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Graphical evidence suggests that the Taylor series centered at 0 for \( {e}^{x} \) converges for all values of \( x \) . To verify this, use the Ratio Test to determine all values of \( x \) for which the Taylor series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{x}^{k}}{k!} \]\n\n(8.5.4)\n\nconverges absolute... | Solution. In previous work, we used the Ratio Test on series of numbers that did not involve a variable; recall, too, that the Ratio Test only applies to series of nonnegative terms. In this example, we have to address the presence of the variable \( x \) . Because we are interested in absolute convergence, we apply th... | Yes |
Determine how well the 10th order Taylor polynomial \( {P}_{10}\left( x\right) \) for \( \sin \left( x\right) \), centered at 0 , approximates \( \sin \left( 2\right) \) . | Solution. To answer this question we use \( f\left( x\right) = \sin \left( x\right), c = 2, a = 0 \), and \( n = {10} \) in the Lagrange error bound formula. To use the bound, we also need to find an appropriate value for \( M \) . Note that the derivatives of \( f\left( x\right) = \sin \left( x\right) \) are all equal... | Yes |
Show that the Taylor series for \( \sin \left( x\right) \) actually converges to \( \sin \left( x\right) \) for all \( x \) . | Solution. Recall from the previous example that since \( f\left( x\right) = \sin \left( x\right) \), we know\n\n\[ \left| {{f}^{\left( n + 1\right) }\left( x\right) }\right| \leq 1 \]\n\nfor any \( n \) and \( x \) . This allows us to choose \( M = 1 \) in the Lagrange error bound formula. Thus,\n\n\[ \left| {{P}_{n}\l... | Yes |
Consider the power series defined by\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{x}^{k}}{{2}^{k}} \]\n\nWhat are \( f\left( 1\right) \) and \( f\left( \frac{3}{2}\right) \) ? Find a general formula for \( f\left( x\right) \) and determine the values for which this power series converges. | Solution. If we evaluate \( f \) at \( x = 1 \) we obtain the series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{1}{{2}^{k}} \]\n\nwhich is a geometric series with ratio \( \frac{1}{2} \) . So we can sum this series and find that\n\n\[ f\left( 1\right) = \frac{1}{1 - \frac{1}{2}} = 2 \]\n\nSimilarly,\n\n\[ f\l... | Yes |
Let \( f\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{x}^{k}}{{k}^{2}} \) . Determine the interval of convergence of this power series. | Solution. First we will draw graphs of some of the partial sums of this power series to get an idea of the interval of convergence. Let\n\n\[ \n{S}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\frac{{x}^{k}}{{k}^{2}} \n\]\n\nfor each \( n \geq 1 \) . Figure 8.6.4 shows plots of \( {S}_{10}\left( x\right) \) ... | Yes |
Find a series expansion centered at \( x = 0 \) for \( \arctan \left( x\right) \), as well as its interval of convergence. | Solution. While we could differentiate \( \arctan \left( x\right) \) repeatedly and look for patterns in the derivative values at \( x = 0 \) in an attempt to find the Maclaurin series for \( \arctan \left( x\right) \) from the definition, it turns out to be far easier to use a known series in an insightful way. In Act... | No |
Find the value of 6 in the number 7,261. | Since 6 is in the tens position of the units period, its value is 6 tens.\n\n6 tens \( = {60} \) | Yes |
Find the value of 9 in the number \( {86},{932},{106},{005} \) . | Since 9 is in the hundreds position of the millions period, its value is 9 hundred millions.\n\n9 hundred millions \( = 9 \) hundred million | Yes |
Find the value of 2 in the number 102,001. | Since 2 is in the ones position of the thousands period, its value is 2 one thousands. 2 one thousands \( = 2 \) thousand | Yes |
Graph the following whole numbers: 3, 5, 9. |  | No |
Specify the whole numbers that are graphed on the following number line. The break in the number line indicates that we are aware of the whole numbers between 0 and 106 , and 107 and 872 , but we are not listing them due to space limitations. | The numbers that have been graphed are \( 0,{106},{873},{874} \) | Yes |
1. Beginning at the right, we can separate this number into distinct periods by inserting a comma between the 2 and 9.\n\n42,958 | 2. Beginning at the left, we read each period individually:\n\n\[ \underset{\text{Units period }}{\underbrace{\mathop{\bigsqcup }\limits^{9}\mathop{\bigsqcup }\limits^{5}\mathop{\bigsqcup }\limits^{8}}} \rightarrow \text{ nine hundred fifty-eight } \]\n\nForty-two thousand, nine hundred fifty-eight. | No |
1. Beginning at the right, we can separate this number into distinct periods by placing commas between the 1 and 3 and the 7 and 9 . 307,991,343 | 2. Beginning at the left, we read each period individually.\n\n\[ \underset{\text{Millions period }}{\underbrace{\mathop{\bigsqcup }\limits^{3}\left| 0\right| \xrightarrow[]{7}\xrightarrow[]{\text{ Three hundred seven million,}}}} \]\n\n\[ \underset{\text{Thousands period }}{\underbrace{\mathop{\bigsqcup }\limits^{9}\m... | Yes |
Read 36000000000001. | 1. Beginning at the right, we can separate this number into distinct periods by placing commas. 36,000,000,001\n\n2. Beginning at the left, we read each period individually.\n\n\[ \xrightarrow[\text{Trillions period}]{}\xrightarrow[]{3}\text{Thirty-six trillion,} \]\n\n\[ \underset{\text{Billions period }}{\underbrace{... | Yes |
Seven thousand, ninety-two. | Using the comma as a period separator, we have  | No |
Fifty billion, one million, two hundred thousand, fourteen. | Using the commas as period separators, we have Fifty billion, \( \rightarrow {50} \) , one million, \( \rightarrow {001} \) , two hundred thousand \( \underline{\text{ fourteen }} \rightarrow {014} \) 50,001,200,014 | Yes |
Round 67 to the nearest ten. | On the number line, 67 is more than halfway from 60 to 70 . The digit immediately to the right of the tens digit, the round-off digit, is the indicator for this.\n\n\n\nthan it is to 6 tens.\n\nThus, 67, rounded to the... | Yes |
Round 3,426 to the nearest ten. | 1. We are rounding to the tens position. Mark the digit in the tens position\n\n3,426\n\n\\( \\uparrow \\)\n\ntens position\n\n2. Observe the digit immediately to the right of the tens position. It is 6 . Since 6 is greater than 5 , we round up by replacing 6 with 0 and adding 1 to the digit in the tens position (the r... | Yes |
Round 9,614,018,007 to the nearest ten million. | 1. We are rounding to the nearest ten million.\n\n9,614,018,007\n\nten millions position\n\n2. Observe the digit immediately to the right of the ten millions position. It is 4 . Since 4 is less than 5 , we round down by replacing 4 and all the digits to its right with zeros. 9,610,000,000\n\nThus, 9,614,018,007 rounded... | Yes |
Round 148,422 to the nearest million. | 1. Since we are rounding to the nearest million, we'll have to imagine a digit in the millions position. We'll write 148,422 as 0,148,422 .\n\n0,148,422\n\n\\(\\uparrow\\)\n\nmillions position\n\n2. The digit immediately to the right is 1 . Since 1 is less than 5 , we'll round down by replacing it and all the digits to... | Yes |
Round 397,000 to the nearest ten thousand. | 1. We are rounding to the nearest ten thousand.\n\n397,000\n\nten thousand position\n\n2. The digit immediately to the right of the ten thousand position is 7 . Since 7 is greater than 5 , we round up by replacing 7 and all the digits to its right with zeros and adding 1 to the digit in the ten thousands position. But ... | Yes |
Add 1459 and 130 | \n\( 9 + 0 = 9 \) .\n\n1459\n\n\( 5 + 3 = 8 \) .\n\n\( + {130} \)\n\n\( 4 + 1 = 5 \) .\n\n1589\n\n\( 1 + 0 = 1 \) . | Yes |
Add 1875 and 358. | 111\n\n1875\n\n+ 358\n\n2233\n\n\\( 5 + 8 = {13}\\; \\) Write 3, carry 1 ten.\n\n\\( 1 + 7 + 5 = {13}\\; \\) Write 3, carry 1 hundred.\n\n\\( 1 + 8 + 3 = {12}\\; \\) Write 2, carry 1 thousand.\n\n\\[ 1 + 1 = 2 \\]\n\nThe sum is 2233. | Yes |
Add 89,208 and 4,946. | \n\\(\\begin{array}{ll} {11} & 1 \\end{array}\\)\n\n89,208\n\n\\(\\begin{array}{r} + 4,{946} \\\\ {94},{154} \\end{array}\\)\n\n\\[8 + 6 = {14}\\]\nWrite 4, carry 1 ten.\n\n\\[1 + 0 + 4 = 5\\]\nWrite the 5 (nothing to carry).\n\n\\[2 + 9 = {11}\\]\nWrite 1, carry one thousand.\n\n\\[1 + 9 + 4 = {14}\\]\nWrite 4, carry ... | Yes |
Add 38 and 95. | \[ \begin{array}{r} + {95} \\ {133} \end{array} \] \( 8 + 5 = {13}\; \) Write 3, carry 1 ten. \( 1 + 3 + 9 = {13}\; \) Write 3, carry 1 hundred. \( 1 + 0 = 1 \) | No |
Find the sum 2648, 1359, and 861. | 2648\n1359\n\( + {861} \)\n4868\n\n\( 8 + 9 + 1 = {18}\; \) Write 8, carry 1 ten.\n\( 1 + 4 + 5 + 6 = {16}\; \) Write 6, carry 1 hundred.\n\( 1 + 6 + 3 + 8 = {18}\; \) Write 8, carry 1 thousand.\n\n\[ 1 + 2 + 1 = 4 \]\n\nThe sum is 4,868. | Yes |
Find the sum of the following numbers. \( \\begin{array}{ll} {132} & 1 \\end{array} \) 878016 9905 38951 \( + {56817} \) 983689 | \n\( 6 + 5 + 1 + 7 = {19}\\; \) Write 9, carry the 1.\n\n\[1 + 1 + 0 + 5 + 1 = 8\\;\\text{Write 8.}\]\n\n\[0 + 9 + 9 + 8 = {26}\]\nWrite 6, carry the 2.\n\n\[2 + 8 + 9 + 8 + 6 = {33}\]\nWrite 3, carry the 3.\n\n\[3 + 7 + 3 + 5 = {18}\]\nWrite 8, carry the 1.\n\n\[1 + 8 = 9\]\nWrite 9.\n\nThe sum is 983,689. | Yes |
\( {106} + {85} + {322} + {406} \) | <table><thead><tr><th colspan=\ | No |
Example 1.34 | 275\n\n\( - {142} \)\n\n133\n\n\( 5 - 2 = 3 \)\n\n\( 7 - 4 = 3 \)\n\n\( 2 - 1 = 1 \) | Yes |
Find the difference between 977 and 235. | Write the numbers vertically, placing the larger number on top. Line up the columns properly.\n\n977\n\n$ - {235} $\n\n742\n\nThe difference between 977 and 235 is 742. | Yes |
In Keys County in 1987, there were 809 cable television installations. In Flags County in 1987, there were 1,159 cable television installations. How many more cable television installations were there in Flags County than in Keys County in 1987? | We need to determine the difference between 1,159 and 809.\n\n11\n\n1,159\n\n\( - \frac{809}{350} \)\n\nThere were 350 more cable television installations in Flags County than in Keys County in 1987. | Yes |
1. Borrow 1 ten from the 4 tens. This leaves 3 tens.\n\n2. Convert the 1 ten to 10 ones.\n\n3. Add 10 ones to 1 one to get 11 ones. We can now perform \( {11} - 8 \) . | 4. Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.\n\n5. Convert the 1 hundred to 10 tens.\n\n6. Add 10 tens to 3 tens to get 13 tens.\n\n7. Now \( {13} - 5 = 8 \).\n\n8. \( 5 - 3 = 2 \). | No |
Example 1.41 | 1. Borrow 1 ten from the 3 tens. This leaves 2 tens.\n2. Convert the 1 ten to 10 ones.\n3. Add 10 ones to 4 ones to get 14 ones. We can now perform 14-5.\n4. Borrow 1 hundred from the 5 hundreds. This leaves 4 hundreds.\n5. Convert the 1 hundred to 10 tens.\n6. Add 10 tens to 2 tens to get 12 tens. We can now perform \... | Yes |
Example 1.43: Borrowing from a single zero.\n503\nConsider the problem\n\( - {37} \)\nSince we do not have a name for \( 3 - 7 \), we must borrow from 0 . | \( {503} = 5 \) hundreds \( + 0 \) tens \( + 3 \) ones\n- 37 3 tens + 7 ones\n\nSince there are no tens to borrow, we must borrow 1 hundred. One hundred \( = {10} \) tens.\n4 hundreds + 10 tens +3 ones\n3 tens +7 ones\n\nWe can now borrow 1 ten from 10 tens (leaving 9 tens). One ten \( = {10} \) ones and 10 ones +3 one... | Yes |
Perform this subtraction.\n\n503\n\n\( - {37} \) | The number 503 contains a single zero\n\n1. The number to the immediate left of 0 is 5 . Decrease 5 by 1 .\n\n\[ 5 - 1 = 4 \]\n\n410\n\n\$63\n\n\( - {37} \)\n\n2. Draw a line through the zero and make it a 10 .\n\n3. Borrow from the 10 and proceed.\n\n9\n\n4,1013\n\n\$63\n\n\[ \begin{array}{r} - {37} \\ {466} \end{arra... | No |
The number 40,000 contains a group of zeros. | 1. The number to the immediate left of the group is 4 . Decrease 4 by 1 .\n\n\( 4 - 1 = 3 \)\n\n2. Make each 0, except the rightmost one, 9. Make the rightmost 0 a 10.\n\n399910\n\n3. Subtract as usual.\n\n39 9910\n\n40,000\n\n- 125\n\n39,875 | No |
The number 8,000,006 contains a group of zeros. | 1. The number to the immediate left of the group is 8 . Decrease 8 by 1. \( 8 - 1 = 7 \)\n\n2. Make each zero, except the rightmost one, 9. Make the rightmost 0 a 10. 7 999 910 8,000,006\n\n- 41,107\n\n3. To perform the subtraction, we'll need to borrow from the ten. 7 999 91016 \$,000,006\n\n\[ \n- {41},{107} \n\]\n\n... | No |
Add the whole numbers  | \( 8 + 5 = {13} \)\n\n\[ 5 + 8 = {13} \]\n\nThe numbers 8 and 5 can be added in any order. Regardless of the order they are added, the sum is 13. | Yes |
Add the whole numbers.  | ( \( {43} \) and 16 are associated.\n\n\( \left( {{43} + {16}}\right) + {27} = {59} + {27} = {86}. \)\n\n\( {43} + \left( {{16} + {27}}\right) = {43} + {43} = {86}. \)\n\n\( {}^{1} \) 16 and 27 are associated. | Yes |
Example 1.50 Add the whole numbers.  | \( {29} + 0 = {29} \)\n\n\[ 0 + {29} = {29} \]\n\nZero added to 29 does not change the identity of 29. | Yes |
Example 2.2\n\n\( {18} + {18} + {18} \) | \( 3 \times {18} \) . Multiplier is 3. Multiplicand is 18. | Yes |
\( 5 \times 6 = {30} \) Write the 0 , carry the 3 . | \( 5 \times 2 = {10} \) Add to 10 the 3 that was carried: \( {10} + 3 = {13} \). Write the 3, carry the 1 . \( 5 \times 5 = {25} \) Add to 25 the 1 that was carried: \( {25} + 1 = 6 \). The product is 2,630 . | No |
Example 2.5\n\n\[ \begin{array}{r} 7\;3 \\ 1,{804} \\ \times \;9 \\ {16},{236} \end{array} \] | \n\( 9 \times 4 = {36} \) Write the 6 , carry the 3 .\n\n\[ 9 \times 0 = 0 \]\nAdd to the 0 the 3 that was carried: \( 0 + 3 = 3 \) . Write the 3 .\n\n\( 9 \times 8 = {72} \) Write the 2 , carry the 7 .\n\nAdd to the 9 the 7 that was carried: \( 9 + 7 = {16} \) .\n\n\( 9 \times 1 = 9 \)\n\nSince there are no more multi... | Yes |
Multiply 326 by 48. | \n\nPart 3: This step is unnecessary since all of the digits in the multiplier have been used.\n\nPart 4: Add the partial products to obtain the total product.\n\n. | Yes |
Multiply 57,847,298 by 38,976. | <table><tr><td colspan=\ | No |
\( \frac{36}{6} \) | Since \( 6 \times 6 = {36} \), \( \frac{36}{6} = 6 \). Thus, there are 6 sixes in 36. | Yes |
9) \( \overline{7}2 \) | Since \( 9 \times 8 = {72} \) ,\n\n8 \( \overset{8}{\overline{){72}}} \)\n\nThus, there are 8 nines in 72 . | No |
\( \frac{19}{0} \) . Since division by 0 does not name a whole number, no quotient exists, and we state \( \frac{19}{0} \) is | undefined | Yes |
Example 2.17\n\n\( 0\overline{){14}} \) . Since division by 0 does not name a defined number, no quotient exists, and we state | \( 0\overline{){14}} \) is undefined | Yes |
9) \( \overline{0} \) . Since division into 0 by any nonzero whole number results in 0, we have \( \frac{0}{9} \) | ## Example 2.19\n\n\( \frac{0}{7} \) . Since division into 0 by any nonzero whole number results in 0, we have \( \frac{0}{7} = 0 \) | No |
Divide 24 by 3. | The display now reads 8, and we conclude that \( {24} \div 3 = 8 \) . | Yes |
Divide 0 by 7. | The display now reads 0, and we conclude that \( 0 \div 7 = 0 \) . | Yes |
Find \( 75 \div 5 \) . | 5)75 Rewrite the problem using a division bracket.\n\n10\n\n5) \( \overline{75} \)\n\nMake an educated guess by noting that one 5 is contained in 75 at most 10 times.\n\nSince 7 is the tens digit, we estimate that 5 goes into 75 at most 10 times.\n\n---\n\n10\n\n5) \( {75} \)\n\n\( - {50} \)\n\n25\n\nNow determine how ... | Yes |
Find \( 2,{232} \div {36} \). | 36) \( {2232} \)\n\nUse the first digit of the divisor and the first two digits of the dividend to make the educated guess.\n\n3 goes into 22 at most 7 times.\n\nTry 7: \( 7 \times {36} = {252} \) which is greater than 223. Reduce the estimate.\n\nTry 6: \( 6 \times {36} = {216} \) which is less than 223 .\n\n\[ \n\tex... | Yes |
Find \( 2,{417},{228} \div {802} \) . | ## 802) \( \overline{2417228} \)\n\nFirst, the educated guess: \( {24} \div 8 = 3 \) . Then \( 3 \times {802} = {2406} \), which is less than 2417 . Use 3 as the guess. Since \( 3 \times {802} = {2406} \), and 2406 has four digits, place the 3 above the fourth digit of the dividend.\n\n\[ \n\text{802) 2417228} \n\]\n\n... | Yes |
\( {3727} \div {49} \) | <table><tr><td>Type</td><td>3727</td></tr><tr><td>Press</td><td>\( \div \)</td></tr><tr><td>Type</td><td>49</td></tr><tr><td>Press</td><td>\( = \)</td></tr></table>\n\nTable 2.9\n\nThe display now reads 76.061224.\n\nThis number is an example of a decimal number (see Section 6.1). When a decimal number results in a cal... | Yes |
Multiply the two whole numbers.  | \( 6 \cdot 7 = {42} \)\n\n\( 7 \cdot 6 = {42} \)\n\nThe numbers 6 and 7 can be multiplied in any order. Regardless of the order they are multiplied, the product is 42. | Yes |
Multiply the whole numbers.  | \( \left( {8 \cdot 3}\right) \cdot {14} = {24} \cdot {14} = {336} \)\n\n\( 8 \cdot \left( {3 \cdot {14}}\right) = 8 \cdot {42} = {336} \) | Yes |
Multiply the whole numbers.  | \( {12} \cdot 1 = {12} \)\n\n\( 1 \cdot {12} = {12} \) | Yes |
\( {62} \cdot {62} \cdot {62} \cdot {62} \cdot {62} \cdot {62} \cdot {62} \cdot {62} \cdot {62} \) | Since the factor 62 appears 9 times, we record this as \( {62}^{9} \) | Yes |
Example 3.4\n\n\( {706}^{3} \) . The exponent 3 is recording 3 factors of 706 in a multiplication. Thus, | \( {706}^{3} = {706} \cdot {706} \cdot {706} \) | Yes |
Use the calculator to find \( \sqrt{121} \) | <table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>121</td><td>121</td></tr><tr><td>Press</td><td>\( \sqrt{x} \)</td><td>11</td></tr></table> | Yes |
Find \( \sqrt[7]{2187} \) . | <table><thead><tr><th></th><th></th><th>Display Reads</th></tr></thead><tr><td>Type</td><td>2187</td><td>2187</td></tr><tr><td>Press</td><td>\( {y}^{x} \)</td><td>2187</td></tr><tr><td>Type</td><td>7</td><td>7</td></tr><tr><td>Press</td><td>\( 1/x \)</td><td>.14285714</td></tr><tr><td>Press</td><td>\( = \)</td><td>3</t... | Yes |
\[ 9 + \left( {3 \cdot 8}\right) \] | Since 3 and 8 are within parentheses, they are to be combined first. \[ 9 + \left( {3 \cdot 8}\right) = 9 + {24} \] \[ = {33} \] Thus, \[ 9 + \left( {3 \cdot 8}\right) = {33} \] | Yes |
[ \left( {{10} \div 0}\right) \cdot 6 ] | Since \( {10} \div 0 \) is undefined, this operation is meaningless, and we attach no value to it. We write, \ | No |
\[ 2 + \left( {8 \cdot 3}\right) - \left( {5 + 6}\right) \] | Combine 8 and 3 first, then combine 5 and 6 . \n\n\( 2 + {24} - {11} \) Now combine left to right. \n\n\( {26} - {11} \) \n\n15 | Yes |
[{10} + \\left\\lbrack {{30} - \\left( {2 \\cdot 9}\\right) }\\right\\rbrack] | Combine 2 and 9 since they occur in the innermost set of parentheses.\n\n\\[{10} + \\left\\lbrack {{30} - {18}}\\right\\rbrack \\text{Now combine 30 and 18.}\n\n\\]\n\n\\[{10} + {12}\\]\n\n22 | Yes |
\( {21} + 3 \cdot {12} \) Multiply first. | \( {21} + {36}\; \) Add.\n\n57 | Yes |
\( {63} - \left( {4 + 6 \cdot 3}\right) + {76} - 4 \) Simplify first within the parenthesis by multiplying, then adding. | \( {63} - \left( {4 + {18}}\right) + {76} - 4 \) \n\n\( {63} - {22} + {76} - 4\; \) Now perform the additions and subtractions, moving left to right. \n\n\( {41} + {76} - 4\; \) Add 41 and 76: \( {41} + {76} = {117} \) . \n\n\[ \n{117} - 4 \n\] \nSubtract 4 from 117: \( \;{117} - 4 = {113} \) . \n\n113 | Yes |
\( 6 \cdot \left( {{3}^{2} + {2}^{2}}\right) + {4}^{2}\; \) Evaluate the exponential forms in the parentheses: \( \;{3}^{2} = 9 \) and \( {2}^{2} = 4 \) | \( 6 \cdot \left( {9 + 4}\right) + {4}^{2} \) Add the 9 and 4 in the parentheses: \( 9 + 4 = {13} \)\n\n\( 6 \cdot \left( {13}\right) + {4}^{2} \) Evaluate the exponential form: \( {4}^{2} = {16} \)\n\n\[ 6 \cdot \left( {13}\right) + {16} \]\nMultiply 6 and 13: \( 6 \cdot {13} = {78} \)\n\n\[ {78} + {16} \]\nAdd 78 and... | Yes |
[ \n\\frac{{6}^{2} + {2}^{2}}{{4}^{2} + 6 \\cdot {2}^{2}} + \\frac{{1}^{3} + {8}^{2}}{{10}^{2} - {19} \\cdot 5} \n] \nRecall that the bar is a grouping symbol. The fraction \\( \\frac{{6}^{2} + {2}^{2}}{{4}^{2} + 6 \\cdot {2}^{2}} \\) is equivalent to \\( \\left( {{6}^{2} + {2}^{2}}\\right) \\div \\left( {{4}^{2} + 6 \... | \[ \n\\frac{{36} + 4}{{16} + 6 \\cdot 4} + \\frac{1 + {64}}{{100} - {19} \\cdot 5} \n\] \n\[ \n\\frac{{36} + 4}{{16} + {24}} + \\frac{1 + {64}}{{100} - {95}} \n\] \n\[ \n\\frac{40}{40} + \\frac{65}{5} \n\] \n\[ \n1 + {13} \n\] \n\[ \n\\text{14} \n\] \n | Yes |
Example 3.21\n\n\( {16}^{4} + {37}^{3} \) | Table 3.7\n\nThus, \( {16}^{4} + {37}^{3} = {116},{189} \) | Yes |
4 is not a factor of 10, since \( {10} \div 4 = {2R2} \) . | (There is a remainder.) | No |
Find all the factors of 24. | Try 1:\n\[ {24} \div 1 = {24} \]\n1 and 24 are factors\n\nTry 2: 24 is even, so 24 is divisible by 2 .\n\[ {24} \div 2 = {12} \]\n2 and 12 are factors\n\nTry 3: \( 2 + 4 = 6 \) and 6 is divisible by 3, so 24 is divisible by 3 .\n\[ {24} \div 3 = 8 \]\n3 and 8 are factors\n\nTry 4:\n\[ {24} \div 4 = 6 \]\n4 and 6 are fa... | Yes |
Find the prime factorization of 60. | Since the last digit of 60 is 0, which is even, 60 is divisible by 2. We will repeatedly divide by 2 until we no longer can. We shall divide as follows: 30 is divisible by 2 again. 15 is not divisible by 2, but it is divisible by 3, the next prime. 5 is not divisible by 3, but it is divisible by 5, the next prime. The ... | Yes |
Find the prime factorization of 441. | 441 is not divisible by 2 since its last digit is not divisible by 2.\n\n441 is divisible by 3 since \( 4 + 4 + 1 = 9 \) and 9 is divisible by 3.\n\n147 is divisible by \( 3\left( {1 + 4 + 7 = {12}}\right) \).\n\n49 is not divisible by 3, nor is it divisible by 5. It is divisible by 7.\n\nThe quotient 1 is finally smal... | Yes |
Find the prime factorization of 31. | 31 is not divisible by 2 Its last digit is not even\n\n\[ \n{31} \div 2 = {15}\mathrm{R}1 \n\]\n\nThe quotient, 15 , is larger than the divisor, 3. Continue.\n\n31 is not divisible by 3 The digits \( 3 + 1 = 4 \), and 4 is not divisible by 3 .\n\n\[ \n{31} \div 3 = {10}\mathrm{R}1 \n\]\n\nThe quotient, 10 , is larger t... | Yes |
Find the GCF of the following numbers.\n\n700, 1,880, and 6,160 | \[{700} = 2 \cdot {350} = 2 \cdot 2 \cdot {175} = 2 \cdot 2 \cdot 5 \cdot {35}\]\n\n\[= \;2 \cdot 2 \cdot 5 \cdot 5 \cdot 7\]\n\n\[= \;{2}^{2} \cdot {5}^{2} \cdot 7\]\n\n\[1,{880} = 2 \cdot {940} = 2 \cdot 2 \cdot {470} = 2 \cdot 2 \cdot 2 \cdot {235}\]\n\n\[= \;2 \cdot 2 \cdot 2 \cdot 5 \cdot {47}\]\n\n1. \[ \; = \;{2... | Yes |
9 and 12 | \[ 9 = 3 \cdot 3 = {3}^{2} \]\n\n1. \n\n\[ {12} = 2 \cdot 6 = 2 \cdot 2 \cdot 3 = {2}^{2} \cdot 3 \]\n\n2. The bases that appear in the prime factorizations are 2 and 3 .\n\n3. The largest exponents appearing on 2 and 3 in the prime factorizations are, respectively, 2 and 2: \[ {2}^{2} \] from 12.\n\n\[ {3}^{2} \] from... | Yes |
90 and 630 | 1. \( \;{630} = \;2 \cdot {315} = 2 \cdot 3 \cdot {105} = 2 \cdot 3 \cdot 3 \cdot {35}\; = 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7 \)\n\n\[ = 2 \cdot {3}^{2} \cdot 5 \cdot 7 \]\n\n2. The bases that appear in the prime factorizations are \( 2,3,5 \), and 7 .\n\n3. The largest exponents that appear on \( 2,3,5 \), and 7 are, r... | Yes |
Find the LCM of the numbers 33, 110, and 484. | 33 = 3 \u00b7 11\n110 = 2 \u00b7 55 = 2 \u00b7 5 \u00b7 11\n484 = 2 \u00b7 242 = 2 \u00b7 2 \u00b7 121 = 2 \u00b7 2 \u00b7 11 \u00b7 11 = 2^2 \u00b7 11^2.\nThe bases that appear in the prime factorizations are 2, 3, 5, and 11.\nThe largest exponents that appear on 2, 3, 5, and 11 are, respectively, 2, 1, 1, and 2:\n2^2... | Yes |
Four hundred seven thousandths. The absence of hyphens indicates that the words seven and thousandths are to be considered individually. | four hundred seven thousandths translates as \( \frac{407}{1000} \) | Yes |
\( \frac{83}{11} \) . Divide 83 by 11. | The improper fraction \( \frac{83}{11} = 7\frac{6}{11} \). | Yes |
\( \frac{104}{4} \) Divide 104 by 4. | \n\n\[ \frac{104}{4} = {26}\frac{0}{4} = {26} \]\n\nThe improper fraction \( \frac{104}{4} = {26} \) . | Yes |
Example 4.17\n\n\( {16}\frac{2}{3} \) | 1. Multiply: \( 3 \cdot {16} = {48} \).\n\n2. Add: \( {48} + 2 = {50} \).\n\n3. Place 50 over \( 3 : \frac{50}{3} \)\n\nThus, \( {16}\frac{2}{3} = \frac{50}{3} \) | Yes |
\( \frac{3}{4} \) and \( \frac{6}{8} \) . Test for equality of the cross products. | \( 3 \cdot 8 \geqq 6 \cdot 4 \)\n\n\( \mathbf{24} \neq \mathbf{24} \) The cross products are equals.\n\nThe fractions \( \frac{3}{4} \) and \( \frac{6}{8} \) are equivalent, so \( \frac{3}{4} = \frac{6}{8} \). | No |
\( \frac{3}{8} \) and \( \frac{9}{16} \) . Test for equality of the cross products. | \( 3 \cdot {16} \triangleq 9 \cdot 8 \)\n\n\( {48} \neq {72}\; \) The cross products are not equal.\n\nThe fractions \( \frac{3}{8} \) and \( \frac{9}{16} \) are not equivalent. | No |
[Example 4.20](The theory/question you extract) | \[ \frac{6}{18} = \frac{\frac{1}{12} \cdot \frac{1}{13}}{\frac{1}{12} \cdot \frac{1}{13} \cdot 3} = \frac{1}{3}1\text{and 3 are relatively prime.} \] | No |
[The theory/question you extract] | [The proof/solution] | No |
\[ \frac{56}{104} = \frac{\frac{1}{12} \cdot \frac{1}{12} \cdot \frac{1}{12} \cdot 7}{\frac{1}{12} \cdot \frac{1}{12} \cdot \frac{1}{12} \cdot {13}} = \frac{7}{13} \] | 7 and 13 are relatively prime (and also truly prime) | No |
\( \frac{8}{15} = \frac{\overline{2} \cdot 2 \cdot 2}{3 \cdot 5} \) No common prime factors, so 8 and 15 are relatively prime. | The fraction \( \frac{8}{15} \) is reduced to lowest terms. | Yes |
\( \frac{25}{30} \) . 5 divides into both 25 and 30 . | \( \frac{5}{\frac{25}{130}} = \frac{5}{6}5 \) and \( 6 \) are relatively prime. | No |
\( \frac{18}{24} \) . Both numbers are even so we can divide by 2 . | \( \frac{9}{\underset{12}{\overline{124}}} \) Now, both 9 and 12 are divisible by 3 . \( \frac{\frac{3}{19}}{\frac{18}{\frac{24}{\frac{12}{4}}}} = \frac{3}{4}3 \) and \( 4 \) are relatively prime. | No |
\( \frac{5}{6} = \frac{45}{?} \) | Divide the original numerator into the new numerator.\n\n\( {45} \div 5 = 9 \) The quotient is 9 . Multiply the original denominator by 9 .\n\n\( \frac{5}{6} = \frac{5 \cdot 9}{6 \cdot 9} = \frac{45}{54} \) The missing denominator is 45 . | No |
\( \frac{3}{4} \cdot \frac{1}{6} = \frac{1}{8} \) | \( \frac{3}{4} \cdot \frac{1}{6} = \frac{3 \cdot 1}{4 \cdot 6} = \frac{3}{24}\; \) Now, reduce.\n\n\[ \n= \frac{\frac{1}{)3}}{\frac{){24}}{8}} = \frac{1}{8} \n\]\n\nThus\n\n\[ \n\frac{3}{4} \cdot \frac{1}{6} = \frac{1}{8} \n\] | Yes |
\( \frac{3}{8} \cdot 4 \) . Write 4 as a fraction by writing \( \frac{4}{1} \) | \[ \frac{3}{8} \cdot \frac{4}{1} = \frac{3 \cdot 4}{8 \cdot 1} = \frac{12}{8} = \frac{3}{2} \] | Yes |
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