\n\n\[ = \;\frac{30}{ - } + \frac{ - }{24} \]\n\n\[ = - {10} + \left( {-2}\right) \]\n\n\[ = \; - {12} \]\n\nThus, when $ a = 6 $ and $ b = - 3,\frac{5a}{b} + \frac{8b}{12} = - {12} $ .

No

$ 3{x}^{2} - {2x} + 1 $, if $ x = 4 $

Replace $ x $ with 4 .\n\n\[ 3{x}^{2} - {2x} + 1 = 3{\left( 4\right) }^{2} - 2\left( 4\right) + 1 \]\n\n\[ = 3 \cdot {16} - 2\left( 4\right) + 1 \]\n\n\[ = \;{48} - 8 + 1 \]\n\n\[ = \;{41} \]\n\nThus, when $ x = 4,3{x}^{2} - {2x} + 1 = {41} $.

Yes

$ - {x}^{2} - 4 $, if $ x = 3 $

Replace $ x $ with 3 . \n\n$ - {x}^{2} - 4 = - {}^{3} - 4 $ Be careful to square only the 3 . The exponent 2 is connected only to 3, not -3\n\n\[ \n= - 9 - 4 \n\] \n\n\[ \n= \; - {13} \n\]

Yes

$ {\left( -x\right) }^{2} - 4 $, if $ x = 3 $.

Replace $ x $ with 3 . \n\n$ {\left( -x\right) }^{2} - 4 = {\left( -3\right) }^{2} - 4 $ The exponent is connected to -3, not 3 as in problem 5 above. \n\n\[ \n= \;9 - 4 \n\] \n\n\[ \n= \; - 5 \n\] \n\nThe exponent is connected to -3 , not 3 as in the problem above.

No

$ - {3a} + {2b} - {5a} + a + {6b} $ . The like terms are

\n\[ \underset{\begin{matrix} {-3 - 5 + 1 = - 7} \\ {-{7a}} \end{matrix}}{\underbrace{-{3a},\; - {5a},\;a}}\underset{\begin{matrix} {2 + 6 = 8} \\ {8b} \end{matrix}}{\underbrace{{2b},\;{6b}}} \] \nThus, $ - {3a} + {2b} - {5a} + a + {6b} = - {7a} + {8b} $

Yes

Example 11.15\n\n$ r - {2s} + {7s} + {3r} - {4r} - {5s} $ . The like terms are

\[ \underset{1 + 3 - 4 = 0}{\underbrace{r,{3r}, - {4r}}}\;\underset{-2 + 7 - 5 = 0}{\underbrace{-{2s},{7s}, - {5s}}} \]\n\n\[ \frac{0r}{{0r} + {0s}} = 0 \]\n\nThus, $ r - {2s} + {7s} + {3r} - {4r} - {5s} = 0 $ .

Yes

Verify that 3 is a solution to $ x + 7 = {10} $ .

When $ x = 3 $ , \[ x + 7 = {10} \] becomes $ 3 + 7 = {10} $ $ \begin{matrix} {10} = {10} & \text{ which is a true statement, verifying that } \\ 3\text{ is a solution to }x + 7 = {10} & \end{matrix} $

Yes

Verify that -6 is a solution to $ {5y} + 8 = - {22} $

When $ y = - 6 $ ,\n\n\[ \n{5y} + 8 = - {22} \n\]\n\nbecomes $ 5\left( {-6}\right) + 8 = - {22} $\n\n\[ \n- {30} + 8 = - {22} \n\]\n\n99. - 99. which is a true statement, verifying that\n\n-6 is a solution to $ {5y} + 8 = - {22} $

Yes

Verify that -2 is a solution to $ {3m} - 2 = - {4m} - {16} $ .

When $ m = - 2 $ ,\n\n\[ \n{3m} - 2 = - {4m} - {16} \n\] \n\nbecomes $ 3\left( {-2}\right) - 2 = - 4\left( {-2}\right) - {16} $\n\n\[ \n- 6 - 2 = \;8 - {16} \n\] \n\nwhich is a true statement, verifying that -2 is a solution to $ {3m} - 2 = - {4m} - {16} $

Yes

$ m - 8 = 5 $ is associated with $ m $ by subtraction. Undo the association by adding 8 to both sides.

\[ m - 8 + 8 = 5 + 8 \] \[ m + 0 = 13 \] \[ m = 13 \] Check: When $ m = 13 $, becomes $ m - 8 = 5 $ $ 13 - 8 = 5 $ $ 5 = 5 $ a true statement. The solution to $ m - 8 = 5 $ is $ m = 13 $.

Yes

Example 11.22 $ - 3 - 5 = y - 2 + 8 $

Before we use the addition/subtraction property, we should simplify as much as possible.\n\n\[ \n- 3 - 5 = y - 2 + 8 \n\]\n\n\[ \n- 8 = y + 6 \n\]\n\n6 is associated with $ y $ by addition. Undo the association by subtracting 6 from both sides.\n\n\[ \n- 8 - 6 = y + 6 - 6 \n\]\n\n\[ \n- {14} = y + 0 \n\]\n\n\[ \n- {14} = y \n\]\n\nThis is equivalent to $ y = - {14} $.\n\nCheck: When $ y = - {14} $, \n\n$ - 3 - 5 = y - 2 + 8 $\n\nbecomes\n\n$ - 3 - 5 \geq - {14} - 2 + 8 $\n\n\[ \n- 8 \leq - 8\text{.} \n\]\n\na true statement.\n\nThe solution to $ - 3 - 5 = y - 2 + 8 $ is $ y = - {14} $.

Yes

- {5a} + 1 + {6a} = - 2

Begin by simplifying the left side of the equation.\n\n\[\n\underset{-5 + 6 = 1}{\underbrace{-{5a} + 1 + {6a}}} = - 2\n\]\n\n$ a + 1 = - {21} $ is associated with $ a $ by addition. Undo the association by subtracting 1 from both sides.\n\n\[\n a + 1 - 1 = - 2 - 1\n\]\n\n\[\n a + 0 = - 3\n\]\n\n\[\n a = - 3\n\]\n\nCheck: When $ a = - 3 $ ,\n\n\[\n- {5a} + 1 + {6a} = - 2\n\]\n\nbecomes\n\n$ - 5\left( {-3}\right) + 1 + 6\left( {-3}\right) \geq - 2 $\n\n\[\n- 2 \neq - 2\text{,}\n\]\n\na true statement.\n\nThe solution to $ - {5a} + 1 + {6a} = - 2 $ is $ a = - 3 $ .

No

$ {7k} - 4 = {6k} + 1 $

\n$ {7k} - 4 = {6k} + 1 $ Since $ {6k} $ represents $ + {6k} $, subtract $ {6k} $ from each side.\n\n\[ \n\underset{7 - 6 = 1}{\underbrace{{7k} - 4 - {6k}}} = \underset{6 - 6 = 0}{\underbrace{{6k} + 1 - {6k}}} \n\]\n\n$ k - 4 = {14} $ is associated with $ k $ by subtraction. Undo the association by adding 4 to both sides.\n\n\[ \nk - 4 + 4 = 1 + 4 \n\]\n\n\[ \nk = 5 \n\]\n\nCheck: When $ k = 5 $ ,\n\n\[ \n{7k} - 4 = {6k} + 1 \n\]\n\nbecomes\n\n$ 7 \cdot 5 - 4 \geqq 6 \cdot 5 + 1 $\n\n$ {35} - 4 \geqq {30} + 1 $\n\n$ {31} \pm {31} $.\n\na true statement.\n\nThe solution to $ {7k} - 4 = {6k} + 1 $ is $ k = 5 $.

Yes

$ - 8 + x = 5 $

\n$ - 8 + x = 5 $ . -8 is associated with $ x $ by addition. Undo the by subtracting -8 from both sides. Subtracting $ - 8 $ we get $ - \left( {-8}\right) = + 8 $ . We actually add 8 to both sides.\n\n$ - 8 + x + 8 = 5 + 8 $\n\n$ x = {13} $\n\nCheck: When $ x = {13} $\n\n$ - 8 + x = 5 $\n\nbecomes\n\n$ - 8 + {13} \geqq 5 $\n\n\[ 5 \leq 5 \]\n\na true statement.\n\nThe solution to $ - 8 + x = 5 $ is $ x = {13} $.

Yes

$ {6y} = {54} $

6 is associated with $ \mathrm{y} $ by multiplication. Undo the association by dividing both sides by 6\n\n\[ \frac{6y}{6} = \frac{54}{6} \]\n\n\[ \frac{\overset{―}{)6}y}{\overset{―}{)6}} = \frac{\frac{9}{\overset{―}{){54}}}}{\overset{―}{)6}} \]\n\n\[ y = 9 \]\n\nCheck: When $ y = 9 $\n\n\[ {6y} = {54} \]\n\nbecomes\n\n$ 6 \cdot 9 \geqq {54} $\n\n$ {54} \leq {54} $\n\na true statement.\n\nThe solution to $ {6y} = {54} $ is $ y = 9 $.

Yes

\[ \frac{x}{-2} = {27} \]

-2 is associated with $ x $ by division. Undo the association by multiplying both sides by -2 .\n\n\[ \left( {-2}\right) \frac{x}{-2} = \left( {-2}\right) {27} \]\n\n\[ \left( \overset{―}{) - 2}\right) \frac{x}{\overset{―}{) - 2}} = \left( { - 2}\right) \;{27} \]\n\n\[ x = - {54} \]\n\nCheck: When $ x = - {54} $ ,\n\n\[ \frac{x}{-2} = {27} \]\n\nbecomes\n\n\[ \frac{-{54}}{-2} \leq {27} \]\n\n$ {27} \leq {27} $\n\na true statement.\n\nThe solution to $ \frac{x}{-2} = {27} $ is $ x = - {54} $

Yes

$ \frac{3a}{7} = 6 $

7 is associated with $ a $ by division. Undo the association by multiplying both sides by 7 .\n\n$ 7 \cdot \frac{3a}{7} = 7 \cdot 6 $\n\nDivide out the 7's.\n\n\[ \overset{―}{)7} \cdot \frac{3a}{\overset{―}{)7}} = {42} \]\n\n\[ {3a} = {42} \]\n\n3 is associated with $ a $ by multiplication. Undo the association by dividing both sides by 3 .\n\n\[ \frac{3a}{3} = \frac{42}{3} \]\n\n\[ \frac{\overset{―}{)3}a}{\overset{―}{)3}} = {14} \]\n\n\[ a = {14} \]\n\nCheck: When $ a = {14} $, \n\n\[ \frac{3a}{7} = 6 \]\n\nbecomes\n\n$ \frac{3 \cdot {14}}{7} \geqq 6 $\n\n\[ \frac{42}{7} = 6 \]\n\n\[ 6 \preccurlyeq 6 \]\n\na true statement.\n\nThe solution to $ \frac{3a}{7} = 6 $ is $ a = {14} $.

Yes

- {8x} = {24}

-8 is associated with $ x $ by multiplication. Undo the association by dividing both sides by -8 .\n\n\[ \frac{-{8x}}{-8} = \frac{24}{-8} \]\n\n\[ \frac{-{8x}}{-8} = \frac{24}{-8} \]\n\n\[ x = - 3 \]\n\nCheck: When $ x = - 3 $ ,\n\n$ - {8x} = {24} $\n\nbecomes\n\n$ - 8\left( {-3}\right) \geqq {24} $\n\n$ {24} \leq {24} $,\n\na true statement.

Yes