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The number of students \( N \) at local college who have had the flu \( t \) months after the semester begins can be modeled by the formula \( N\left( t\right) = {500} - \frac{450}{1 + {3t}} \) for \( t \geq 0 \). | ## Solution.\n\n1. \( N\left( 0\right) = {500} - \frac{450}{1 + 3\left( 0\right) } = {50} \) . This means that at the beginning of the semester,50 students\n\nhave had the flu.\n\n2. We set \( N\left( t\right) = {300} \) to get \( {500} - \frac{450}{1 + {3t}} = {300} \) and solve. Isolating the fraction gives \( \frac{... | Yes |
Theorem 4.3. Determination of Slant Asymptotes: Suppose \( r \) is a rational function and \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \), where the degree of \( p \) is exactly one more than the degree of \( q \) . Then the graph of \( y = r\left( x\right) \) has the slant asymptote \( y = L\left... | In the same way that Theorem 4.2 gives us an easy way to see if the graph of a rational function \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \) has a horizontal asymptote by comparing the degrees of the numerator and denominator, Theorem 4.3 gives us an easy way to check for slant asymptotes. Unli... | Yes |
Find the slant asymptotes of the graphs of the following functions if they exist. Verify your answers using a graphing calculator and describe the behavior of the graph near them using proper notation. | 1. The degree of the numerator is 2 and the degree of the denominator is 1 , so Theorem 4.3 guarantees us a slant asymptote. To find it, we divide \( 1 - x = - x + 1 \) into \( {x}^{2} - {4x} + 2 \) and get a quotient of \( - x + 3 \), so our slant asymptote is \( y = - x + 3 \) . We confirm this graphically, and we se... | Yes |
1. Solve \( \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \) . | To solve the equation, we clear denominators\n\n\[ \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \]\n\n\[ \left( \frac{{x}^{3} - {2x} + 1}{x - 1}\right) \cdot 2\left( {x - 1}\right) = \left( {\frac{1}{2}x - 1}\right) \cdot 2\left( {x - 1}\right) \]\n\n\[ 2{x}^{3} - {4x} + 2 = {x}^{2} - {3x} + 2 \]\n\n\[ 2{x}^{3} ... | Yes |
Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own? | Solution. The key relationship between work and time which we use in this problem is:\namount of work done \( = \) rate of work \( \cdot \) time spent working\n\nWe are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor's case then:\namount of work Taylor does \( = \) rate of Taylor working \( \... | Yes |
1. Find an expression for the average cost function \( \bar{C}\left( x\right) \) . | 1. From \( \bar{C}\left( x\right) = \frac{C\left( x\right) }{x} \), we obtain \( \bar{C}\left( x\right) = \frac{{80x} + {150}}{x} \) . The domain of \( C \) is \( x \geq 0 \), but since \( x = 0 \) causes problems for \( \bar{C}\left( x\right) \), we get our domain to be \( x > 0 \), or \( \left( {0,\infty }\right) \) ... | Yes |
A box with a square base and no top is to be constructed so that it has a volume of 1000 cubic centimeters. Let \( x \) denote the width of the box, in centimeters as seen below. | ## Solution.\n\n1. We are told that the volume of the box is 1000 cubic centimeters and that \( x \) represents the width, in centimeters. From geometry, we know Volume \( = \) width \( \times \) height \( \times \) depth. Since the base of the box is a square, the width and the depth are both \( x \) centimeters. Usin... | Yes |
1. Hooke’s Law: The force \( F \) exerted on a spring is directly proportional the extension \( x \) of the spring. | Applying the definition of direct variation, we get \( F = {kx} \) for some constant \( k \) . | Yes |
1. Use your calculator to generate a scatter diagram for these data using \( V \) as the independent variable and \( P \) as the dependent variable. Does it appear from the graph that \( P \) is inversely proportional to \( V \) ? Explain. | 1. If \( P \) really does vary inversely with \( V \), then \( P = \frac{k}{V} \) for some constant \( k \) . From the data plot, the points do seem to lie along a curve like \( y = \frac{k}{x} \) . | No |
1. \( \left( {g \circ f}\right) \left( 1\right) \) | Using Definition 5.1, \( \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) \) . We find \( f\left( 1\right) = - 3 \), so\n\n\[ \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) = g\left( {-3}\right) = 2 \]\n | Yes |
Theorem 5.1. Properties of Function Composition: Suppose \( f, g \), and \( h \) are functions.\n\n- \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f \), provided the composite functions are defined. | By repeated applications of Definition 5.1, we find \( \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) = h\left( {\left( {g \circ f}\right) \left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) \) . Similarly, \( \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\rig... | No |
The surface area \( S \) of a sphere is a function of its radius \( r \) and is given by the formula \( S\left( r\right) = {4\pi }{r}^{2} \) . Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula \( r\left( t\right) = 3{t}^{2} \), where \( t \) is measured in seco... | Solution. If we look at the functions \( S\left( r\right) \) and \( r\left( t\right) \) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a specific time, \( t \), we could find the radius at that time, \( r\left( t\right) \... | Yes |
1. \( F\left( x\right) = \left| {{3x} - 1}\right| \) | 1. Our goal is to express the function \( F \) as \( F = g \circ f \) for functions \( g \) and \( f \) . From Definition 5.1, we know \( F\left( x\right) = g\left( {f\left( x\right) }\right) \), and we can think of \( f\left( x\right) \) as being the ’inside’ function and \( g \) as being the ’outside’ function. Looki... | Yes |
Theorem 5.2. Properties of Inverse Functions: Suppose \( f \) and \( g \) are inverse functions.\n\n- The range \( {}^{a} \) of \( f \) is the domain of \( g \) and the domain of \( f \) is the range of \( g \)\n\n- \( f\left( a\right) = b \) if and only if \( g\left( b\right) = a \)\n\n- \( \left( {a, b}\right) \) is ... | Theorem 5.2 is a consequence of Definition 5.2 and the Fundamental Graphing Principle for Functions. We note the third property in Theorem 5.2 tells us that the graphs of inverse functions are reflections about the line \( y = x \) . For a proof of this, see Example 1.1.7 in Section 1.1 and Exercise 72 in Section 2.1. | No |
Theorem 5.3. Uniqueness of Inverse Functions and Their Graphs : Suppose \( f \) is an invertible function.\n\n- There is exactly one inverse function for \( f \), denoted \( {f}^{-1} \) (read \( f \) -inverse)\n\n- The graph of \( y = {f}^{-1}\left( x\right) \) is the reflection of the graph of \( y = f\left( x\right) ... | The notation \( {f}^{-1} \) is an unfortunate choice since you’ve been programmed since Elementary Algebra to think of this as \( \frac{1}{f} \) . This is most definitely not the case since, for instance, \( f\left( x\right) = {3x} + 4 \) has as its inverse \( {f}^{-1}\left( x\right) = \frac{x - 4}{3} \), which is cert... | Yes |
Determine if the following functions are one-to-one in two ways: (a) analytically using Definition 5.3 and (b) graphically using the Horizontal Line Test.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \) 2. \( g\left( x\right) = \frac{2x}{1 - x} \)\n\n3. \( h\left( x\right) = {x}^{2} - {2x} + 4 \) 4. \( F = \{ \left( ... | 1. (a) To determine if \( f \) is one-to-one analytically, we assume \( f\left( c\right) = f\left( d\right) \) and attempt to deduce that \( c = d \) .\n\n\[ f\left( c\right) = f\left( d\right) \]\n\n\[ \frac{1 - {2c}}{5} = \frac{1 - {2d}}{5} \]\n\n\[ 1 - {2c} = 1 - {2d} \]\n\n\[ - {2c} = - {2d} \]\n\n\[ c = d\checkmar... | Yes |
Find the inverse of the following one-to-one functions. Check your answers analytically using function composition and graphically.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \) | Solution.\n\n1. As we mentioned earlier, it is possible to think our way through the inverse of \( f \) by recording the steps we apply to \( x \) and the order in which we apply them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algori... | Yes |
Graph the following functions to show they are one-to-one and find their inverses. Check your answers analytically using function composition and graphically.\n\n1. \( j\left( x\right) = {x}^{2} - {2x} + 4, x \leq 1 \) | The function \( j \) is a restriction of the function \( h \) from Example 5.2.1. Since the domain of \( j \) is restricted to \( x \leq 1 \), we are selecting only the ’left half’ of the parabola. We see that the graph of \( j \) passes the Horizontal Line Test and thus \( j \) is invertible.\n\n![800adaac-5958-4055-8... | Yes |
1. Explain why \( p \) is one-to-one and find a formula for \( {p}^{-1}\left( x\right) \). State the restricted domain. | 1. We leave to the reader to show the graph of \( p\left( x\right) = - {1.5x} + {250},0 \leq x \leq {166} \), is a line segment from \( \left( {0,{250}}\right) \) to \( \left( {{166},1}\right) \), and as such passes the Horizontal Line Test. Hence, \( p \) is one-to-one. We find the expression for \( {p}^{-1}\left( x\r... | No |
Theorem 5.6. Properties of Radicals: Let \( x \) and \( y \) be real numbers and \( m \) and \( n \) be natural numbers. If \( \sqrt[n]{x},\sqrt[n]{y} \) are real numbers, then\n\n- Product Rule: \( \sqrt[n]{xy} = \sqrt[n]{x}\sqrt[n]{y} \)\n\n- Powers of Radicals: \( \sqrt[n]{{x}^{m}} = {\left( \sqrt[n]{x}\right) }^{m}... | The proof of Theorem 5.6 is based on the definition of the principal roots and properties of exponents. To establish the product rule, consider the following. If \( n \) is odd, then by definition \( \sqrt[n]{xy} \) is the unique real number such that \( {\left( \sqrt[n]{xy}\right) }^{n} = {xy} \) . Given that \( {\lef... | No |
For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator. | 1. As far as domain is concerned, \( f\left( x\right) \) has no denominators and no even roots, which means its domain is \( \left( {-\infty ,\infty }\right) \) . To create the sign diagram, we find the zeros of \( f \) .\n\n\[ f\left( x\right) = 0 \]\n\n\[ {3x}\sqrt[3]{2 - x} = 0 \]\n\n\[ {3x} = 0\text{ or }\sqrt[3]{2... | Yes |
1. \( {x}^{2/3} < {x}^{4/3} - 6 \) | Solution. 1. To solve \( {x}^{2/3} < {x}^{4/3} - 6 \), we get 0 on one side and attempt to solve \( {x}^{4/3} - {x}^{2/3} - 6 > 0 \) . We set \( r\left( x\right) = {x}^{4/3} - {x}^{2/3} - 6 \) and note that since the denominators in the exponents are 3, they correspond to cube roots, which means the domain of \( r \) i... | Yes |
1. Express the total cost \( C \) of connecting the Junction Box to the Outpost as a function of \( x \) , the number of miles the cable is run along Route 117 before heading off road directly towards the Outpost. Determine a reasonable applied domain for the problem. | 1. The cost is broken into two parts: the cost to run cable along Route 117 at \( \$ {15} \) per mile, and the cost to run it off road at \( \$ {20} \) per mile. Since \( x \) represents the miles of cable run along Route 117, the cost for that portion is \( {15x} \) . From the diagram, we see that the number of miles ... | Yes |
Find and interpret \( V\left( 0\right) \). | To find \( V\left( 0\right) \), we replace \( x \) with 0 to obtain \( V\left( 0\right) = {25}{\left( \frac{4}{5}\right) }^{0} = {25} \). Since \( x \) represents the age of the car in years, \( x = 0 \) corresponds to the car being brand new. Since \( V\left( x\right) \) is measured in thousands of dollars, \( V\left(... | Yes |
1. Find and interpret \( T\left( 0\right) \) . | 1. To find \( T\left( 0\right) \), we replace every occurrence of the independent variable \( t \) with 0 to obtain \( T\left( 0\right) = {70} + {90}{e}^{-{0.1}\left( 0\right) } = {160} \) . This means that the coffee was served at \( {160}^{ \circ }\mathrm{F} \) . | Yes |
Theorem 6.2. Properties of Logarithmic Functions: Suppose \( f\left( x\right) = {\log }_{b}\left( x\right) \) . | - The domain of \( f \) is \( \left( {0,\infty }\right) \) and the range of \( f \) is \( \left( {-\infty ,\infty }\right) \) . \n- \( \left( {1,0}\right) \) is on the graph of \( f \) and \( x = 0 \) is a vertical asymptote of the graph of \( f \) . \n- \( f \) is one-to-one, continuous and smooth \n- \( {b}^{a} = c \... | Yes |
Simplify the following.\n\n1. \( {\log }_{3}\left( {81}\right) \) 2. \( {\log }_{2}\left( \frac{1}{8}\right) \) 3. \( {\log }_{\sqrt{5}}\left( {25}\right) \) 4. \( \ln \left( \sqrt[3]{{e}^{2}}\right) \)\n\n5. \( \log \left( {0.001}\right) \) 6. \( {2}^{{\log }_{2}\left( 8\right) } \) 7. \( {117}^{-{\log }_{117}\left( 6... | Solution.\n\n1. The number \( {\log }_{3}\left( {81}\right) \) is the exponent we put on 3 to get 81 . As such, we want to write 81 as a power of 3 . We find \( {81} = {3}^{4} \), so that \( {\log }_{3}\left( {81}\right) = 4 \) .\n\n2. To find \( {\log }_{2}\left( \frac{1}{8}\right) \), we need rewrite \( \frac{1}{8} \... | Yes |
Find the domain of the following functions. Check your answers graphically using the calculator.\n\n1. \( f\left( x\right) = 2\log \left( {3 - x}\right) - 1 \) | ## Solution.\n\n1. We set \( 3 - x > 0 \) to obtain \( x < 3 \), or \( \left( {-\infty ,3}\right) \) . The graph from the calculator below verifies this. Note that we could have graphed \( f \) using transformations. Taking a cue from Theorem 1.7, we rewrite \( f\left( x\right) = 2{\log }_{10}\left( {-x + 3}\right) - 1... | Yes |
1. Graph \( f \) using transformations and state the domain and range of \( f \) . | If we identify \( g\left( x\right) = {2}^{x} \), we see \( f\left( x\right) = g\left( {x - 1}\right) - 3 \) . We pick the points \( \left( {-1,\frac{1}{2}}\right) ,\left( {0,1}\right) \) and \( \left( {1,2}\right) \) on the graph of \( g \) along with the horizontal asymptote \( y = 0 \) to track through the transforma... | Yes |
Product Rule: \( g\left( {uw}\right) = g\left( u\right) + g\left( w\right) \) . In other words, \( {\log }_{b}\left( {uw}\right) = {\log }_{b}\left( u\right) + {\log }_{b}\left( w\right) \) | Let \( a = {\log }_{b}\left( {uw}\right), c = {\log }_{b}\left( u\right) \), and \( d = {\log }_{b}\left( w\right) \). Then, by definition, \( {b}^{a} = {uw},{b}^{c} = u \) and \( {b}^{d} = w \). Hence, \( {b}^{a} = {uw} = {b}^{c}{b}^{d} = {b}^{c + d} \), so that \( {b}^{a} = {b}^{c + d} \). By the one-to-one property ... | Yes |
Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers.\n\n1. \( {\log }_{2}\left( \frac{8}{x}\right) \) | To expand \( {\log }_{2}\left( \frac{8}{x}\right) \), we use the Quotient Rule identifying \( u = 8 \) and \( w = x \) and simplify.\n\n\[ {\log }_{2}\left( \frac{8}{x}\right) = {\log }_{2}\left( 8\right) - {\log }_{2}\left( x\right) \;\text{ Quotient Rule } \]\n\n\[ = 3 - {\log }_{2}\left( x\right) \;\text{Since}{2}^{... | Yes |
Use the properties of logarithms to write the following as a single logarithm. | 1. The difference of logarithms requires the Quotient Rule: \( {\log }_{3}\left( {x - 1}\right) - {\log }_{3}\left( {x + 1}\right) = {\log }_{3}\left( \frac{x - 1}{x + 1}\right) \) . | Yes |
Theorem 6.7. (Change of Base Formulas) Let \( a, b > 0, a, b \neq 1 \) .\n\n- \( {a}^{x} = {b}^{x{\log }_{b}\left( a\right) } \) for all real numbers \( x \) .\n\n- \( {\log }_{a}\left( x\right) = \frac{{\log }_{b}\left( x\right) }{{\log }_{b}\left( a\right) } \) for all real numbers \( x > 0 \) . | The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with \( {b}^{x{\log }_{b}\left( a\right) } \) and use the Power Rule in the exponent to rewrite \( x{\log }_{b}\left( a\right) \) as \( {\log }_{b}\left( {a}^{x}\right) \) and then apply one of the Invers... | Yes |
Example 6.2.3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate. | ## Solution.\n\n1. We apply the Change of Base formula with \( a = 3 \) and \( b = {10} \) to obtain \( {3}^{2} = {10}^{2\log \left( 3\right) } \) . Typing the latter in the calculator produces an answer of 9 as required.\n\n2. Here, \( a = 2 \) and \( b = e \) so we have \( {2}^{x} = {e}^{x\ln \left( 2\right) } \) . T... | Yes |
1. \( {2}^{3x} = {16}^{1 - x} \) | Since 16 is a power of 2, we can rewrite \( {2}^{3x} = {16}^{1 - x} \) as \( {2}^{3x} = {\left( {2}^{4}\right) }^{1 - x} \) . Using properties of exponents, we get \( {2}^{3x} = {2}^{4\left( {1 - x}\right) } \) . Using the one-to-one property of exponential functions, we get \( {3x} = 4\left( {1 - x}\right) \) which gi... | Yes |
1. \( {2}^{{x}^{2} - {3x}} - {16} \geq 0 \) | Since we already have 0 on one side of the inequality, we set \( r\left( x\right) = {2}^{{x}^{2} - {3x}} - {16} \) . The domain of \( r \) is all real numbers, so in order to construct our sign diagram, we need to find the zeros of \( r \) . Setting \( r\left( x\right) = 0 \) gives \( {2}^{{x}^{2} - {3x}} - {16} = 0 \)... | Yes |
Recall from Example 6.1.2 that the temperature of coffee \( T \) (in degrees Fahrenheit) \( t \) minutes after it is served can be modeled by \( T\left( t\right) = {70} + {90}{e}^{-{0.1t}} \) . When will the coffee be warmer than \( {100}^{ \circ }\mathrm{F} \) ? | Solution. We need to find when \( T\left( t\right) > {100} \), or in other words, we need to solve the inequality \( {70} + {90}{e}^{-{0.1t}} > {100} \) . Getting 0 on one side of the inequality, we have \( {90}{e}^{-{0.1t}} - {30} > 0 \), and we set \( r\left( t\right) = {90}{e}^{-{0.1t}} - {30} \) . The domain of \( ... | Yes |
The function \( f\left( x\right) = \frac{5{e}^{x}}{{e}^{x} + 1} \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator. | Solution. We start by writing \( y = f\left( x\right) \), and interchange the roles of \( x \) and \( y \) . To solve for \( y \), we first clear denominators and then isolate the exponential function.\n\n\[ y = \frac{5{e}^{x}}{{e}^{x} + 1} \]\n\n\[ x = \frac{5{e}^{y}}{{e}^{y} + 1}\;\text{ Switch }x\text{ and }y \]\n\n... | No |
1. \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \) | Since we have the same base on both sides of the equation \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \) , we equate what’s inside the logs to get \( 1 - {3x} = {x}^{2} - 3 \) . Solving \( {x}^{2} + {3x} - 4 = 0 \) gives \( x = - 4 \) and \( x = 1 \) . To check these answers usin... | Yes |
1. \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \) | We start solving \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \) by getting 0 on one side of the inequality: \( \frac{1}{\ln \left( x\right) + 1} - 1 \leq 0 \) . Getting a common denominator yields \( \frac{1}{\ln \left( x\right) + 1} - \frac{\ln \left( x\right) + 1}{\ln \left( x\right) + 1} \leq 0 \) which reduces to \... | Yes |
In order to successfully breed Ippizuti fish the \( \mathrm{{pH}} \) of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator. | Solution. Recall from Exercise 77 in Section 6.1 that \( \mathrm{{pH}} = - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) where \( \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) is the hydrogen ion concentration in moles per liter. We require \( {7.8} \leq - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq... | Yes |
The function \( f\left( x\right) = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator. | Solution. We first write \( y = f\left( x\right) \) then interchange the \( x \) and \( y \) and solve for \( y \) .\n\n\[ y = f\left( x\right) \]\n\n\[ y = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \]\n\n\[ x = \frac{\log \left( y\right) }{1 - \log \left( y\right) } \]\nInterchange \( x \) and \( y \) .\... | Yes |
1. Express the amount \( A \) in the account as a function of the term of the investment \( t \) in years. | 1. Substituting \( P = {2000}, r = {0.07125} \), and \( n = {12} \) (since interest is compounded monthly) into Equation 6.2 yields \( A\left( t\right) = {2000}{\left( 1 + \frac{0.07125}{12}\right) }^{12t} = {2000}{\left( {1.0059375}\right) }^{12t} \) . | Yes |
In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula fo... | Solution. We begin with \( N\left( t\right) = {N}_{0}{e}^{kt} \) . Since \( N \) is to give the number of cells in thousands, we have \( {N}_{0} = {12} \), so \( N\left( t\right) = {12}{e}^{kt} \) . In order to complete the formula, we need to determine the growth rate \( k \) . We know that after one week, the number ... | Yes |
Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life \( {}^{10} \) of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, find a function w... | Solution. Since we start with 5 grams initially, Equation 6.5 gives \( A\left( t\right) = 5{e}^{kt} \) . Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, \( A\left( 8\right) = {2.5} \) which means \( 5{e}^{8k} = {2.5} \) . Solving, we get \( k = \frac... | Yes |
1. Assuming the temperature of the roast follows Newton's Law of Warming, find a formula for the temperature of the roast \( T \) as a function of its time in the oven, \( t \), in hours. | 1. The initial temperature of the roast is \( {40}^{ \circ }\mathrm{F} \), so \( {T}_{0} = {40} \) . The environment in which we are placing the roast is the \( {350}^{ \circ }\mathrm{F} \) oven, so \( {T}_{a} = {350} \) . Newton’s Law of Warming tells us \( T\left( t\right) = {350} + \left( {{40} - {350}}\right) {e}^{... | Yes |
The number of people \( N \), in hundreds, at a local community college who have heard the rumor 'Carl is afraid of Virginia Woolf' can be modeled using the logistic equation\n\n\[ N\left( t\right) = \frac{84}{1 + {2799}{e}^{-t}} \]\n\nwhere \( t \geq 0 \) is the number of days after April 1,2009.\n\n1. Find and interp... | ## Solution.\n\n1. We find \( N\left( 0\right) = \frac{84}{1 + {2799}{e}^{0}} = \frac{84}{2800} = \frac{3}{100} \). Since \( N\left( t\right) \) measures the number of people who have heard the rumor in hundreds, \( N\left( 0\right) \) corresponds to 3 people. Since \( t = 0 \) corresponds to April 1,2009, we may concl... | Yes |
1. If a 7 character case-sensitive \( {}^{16} \) password is comprised of letters and numbers only, find the associated information entropy. | 1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as different. There are 10 digits (0 through 9) for a total of \( N = {62} \) symbols. Since the password is to be 7 characters long, \( L = 7 \) . Thus, \( H = 7{\log }_{2}\left( {62}\right) = \frac{7\ln \left( {62}\right) }{\ln \l... | Yes |
Find the partial pressure of carbon dioxide in arterial blood if the \( \mathrm{{pH}} \) is 7.4. | Solution. We set \( \mathrm{{pH}} = {7.4} \) and get \( {7.4} = {6.1} + \log \left( \frac{800}{x}\right) \), or \( \log \left( \frac{800}{x}\right) = {1.3} \) . Solving, we find \( x = \frac{800}{{10}^{1 \cdot 3}} \approx {40.09} \) . Hence, the partial pressure of carbon dioxide in the blood is about 40 torr. | Yes |
Write the standard equation of the circle with center \( \left( {-2,3}\right) \) and radius 5. | Solution. Here, \( \left( {h, k}\right) = \left( {-2,3}\right) \) and \( r = 5 \), so we get\n\n\[ \n{\left( x - \left( -2\right) \right) }^{2} + {\left( y - 3\right) }^{2} = {\left( 5\right) }^{2} \n\]\n\n\[ \n{\left( x + 2\right) }^{2} + {\left( y - 3\right) }^{2} = {25} \n\] | Yes |
Graph \( {\left( x + 2\right) }^{2} + {\left( y - 1\right) }^{2} = 4 \) . Find the center and radius. | Solution. From the standard form of a circle, Equation 7.1, we have that \( x + 2 \) is \( x - h \), so \( h = - 2 \) and \( y - 1 \) is \( y - k \) so \( k = 1 \) . This tells us that our center is \( \left( {-2,1}\right) \) . Furthermore, \( {r}^{2} = 4 \), so \( r = 2 \) . Thus we have a circle centered at \( \left(... | Yes |
Complete the square to find the center and radius of \( 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \) . | \[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \] \[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} = 4 \] add 4 to both sides \[ 3\left( {{x}^{2} - {2x}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y}\right) = 4 \] factor out leading coefficients \[ 3\left( {{x}^{2} - {2x} + \underline{1}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y + \unde... | Yes |
Write the standard equation of the circle which has \( \left( {-1,3}\right) \) and \( \left( {2,4}\right) \) as the endpoints of a diameter. | Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields\n\n\n\nSince the given points are endpoints of a diameter, we know ... | Yes |
Find the points on the unit circle with \( y \) -coordinate \( \frac{\sqrt{3}}{2} \) . | We replace \( y \) with \( \frac{\sqrt{3}}{2} \) in the equation \( {x}^{2} + {y}^{2} = 1 \) to get\n\n\[ \n{x}^{2} + {y}^{2} = 1 \n\]\n\n\[ \n{x}^{2} + {\left( \frac{\sqrt{3}}{2}\right) }^{2} = 1 \n\]\n\n\[ \n\frac{3}{4} + {x}^{2} = 1 \n\]\n\n\[ \n{x}^{2} = \frac{1}{4} \n\]\n\n\[ \nx = \pm \sqrt{\frac{1}{4}} \n\]\n\n\... | Yes |
Graph \( {\left( x + 1\right) }^{2} = - 8\left( {y - 3}\right) \) . Find the vertex, focus, and directrix. | Solution. We recognize this as the form given in Equation 7.2. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 3 \) so \( k = 3 \) . Hence, the vertex is \( \left( {-1,3}\right) \) . We also see that \( {4p} = - 8 \) so \( p = - 2 \) . Since \( p < 0 \), the focus will be below the vertex ... | Yes |
Find the standard form of the parabola with focus \( \left( {2,1}\right) \) and directrix \( y = - 4 \) . | Solution. Sketching the data yields,\n\n\n\n---\n\n\n\nFrom the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don’t see that immediately.) Hence, the vertex lies below the focus ... | Yes |
Graph \( {\left( y - 2\right) }^{2} = {12}\left( {x + 1}\right) \) . Find the vertex, focus, and directrix. | Solution. We recognize this as the form given in Equation 7.3. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 2 \) so \( k = 2 \) . Hence, the vertex is \( \left( {-1,2}\right) \) . We also see that \( {4p} = {12} \) so \( p = 3 \) . Since \( p > 0 \), the focus will be the right of the v... | Yes |
Consider the equation \( {y}^{2} + {4y} + {8x} = 4 \) . Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. | Solution. We need a perfect square (in this case, using \( y \) ) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the \( x \) ) on the other.\n\n\[ {y}^{2} + {4y} + {8x} = 4 \]\n\n\[ {y}^{2} + {4y} = - {8x} + 4 \]\n\n\[ {y}^{2} + {4y} + 4 = - {8x} + 4 + 4\... | Yes |
Example 7.3.5. A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be? | Solution. One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we’ll assume the vertex is \( \left( {0,0}\right) \) and the parabola opens upwards. Our standard form for such a parabola is \( {x}^{2} = {4py} \) . Since the focus is 2 units above the... | Yes |
Graph \( \frac{{\left( x + 1\right) }^{2}}{9} + \frac{{\left( y - 2\right) }^{2}}{25} = 1 \) . Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. | Solution. We see that this equation is in the standard form of Equation 7.4. Here \( x - h \) is \( x + 1 \) so \( h = - 1 \), and \( y - k \) is \( y - 2 \) so \( k = 2 \) . Hence, our ellipse is centered at \( \left( {-1,2}\right) \) . We see that \( {a}^{2} = 9 \) so \( a = 3 \), and \( {b}^{2} = {25} \) so \( b = 5... | Yes |
Find the equation of the ellipse with foci \( \\left( {2,1}\\right) \) and \( \\left( {4,1}\\right) \) and vertex \( \\left( {0,1}\\right) \) . | Solution. Plotting the data given to us, we have\n\n\n\nFrom this sketch, we know that the major axis is horizontal, meaning \( a > b \) . Since the center is the midpoint of the foci, we know it is \( \\left( {3,1}\... | Yes |
Graph \( {x}^{2} + 4{y}^{2} - {2x} + {24y} + {33} = 0 \) . Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. | Solution. Since we have a sum of squares and the squared terms have unequal coefficients, it's a good bet we have an ellipse on our hands. \( {}^{2} \) We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1 . \n\n\[ \n{x}^{2} + 4{y}^{2} - {2x} + {24y} + {33} = 0 \n\] \n\n... | Yes |
Find the equation of the ellipse whose vertices are \( \left( {\pm 5,0}\right) \) with eccentricity \( e = \frac{1}{4} \) . | Solution. As before, we plot the data given to us\n\n\n\nFrom this sketch, we know that the major axis is horizontal, meaning \( a > b \) . With the vertices located at \( \left( {\pm 5,0}\right) \), we get \( a = 5 ... | Yes |
Example 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room is 40 feet high at the center and 100 feet wide at the floor, how far from the outer wall should eac... | Solution. Graphing the data yields\n\n\n\n100 units wide\n\nIt’s most convenient to imagine this ellipse centered at \( \left( {0,0}\right) \) . Since the ellipse is 100 units wide and 40 units tall, we get \( a = {5... | Yes |
Graph the equation \( \frac{{\left( x - 2\right) }^{2}}{4} - \frac{{y}^{2}}{25} = 1 \) . Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. | Solution. We first see that this equation is given to us in the standard form of Equation 7.6. Here \( x - h \) is \( x - 2 \) so \( h = 2 \), and \( y - k \) is \( y \) so \( k = 0 \) . Hence, our hyperbola is centered at \( \left( {2,0}\right) \) . We see that \( {a}^{2} = 4 \) so \( a = 2 \), and \( {b}^{2} = {25} \... | Yes |
Find the equation of the hyperbola with asymptotes \( y = \pm {2x} \) and vertices \( \left( {\pm 5,0}\right) \) . | Plotting the data given to us, we have\n\n\n\nThis graph not only tells us that the branches of the hyperbola open to the left and to the right, it also tells us that the center is \( \left( {0,0}\right) \) . Hence, ... | Yes |
Consider the equation \( 9{y}^{2} - {x}^{2} - {6x} = {10} \). Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci, and the equations of the asymptotes. | Solution. We need only complete the square on \( x \):\n\n\[ 9{y}^{2} - {x}^{2} - {6x} = {10} \]\n\n\[ 9{y}^{2} - 1\left( {{x}^{2} + {6x}}\right) = {10} \]\n\n\[ 9{y}^{2} - \left( {{x}^{2} + {6x} + 9}\right) = {10} - 1\left( 9\right) \]\n\n\[ 9{y}^{2} - {\left( x + 3\right) }^{2} = 1 \]\n\n\[ \frac{{y}^{2}}{\frac{1}{9}... | Yes |
Jeff is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be l... | Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeff by multiplying\n\n\[ \n{760}\frac{\text{miles}}{\text{hour}} \times \frac{1\text{ hour}}{{3600}\text{ seconds}} \times 9\text{ seconds} = {1.9... | Yes |
By a stroke of luck, Kai was also camping in the woods during the events of the previous example. He was located 6 miles due north of Jeff and heard the Sasquatch call 18 seconds after Jeff did. Use this added information to locate Sasquatch. | Kai and Jeff are now the foci of a second hyperbola where the fixed distance \( d \) can be determined as before\n\n\[ \n{760}\frac{\\text{miles}}{\\text{hour}} \\times \\frac{1\\text{ hour}}{{3600}\\text{ seconds}} \\times {18}\\text{ seconds} = {3.8}\\text{ miles} \n\]\n\nSince Jeff was positioned at \( \\left( {-5,0... | Yes |
Solve the following systems of equations. Check your answer algebraically and graphically.\n\n1. \( \begin{cases} {2x} - y & = 1 \\ y & = 3 \end{cases} \) | Our first system is nearly solved for us. The second equation tells us that \( y = 3 \) . To find the corresponding value of \( x \), we substitute this value for \( y \) into the the first equation to obtain \( {2x} - 3 = 1 \), so that \( x = 2 \) . Our solution to the system is \( \left( {2,3}\right) \) . To check th... | Yes |
Theorem 8.1. Given a system of equations, the following moves will result in an equivalent system of equations.\n\n- Interchange the position of any two equations.\n\n- Replace an equation with a nonzero multiple of itself. \( {}^{a} \)\n\n- Replace an equation with itself plus a nonzero multiple of another equation.\n... | We have seen plenty of instances of the second and third moves in Theorem 8.1 when we solved the systems in Example 8.1.1. The first move, while it obviously admits an equivalent system, seems silly. Our perception will change as we consider more equations and more variables in this, and later sections. | No |
Theorem 8.2. Row Operations: Given an augmented matrix for a system of linear equations, the following row operations produce an augmented matrix which corresponds to an equivalent system of linear equations. | - Interchange any two rows.\n\n- Replace a row with a nonzero multiple of itself. \( {}^{a} \)\n\n- Replace a row with itself plus a nonzero multiple of another row. \( {}^{b} \)\n\n\( {}^{a} \) That is, the row obtained by multiplying each entry in the row by the same nonzero number.\n\n\( {}^{b} \) Where we add entri... | Yes |
Find the quadratic function passing through the points \( \left( {-1,3}\right) ,\left( {2,4}\right) ,\left( {5, - 2}\right) \) . | Solution. According to Definition 2.5, a quadratic function has the form \( f\left( x\right) = a{x}^{2} + {bx} + c \) where \( a \neq 0 \) . Our goal is to find \( a, b \) and \( c \) so that the three given points are on the graph of \( f \) . If \( \left( {-1,3}\right) \) is on the graph of \( f \), then \( f\left( {... | Yes |
Theorem 8.4. Properties of Scalar Multiplication\n\n- Associative Property: For every \( m \times n \) matrix \( A \) and scalars \( k \) and \( r,\left( {kr}\right) A = k\left( {rA}\right) \) . | As with the other results in this section, Theorem 8.4 can be proved using the definitions of scalar multiplication and matrix addition. For example, to prove that \( k\left( {A + B}\right) = {kA} + {kB} \) for a scalar \( k \) and \( m \times n \) matrices \( A \) and \( B \), we start by adding \( A \) and \( B \), t... | No |
2. If a point \( P \) is on the hyperbola \( {x}^{2} - {y}^{2} = 4 \), show that the point \( {RP} \) is on the curve \( y = \frac{2}{x} \) . | For a generic point \( P\left( {x, y}\right) \) on the hyperbola \( {x}^{2} - {y}^{2} = 4 \), we have\n\n\[ \n{RP} = \left\lbrack \begin{array}{rr} \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack ... | Yes |
1. If \( A \) is invertible then \( {A}^{-1} \) is unique. | To establish the first property, we assume that \( A \) is invertible and suppose the matrices \( B \) and \( C \) act as inverses for \( A \) . That is, \( {BA} = {AB} = {I}_{n} \) and \( {CA} = {AC} = {I}_{n} \) . We need to show that \( B \) and \( C \) are, in fact, the same matrix. To see this, we note that \( B =... | Yes |
Consider the circuit diagram below. \( {}^{7} \) We have two batteries with source voltages \( V{B}_{1} \) and \( V{B}_{2} \), measured in volts \( V \), along with six resistors with resistances \( {R}_{1} \) through \( {R}_{6} \), measured in kiloohms, \( {k\Omega } \) . Using Ohm’s Law and Kirchhoff’s Voltage Law, w... | The system of linear equations associated with this circuit is\n\n\[ \n\begin{cases} \left( {{R}_{1} + {R}_{3}}\right) {i}_{1} - {R}_{3}{i}_{2} - {R}_{1}{i}_{4} & = V{B}_{1} \\ - {R}_{3}{i}_{1} + \left( {{R}_{2} + {R}_{3} + {R}_{4}}\right) {i}_{2} - {R}_{4}{i}_{3} - {R}_{2}{i}_{4} & = 0 \\ - {R}_{4}{i}_{2} + \left( {{R... | Yes |
Theorem 8.7. Properties of the Determinant: Let \( A = {\left\lbrack {a}_{ij}\right\rbrack }_{n \times n} \) . | Unfortunately, while we can easily demonstrate the results in Theorem 8.7, the proofs of most of these properties are beyond the scope of this text. We could prove these properties for generic \( 2 \times 2 \) or even \( 3 \times 3 \) matrices by brute force computation, but this manner of proof belies the elegance and... | No |
Theorem 8.8. Cramer’s Rule: Suppose \( {AX} = B \) is the matrix form of a system of \( n \) linear equations in \( n \) unknowns where \( A \) is the coefficient matrix, \( X \) is the unknowns matrix, and \( B \) is the constant matrix. If \( \det \left( A\right) \neq 0 \), then the corresponding system is consistent... | \[ {x}_{j} = \frac{\det \left( {A}_{j}\right) }{\det \left( A\right) } \] where \( {A}_{j} \) is the matrix \( A \) whose \( j \) th column has been replaced by the constants in \( B \). | Yes |
Solve \( \begin{cases} 2{x}_{1} - 3{x}_{2} & = 4 \\ 5{x}_{1} + {x}_{2} & = - 2 \end{cases} \) for \( {x}_{1} \) and \( {x}_{2} \) | Writing this system in matrix form, we find\n\n\[ A = \left\lbrack \begin{array}{rr} 2 & - 3 \\ 5 & 1 \end{array}\right\rbrack \;X = \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right\rbrack \;B = \left\lbrack \begin{array}{r} 4 \\ - 2 \end{array}\right\rbrack \]\n\nTo find the matrix \( {A}_{1} \), we ... | Yes |
Theorem 8.10. Suppose \( R\left( x\right) = \frac{N\left( x\right) }{D\left( x\right) } \) is a rational function where the degree of \( N\left( x\right) \) less than the degree of \( D\left( x\right) \) and \( N\left( x\right) \) and \( D\left( x\right) \) have no common factors. \( {}^{a} \n\n- If \( \alpha \) is a r... | The proof of Theorem 8.10 is best left to a course in Abstract Algebra. | No |
Theorem 8.11. Suppose\n\n\[ \n{a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} = {b}_{m}{x}^{m} + {m}_{m - 1}{x}^{m - 1} + \cdots + {b}_{2}{x}^{2} + {b}_{1}x + {b}_{0} \n\]\n\nfor all \( x \) in an open interval \( I \) . Then \( n = m \) and \( {a}_{i} = {b}_{i} \) for all \( i = ... | Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Define \( p\left( x\right) \) to be the difference of the left hand side of the equation in Theorem 8.11 and the right hand side. Then \( p\left( x\right) = 0 \) for all \( x \) in the open interval \( I \) . If \( p\left( x\right) \) were a... | Yes |
Resolve the following rational functions into partial fractions.\n\n1. \( R\left( x\right) = \frac{x + 5}{2{x}^{2} - x - 1} \) | We begin by factoring the denominator to find \( 2{x}^{2} - x - 1 = \left( {{2x} + 1}\right) \left( {x - 1}\right) \) . We get \( x = - \frac{1}{2} \) and \( x = 1 \) are both zeros of multiplicity one and thus we know\n\n\[ \frac{x + 5}{2{x}^{2} - x - 1} = \frac{x + 5}{\left( {{2x} + 1}\right) \left( {x - 1}\right) } ... | Yes |
1. \( \\begin{cases} {x}^{2} + {y}^{2} & = 4 \\ 4{x}^{2} + 9{y}^{2} & = {36} \\end{cases} \) | Since both equations contain \( {x}^{2} \) and \( {y}^{2} \) only, we can eliminate one of the variables as we did in Section 8.1.\n\n\[ \n\\left\\{ \\begin{matrix} \\left( {E1}\\right) & {x}^{2} + {y}^{2} & = & 4 \\ \\left( {E2}\\right) & 4{x}^{2} + 9{y}^{2} & = & {36} \\end{matrix}\\right. \\;\\xrightarrow[{-{4E1} + ... | No |
Solve the following systems of equations. Verify your answers algebraically and graphically, as appropriate.\n\n1. \( \left\{ \begin{array}{l} {x}^{2} + {2xy} - {16} = 0 \\ {y}^{2} + {2xy} - {16} = 0 \end{array}\right. \) | At first glance, it doesn't appear as though elimination will do us any good since it's clear that we cannot completely eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it ... | Yes |
Sketch the solution to the following nonlinear inequalities in the plane.\n\n1. \( {y}^{2} - 4 \leq x < y + 2 \) | ## Solution.\n\n1. The inequality \( {y}^{2} - 4 \leq x < y + 2 \) is a compound inequality. It translates as \( {y}^{2} - 4 \leq x \) and \( x < y + 2 \) . As usual, we solve each inequality and take the set theoretic intersection to determine the region which satisfies both inequalities. To solve \( {y}^{2} - 4 \leq ... | Yes |
Write the first four terms of the following sequences.\n\n1. \( {a}_{n} = \frac{{5}^{n - 1}}{{3}^{n}}, n \geq 1 \) | Since we are given \( n \geq 1 \), the first four terms of the sequence are \( {a}_{1},{a}_{2},{a}_{3} \) and \( {a}_{4} \) . Since the notation \( {a}_{1} \) means the same thing as \( a\left( 1\right) \), we obtain our first term by replacing every occurrence of \( n \) in the formula for \( {a}_{n} \) with \( n = 1 ... | Yes |
Determine if the following sequences are arithmetic, geometric or neither. If arithmetic, find the common difference \( d \) ; if geometric, find the common ratio \( r \). 1. \( {a}_{n} = \frac{{5}^{n - 1}}{{3}^{n}}, n \geq 1 \) | From Example 9.1.1, we know that the first four terms of this sequence are \( \frac{1}{3},\frac{5}{9},\frac{25}{27} \) and \( \frac{125}{81} \). To see if this is an arithmetic sequence, we look at the successive differences of terms. We find that \( {a}_{2} - {a}_{1} = \frac{5}{9} - \frac{1}{3} = \frac{2}{9} \) and \(... | Yes |
Find an explicit formula for the \( {n}^{\text{th }} \) term of the following sequences.\n\n1. \( {0.9},{0.09},{0.009},{0.0009},\ldots \) | Although this sequence may seem strange, the reader can verify it is actually a geometric sequence with common ratio \( r = {0.1} = \frac{1}{10} \) . With \( a = {0.9} = \frac{9}{10} \), we get \( {a}_{n} = \frac{9}{10}{\left( \frac{1}{10}\right) }^{n - 1} \) for \( n \geq 0 \) . Simplifying, we get \( {a}_{n} = \frac{... | No |
If monthly payments of \$50 are made, find the value of the annuity in 30 years. | We have \( r = {0.06} \) and \( n = {12} \) so that \( i = \frac{r}{n} = \frac{0.06}{12} = {0.005} \) . With \( P = {50} \) and \( t = {30} \) ,\n\n\[ A = \frac{{50}\left( {{\left( 1 + {0.005}\right) }^{\left( {12}\right) \left( {30}\right) } - 1}\right) }{0.005} \approx {50225.75} \]\n\nOur final answer is \$50,225.75... | Yes |
Theorem 9.2. Geometric Series: Given the sequence \( {a}_{k} = a{r}^{k - 1} \) for \( k \geq 1 \), where \( \left| r\right| < 1 \), \[ a + {ar} + a{r}^{2} + \ldots = \mathop{\sum }\limits_{{k = 1}}^{\infty }a{r}^{k - 1} = \frac{a}{1 - r} \] If \( \left| r\right| \geq 1 \), the sum \( a + {ar} + a{r}^{2} + \ldots \) is ... | The justification of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the first \( n \) terms of a geometric sequence and examining the formula as \( n \rightarrow \infty \) . Assuming \( \left| r\right| < 1 \) means \( - 1 < r < 1 \), so \( {r}^{n} \rightarrow 0 \) as \( n \righta... | Yes |
1. The sum formula for arithmetic sequences: \( \mathop{\sum }\limits_{{j = 1}}^{n}\left( {a + \left( {j - 1}\right) d}\right) = \frac{n}{2}\left( {{2a} + \left( {n - 1}\right) d}\right) \) . | We set \( P\left( n\right) \) to be the equation we are asked to prove. For \( n = 1 \), we compare both sides of the equation given in \( P\left( n\right) \)\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{1}\left( {a + \left( {j - 1}\right) d}\right) \overset{?}{ = }\frac{1}{2}\left( {{2a} + \left( {1 - 1}\right) d}\right) \]... | Yes |
Theorem 9.3. For natural numbers \( n \) and \( j \) with \( n \geq j \) , \[ \left( \begin{matrix} n \\ j - 1 \end{matrix}\right) + \left( \begin{array}{l} n \\ j \end{array}\right) = \left( \begin{matrix} n + 1 \\ j \end{matrix}\right) \] | \[ \left( \begin{matrix} n \\ j - 1 \end{matrix}\right) + \left( \begin{array}{l} n \\ j \end{array}\right) = \frac{n!}{\left( {j - 1}\right) !\left( {n - \left( {j - 1}\right) }\right) !} + \frac{n!}{j!\left( {n - j}\right) !} \] \[ = \frac{n!}{\left( {j - 1}\right) !\left( {n - j + 1}\right) !} + \frac{n!}{j!\left( {... | Yes |
Theorem 9.4. Binomial Theorem: For nonzero real numbers \( a \) and \( b \) ,\n\n\[ \n{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {a}^{n - j}{b}^{j} \n\]\n\nfor all natural numbers \( n \) . | To prove the Binomial Theorem, we let \( P\left( n\right) \) be the expansion formula given in the statement of the theorem and we note that \( P\left( 1\right) \) is true since\n\n\[ \n{\left( a + b\right) }^{1}\overset{?}{ = }\mathop{\sum }\limits_{{j = 0}}^{1}\left( \begin{array}{l} 1 \\ j \end{array}\right) {a}^{1 ... | Yes |
Use the Binomial Theorem to find the following.\n\n1. \( {\left( x - 2\right) }^{4} \) 2. \( {2.1}^{3} \)\n\n3. The term containing \( {x}^{3} \) in the expansion \( {\left( 2x + y\right) }^{5} \) | 1. Since \( {\left( x - 2\right) }^{4} = {\left( x + \left( -2\right) \right) }^{4} \), we identify \( a = x, b = - 2 \) and \( n = 4 \) and obtain\n\n\[ \n{\left( x - 2\right) }^{4} = \mathop{\sum }\limits_{{j = 0}}^{4}\left( \begin{array}{l} 4 \\ j \end{array}\right) {x}^{4 - j}{\left( -2\right) }^{j} \n\]\n\n\[ \n= ... | No |
1. Convert \( \alpha \) to the DMS system. Round your answer to the nearest second. | To convert \( \alpha \) to the DMS system, we start with \( {111.371}^{ \circ } = {111}^{ \circ } + {0.371}^{ \circ } \) . Next we convert \( {0.371}^{ \circ }\left( \frac{{60}^{\prime }}{{1}^{ \circ }}\right) = {22.26}^{\prime } \) . Writing \( {22.26}^{\prime } = {22}^{\prime } + {0.26}^{\prime } \), we convert \( {0... | Yes |
Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. | 1. To graph \( \alpha = {60}^{ \circ } \), we draw an angle with its initial side on the positive \( x \) -axis and rotate counter-clockwise \( \frac{{60}^{ \circ }}{{360}^{ \circ }} = \frac{1}{6} \) of a revolution. We see that \( \alpha \) is a Quadrant I angle. To find angles which are coterminal, we look for angles... | Yes |
Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. | 1. The angle \( \alpha = \frac{\pi }{6} \) is positive, so we draw an angle with its initial side on the positive \( x \) -axis and rotate counter-clockwise \( \frac{\left( \pi /6\right) }{2\pi } = \frac{1}{12} \) of a revolution. Thus \( \alpha \) is a Quadrant I angle. Coterminal angles \( \theta \) are of the form \... | Yes |
Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. | 1. The arc associated with \( t = \frac{3\pi }{4} \) is the arc on the Unit Circle which subtends the angle \( \frac{3\pi }{4} \) in radian measure. Since \( \frac{3\pi }{4} \) is \( \frac{3}{8} \) of a revolution, we have an arc which begins at the point \( \left( {1,0}\right) \) proceeds counter-clockwise up to midwa... | Yes |
Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at \( {41.6... | Solution. To use the formula \( v = {r\omega } \), we first need to compute the angular velocity \( \omega \) . The earth makes one revolution in 24 hours, and one revolution is \( {2\pi } \) radians, so \( \omega = \frac{{2\pi }\text{ radians }}{{24}\text{ hours }} = \frac{\pi }{{12}\text{ hours }} \), where, once aga... | Yes |
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