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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 & {7... | 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) } \c... | 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) } = \\t... | 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... | 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 & \down... | 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>... | 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,{... | 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... | 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 a... | 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... | 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 af... | 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{al... | 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}... | 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} + {... | 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} ... | 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... | 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}... | 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 t... | 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= \;... | Yes |
Convert 3 kilograms to grams. | (a) \( 3\mathrm{\;{kg}} \) can be written as \( {3.0}\mathrm{\;{kg}} \) . Then, \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 con... | 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\... | 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... | 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... | 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\... | 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{... | 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 ... | 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}\ma... | 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... | 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\[ 4\mathrm{{ft}} = \frac{4\mathrm{{ft}}}{1} \cdot \frac{{12}\mathrm{{in}}.}{1\mathrm{{ft}}} \]... | 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{\;... | 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 appr... | 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}... | 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\( - 3 \leq x < 2 \)  | 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)... | 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 \( - {12} + {25} \) . \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)... | 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\[ = \... | 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><t... | 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><... | 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} = -... | 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- {1... | 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\n... | 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... | 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\(... | 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\nbecome... | 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 \( x = - {54} \) ,\... | 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 b... | 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} ... | Yes |
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