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Solve \( y = {\log }_{4}\left( {64}\right) \) without using a calculator. | Solution First we rewrite the logarithm in exponential form: \( {4}^{y} = {64} \) . Next, we ask,\ | No |
Evaluate \( y = {\log }_{3}\left( \frac{1}{27}\right) \) without using a calculator. | Solution First we rewrite the logarithm in exponential form: \( {3}^{y} = \frac{1}{27} \) . Next, we ask,\ | No |
What is the domain of \( f\left( x\right) = {\log }_{2}\left( {x + 3}\right) \) ? | The logarithmic function is defined only when the input is positive, so this function is defined when \( x + 3 > 0 \) .\n\nSolving this inequality,\n\n\( x + 3 > 0\; \) The input must be positive.\n\n\( x > - 3\; \) Subtract 3.\n\nThe domain of \( f\left( x\right) = {\log }_{2}\left( {x + 3}\right) \) is \( \left( {-3,... | Yes |
What is the domain of \( f\left( x\right) = \log \left( {5 - {2x}}\right) \) ? | Solution The logarithmic function is defined only when the input is positive, so this function is defined when \( 5 - {2x} > 0 \) .\n\nSolving this inequality,\n\n\( 5 - {2x} > 0\; \) The input must be positive.\n\n\( - {2x} > - 5\; \) Subtract 5.\n\n\[ x < \frac{5}{2}\;\text{Divide by -2 and switch the inequality.} \]... | Yes |
Graphing a Logarithmic Function with the Form \( f\left( x\right) = {\log }_{b}\left( x\right) \). \( \operatorname{Graph}f\left( x\right) = {\log }_{5}\left( x\right) \). State the domain, range, and asymptote. | Solution Before graphing, identify the behavior and key points for the graph.\n\n- Since \( b = 5 \) is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote \( x = 0 \), and the right tail will increase slowly without bound.\n\n- The \( x \) -intercept is... | Yes |
Sketch the horizontal shift \( f\left( x\right) = {\log }_{3}\left( {x - 2}\right) \) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. | Solution Since the function is \( f\left( x\right) = {\log }_{3}\left( {x - 2}\right) \), we notice \( x + \left( {-2}\right) = x - 2 \) .\n\nThus \( c = - 2 \), so \( c < 0 \). This means we will shift the function \( f\left( x\right) = {\log }_{3}\left( x\right) \) right 2 units.\n\nThe vertical asymptote is \( x = -... | Yes |
Expand \( {\log }_{2}\left( {x}^{5}\right) \) . | The argument is already written as a power, so we identify the exponent,5, and the base, \( x \), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\n\n\[ \n{\log }_{2}\left( {x}^{5}\right) = 5{\log }_{2}\left( x\right) \n\] | Yes |
Expand \( {\log }_{3}\left( {25}\right) \) using the power rule for logs. | Solution Expressing the argument as a power, we get \( {\log }_{3}\left( {25}\right) = {\log }_{3}\left( {5}^{2}\right) \). Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. \[ {\log }_{3}\left( {5}^{2}\right) = 2{\log }... | Yes |
Expand \( \log \left( \sqrt{x}\right) \) . | \[ \log \left( \sqrt{x}\right) = \log {\left( x\right) }^{\frac{1}{2}} \] \[ = \frac{1}{2}\log \left( x\right) \] | Yes |
Expand \( {\log }_{6}\left( \frac{{64}{x}^{3}\left( {{4x} + 1}\right) }{\left( 2x - 1}\right) }\right) \). | Solution We can expand by applying the Product and Quotient Rules.\n\n\[ {\log }_{6}\left( \frac{{64}{x}^{3}\left( {{4x} + 1}\right) }{\left( 2x - 1}\right) }\right) = {\log }_{6}\left( {64}\right) + {\log }_{6}\left( {x}^{3}\right) + {\log }_{6}\left( {{4x} + 1}\right) - {\log }_{6}\left( {{2x} - 1}\right) \]\n\n\[ = ... | Yes |
Write \( {\log }_{3}\left( 5\right) + {\log }_{3}\left( 8\right) - {\log }_{3}\left( 2\right) \) as a single logarithm. | Solution Using the product and quotient rules\n\n\[{\log }_{3}\left( 5\right) + {\log }_{3}\left( 8\right) = {\log }_{3}\left( {5 \cdot 8}\right) = {\log }_{3}\left( {40}\right)\]\n\nThis reduces our original expression to\n\n\[{\log }_{3}\left( {40}\right) - {\log }_{3}\left( 2\right)\]\n\nThen, using the quotient rul... | Yes |
Condense \( {\log }_{2}\left( {x}^{2}\right) + \frac{1}{2}{\log }_{2}\left( {x - 1}\right) - 3{\log }_{2}\left( {\left( x + 3\right) }^{2}\right) \) . | Solution We apply the power rule first:\n\n\[ \n{\log }_{2}\left( {x}^{2}\right) + \frac{1}{2}{\log }_{2}\left( {x - 1}\right) - 3{\log }_{2}\left( {\left( x + 3\right) }^{2}\right) = {\log }_{2}\left( {x}^{2}\right) + {\log }_{2}\left( \sqrt{x - 1}\right) - {\log }_{2}\left( {\left( x + 3\right) }^{6}\right) \n\] \n\n... | Yes |
Rewrite \( 2\log \left( x\right) - 4\log \left( {x + 5}\right) + \frac{1}{x}\log \left( {{3x} + 5}\right) \) as a single logarithm. | Solution We apply the power rule first:\n\n\[ 2\log \left( x\right) - 4\log \left( {x + 5}\right) + \frac{1}{x}\log \left( {{3x} + 5}\right) = \log \left( {x}^{2}\right) - \log \left( {\left( x + 5\right) }^{4}\right) + \log \left( {\left( 3x + 5\right) }^{{x}^{-1}}\right) \]\n\nNext we apply the product rule to the su... | Yes |
Solve \( {2}^{x - 1} = {2}^{{2x} - 4} \) . | \[ \n{2}^{x - 1} = {2}^{{2x} - 4} \n\] \nThe common base is 2 . \n\n\[ \nx - 1 = {2x} - 4 \n\] \nBy the one-to-one property the exponents must be equal. \n\n\[ \nx = 3 \n\] \nSolve for \( x \) . | Yes |
Solve \( {8}^{x + 2} = {16}^{x + 1} \) . | \[ \n{8}^{x + 2} = {16}^{x + 1} \n\] \n\[ \n{\left( {2}^{3}\right) }^{x + 2} = {\left( {2}^{4}\right) }^{x + 1} \n\] \nWrite 8 and 16 as powers of 2 . \n\[ \n{2}^{{3x} + 6} = {2}^{{4x} + 4} \n\] \nTo take a power of a power, multiply exponents . \n\[ \n{3x} + 6 = {4x} + 4 \n\] \nUse the one-to-one property to set the e... | Yes |
Solve \( {3}^{x + 1} = - 2 \) . | This equation has no solution. There is no real value of \( x \) that will make the equation a true statement because any power of a positive number is positive. | Yes |
Solve \( {5}^{x + 2} = {4}^{x} \) . | Solution \( {5}^{x + 2} = {4}^{x}\; \) There is no easy way to get the powers to have the same base . \n\n\( \ln \left( {5}^{x + 2}\right) = \ln \left( {4}^{x}\right) \; \) Take ln of both sides. \n\n\( \left( {x + 2}\right) \ln \left( 5\right) = x\ln \left( 4\right) \; \) Use laws of logs. \n\n\( x\ln \left( 5\right) ... | Yes |
Solve \( {100} = {20}{e}^{2t} \) . | Solution\n\[ \n{100} = {20}{e}^{2t} \n\]\n\n\[ \n5 = {e}^{2t} \n\]\nDivide by the coefficient of the power .\n\n\[ \n\ln \left( 5\right) = {2t} \n\]\nTake ln of both sides. Use the fact that \( \ln \left( x\right) \) and \( {e}^{x} \) are inverse functions.\n\n\( t = \frac{\ln \left( 5\right) }{2} \) Divide by the coef... | Yes |
Solve \( 4{e}^{2x} + 5 = {12} \) . | \[ 4{e}^{2x} + 5 = {12} \] \n\n\( 4{e}^{2x} = 7\; \) Combine like terms. \n\n\( {e}^{2x} = \frac{7}{4}\; \) Divide by the coefficient of the power. \n\n\( {2x} = \ln \left( \frac{7}{4}\right) \; \) Take ln of both sides. \n\n\[ x = \frac{1}{2}\ln \left( \frac{7}{4}\right) \;\text{ Solve for }x. \] | Yes |
Solve \( {\mathrm{e}}^{2x} - {e}^{x} = {56} \) . | \[ {e}^{2x} - {\mathrm{e}}^{x} = {56} \]\n\n\[ {e}^{2x} - {e}^{x} - {56} = 0\;\text{Get one side of the equation equal to zero.} \]\n\n\( \left( {{e}^{x} + 7}\right) \left( {{e}^{x} - 8}\right) = 0\; \) Factor by the FOIL method.\n\n\( {e}^{x} + 7 = 0 \) or \( {e}^{x} - 8 = 0\; \) If a product is zero, then one factor ... | Yes |
Solve \( 2\ln \left( x\right) + 3 = 7 \) . | \[ 2\ln \left( x\right) + 3 = 7 \] \[ 2\ln \left( x\right) = 4 \] Subtract 3. \[ \ln \left( x\right) = 2 \] Divide by 2. \[ x = {e}^{2} \] Rewrite in exponential form. | Yes |
Solve \( 2\ln \left( {6x}\right) = 7 \) | Solution \( 2\ln \left( {6x}\right) = 7 \)\n\n\[ \ln \left( {6x}\right) = \frac{7}{2} \]\nDivide by 2.\n\n\[ {6x} = {e}^{\frac{7}{2}} \]\nUse the definition of \( \ln \).\n\n\[ x = \frac{1}{6}{e}^{\frac{7}{2}} \]\nDivide by 6. | Yes |
Solve \( \ln \left( x\right) = 3 \) . | \[ \ln \left( x\right) = 3 \] \[ x = {e}^{3} \] Use the definition of the natural logarithm. | Yes |
Solve \( \ln \left( {x}^{2}\right) = \ln \left( {{2x} + 3}\right) \) . | \[ \ln \left( {x}^{2}\right) = \ln \left( {{2x} + 3}\right) \] \[ {x}^{2} = {2x} + 3 \] Use the one-to-one property of the logarithm. \[ {x}^{2} - {2x} - 3 = 0 \] Get zero on one side before factoring. \[ \left( {x - 3}\right) \left( {x + 1}\right) = 0 \] Factor using FOIL. \[ x - 3 = 0\text{or}x + 1 = 0 \] If a produc... | Yes |
How long will it take for ten percent of a 1,000-gram sample of uranium-235 to decay? | \[ y = {1000}{e}^{\frac{\ln \left( {0.5}\right) }{{703},{800},{000}}t} \] \[ {900} = {1000}{e}^{\frac{\ln \left( {0.5}\right) }{{703},{800},{000}}t} \] After \( {10}\% \) decays,900 grams are left. \[ {0.9} = {e}^{\frac{\ln \left( {0.5}\right) }{{703},{800},{000}}t} \] Divide by 1000. \[ \ln \left( {0.9}\right) = \ln \... | Yes |
Find an angle \( \beta \) that is coterminal with \( \frac{19\pi }{4} \), where \( 0 \leq \beta < {2\pi } \) . | Solution When working in degrees, we found coterminal angles by adding or subtracting 360 degrees, a full rotation. Likewise, in radians, we can find coterminal angles by adding or subtracting full rotations of \( {2\pi } \) radians:\n\n\[ \n\frac{19\pi }{4} - {2\pi } = \frac{19\pi }{4} - \frac{8\pi }{4} \n\]\n\n\[ \n=... | Yes |
If \( \sin \left( t\right) = \frac{3}{7} \) and \( t \) is in the second quadrant, find \( \cos \left( t\right) \) . | If we drop a vertical line from the point on the unit circle corresponding\nto \( t \), we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See Figure 8.\n\nSubstituting the known value for sine into the Pythagorean Identity,\n\n\[{\cos }^{2}\le... | Yes |
Simplify \( \frac{\sec t}{\tan t} \) . | Solution We can simplify this by rewriting both functions in terms of sine and cosine.\n\n\[ \frac{\sec t}{\tan t} = \frac{\frac{1}{\cos t}}{\frac{\sin t}{\cos t}}\;\text{To divide the functions, we multiply by the reciprocal.} \]\n\n\[ = \frac{1\cos t}{\cos t\sin t}\;\text{Divide out the cosines.} \]\n\n\[ = \frac{1}{... | Yes |
Evaluate the cosecant of \( \frac{5\pi }{7} \) . | For a scientific calculator, enter information as follows:\n\n\[ 1/\left( {5 \times \pi /7}\right) \mathrm{{SIN}} = \]\n\n\[ \csc \left( \frac{5\pi }{7}\right) \approx {1.279} \] | Yes |
Using the triangle shown in Figure 6, evaluate \( \sin \alpha ,\cos \alpha \) , \( \tan \alpha \) , \( \sec \alpha ,\csc \alpha \), and \( \cot \alpha \) . | \[ \sin \alpha = \frac{\text{ opposite }\alpha }{\text{ hypotenuse }} = \frac{4}{5} \]\n\[ \cos \alpha = \frac{\text{ adjacent to }\alpha }{\text{ hypotenuse }} = \frac{3}{5} \]\n\[ \tan \alpha = \frac{\text{ opposite }\alpha }{\text{ adjacent to }\alpha } = \frac{4}{3} \]\n\[ \sec \alpha = \frac{\text{ hypotenuse }}{\... | Yes |
Find the exact value of the trigonometric functions of \( \frac{\pi }{3} \), using side lengths. | \[ \sin \left( \frac{\pi }{3}\right) = \frac{\mathrm{{opp}}}{\mathrm{{hyp}}} = \frac{\sqrt{3}s}{2s} = \frac{\sqrt{3}}{2} \] \[ \cos \left( \frac{\pi }{3}\right) = \frac{\mathrm{{adj}}}{\mathrm{{hyp}}} = \frac{s}{2s} = \frac{1}{2} \] \[ \tan \left( \frac{\pi }{3}\right) = \frac{\mathrm{{opp}}}{\mathrm{{adj}}} = \frac{\s... | Yes |
If \( \sin t = \frac{5}{12} \), find \( \cos \left( {\frac{\pi }{2} - t}\right) \) . | Solution According to the cofunction identities for sine and cosine,\n\n\[ \sin t = \cos \left( {\frac{\pi }{2} - t}\right) \]\n\nSo\n\n\[ \cos \left( {\frac{\pi }{2} - t}\right) = \frac{5}{12} \] | Yes |
Determine the period of the function \( f\left( x\right) = \sin \left( {\frac{\pi }{6}x}\right) \) . | Solution Let’s begin by comparing the equation to the general form \( y = A\sin \left( {Bx}\right) \) .\n\nIn the given equation, \( B = \frac{\pi }{6} \), so the period will be\n\n\[ P = \frac{2\pi }{\left| B\right| } \]\n\n\[ = \frac{2\pi }{\frac{\pi }{6}} \]\n\n\[ = {2\pi } \cdot \frac{6}{\pi } \]\n\n\[ = {12} \] | Yes |
Determine the direction and magnitude of the vertical shift for \( f\left( x\right) = \cos \left( x\right) - 3 \) . | Solution Let’s begin by comparing the equation to the general form \( y = A\cos \left( {{Bx} - C}\right) + D \) . In the given equation, \( D = - 3 \) so the shift is 3 units downward. | Yes |
Determine the midline, amplitude, period, and phase shift of the function \( y = 3\sin \left( {2x}\right) + 1 \) . | Solution Let’s begin by comparing the equation to the general form \( y = A\sin \left( {{Bx} - C}\right) + D \) .\n\n\( A = 3 \), so the amplitude is \( \left| A\right| = 3 \) .\n\nNext, \( B = 2 \), so the period is \( P = \frac{2\pi }{\left| B\right| } = \frac{2\pi }{2} = \pi \) .\n\nThere is no added constant inside... | Yes |
Sketch a graph of \( f\left( x\right) = - 2\sin \left( \frac{\pi x}{2}\right) \) . | Solution Let’s begin by comparing the equation to the form \( y = A\sin \left( {Bx}\right) \). Step 1. We can see from the equation that \( A = - 2 \), so the amplitude is 2 . \[ \left| A\right| = 2 \] Step 2. The equation shows that \( B = \frac{\pi }{2} \), so the period is \[ P = \frac{2\pi }{\frac{\pi }{2}} \] \[ =... | Yes |
Sketch a graph of \( f\left( x\right) = 3\sin \left( {\frac{\pi }{4}x - \frac{\pi }{4}}\right) \) . | Solution\n\nStep 1. The function is already written in general form: \( f\left( x\right) = 3\sin \left( {\frac{\pi }{4}x - \frac{\pi }{4}}\right) \) . This graph will have the shape of a sine function, starting at the midline and increasing to the right.\n\nStep 2. \( \left| A\right| = \left| 3\right| = 3 \) . The ampl... | Yes |
Given \( y = - 2\cos \left( {\frac{\pi }{2}x + \pi }\right) + 3 \), determine the amplitude, period, phase shift, and horizontal shift. Then graph the function. | Solution Begin by comparing the equation to the general form and use the steps outlined in Example 9.\n\n\[ y = A\cos \left( {{Bx} - C}\right) + D \]\n\nStep 1. The function is already written in general form.\n\nStep 2. Since \( A = - 2 \), the amplitude is \( \left| A\right| = 2 \) .\n\nStep 3. \( \left| B\right| = \... | Yes |
Determining a Rider's Height on a Ferris Wheel\n\nThe London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. | Solution With a diameter of \( {135}\mathrm{\;m} \), the wheel has a radius of \( {67.5}\mathrm{\;m} \) . The height will oscillate with amplitude \( {67.5}\mathrm{\;m} \) above and below the center.\n\nPassengers board \( 2\mathrm{\;m} \) above ground level, so the center of the wheel must be located \( {67.5} + 2 = {... | Yes |
Find a formula for the function graphed in Figure 4. | Solution The graph has the shape of a tangent function.\n\nStep 1. One cycle extends from -4 to 4, so the period is \( P = 8 \) . Since \( P = \frac{\pi }{\left| B\right| } \), we have \( B = \frac{\pi }{P} = \frac{\pi }{8} \) .\n\nStep 2. The equation must have the form \( f\left( x\right) = A\tan \left( {\frac{\pi }{... | Yes |
Graph one period of \( f\left( x\right) = {2.5}\sec \left( {0.4x}\right) \) . | Solution\n\nStep 1. The given function is already written in the general form, \( y = A\sec \left( {Bx}\right) \) .\n\nStep 2. \( A = {2.5} \) so the stretching factor is 2.5 .\n\nStep 3. \( B = {0.4} \) so \( P = \frac{2\pi }{0.4} = {5\pi } \) . The period is \( {5\pi } \) units.\n\nStep 4. Sketch the graph of the fun... | Yes |
Graph one period of \( f\left( x\right) = - 3\csc \left( {4x}\right) \) . | Solution\n\nStep 1. The given function is already written in the general form, \( y = A\csc \left( {Bx}\right) \) .\n\nStep 2. \( \left| A\right| = \left| {-3}\right| = 3 \), so the stretching factor is 3 .\n\nStep 3. \( B = 4 \), so \( P = \frac{2\pi }{4} = \frac{\pi }{2} \) . The period is \( \frac{\pi }{2} \) units.... | Yes |
Sketch a graph of \( y = 2\csc \left( {\frac{\pi }{2}x}\right) + 1 \) . What are the domain and range of this function? | Solution\n\nStep 1. Express the function given in the form \( y = 2\csc \left( {\frac{\pi }{2}x}\right) + 1 \) .\n\nStep 2. Identify the stretching/compressing factor, \( \left| A\right| = 2 \) .\n\nStep 3. The period is \( \frac{2\pi }{\left| B\right| } = \frac{2\pi }{\frac{\pi }{2}} = \frac{2\pi }{1} \cdot \frac{2}{\... | Yes |
Evaluate each of the following.\na. \( {\sin }^{-1}\left( \frac{1}{2}\right) \) b. \( {\sin }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) \) c. \( {\cos }^{-1}\left( {-\frac{\sqrt{3}}{2}}\right) \) d. \( {\tan }^{-1}\left( 1\right) \) | ## Solution\na. Evaluating \( {\sin }^{-1}\left( \frac{1}{2}\right) \) is the same as determining the angle that would have a sine value of \( \frac{1}{2} \) . In other words, what angle \( x \) would satisfy \( \sin \left( x\right) = \frac{1}{2} \) ? There are multiple values that would satisfy this relationship, such... | Yes |
Solve the triangle in Figure 8 for the angle \( \theta \) . | Solution Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function. \[ \cos \theta = \frac{9}{12} \] \[ \theta = {\cos }^{-1}\left( \frac{9}{12}\right) \] Apply definition of the inverse. \( \theta \approx {0.7227} \) or about \( {41.4096}^{ \circ }\; \) Evaluat... | Yes |
Evaluate \( {\sin }^{-1}\left( {\cos \left( \frac{13\pi }{6}\right) }\right) \) | a. Here, we can directly evaluate the inside of the composition.\n\n\[ \cos \left( \frac{13\pi }{6}\right) = \cos \left( {\frac{\pi }{6} + {2\pi }}\right) \]\n\n\[ = \cos \left( \frac{\pi }{6}\right) \]\n\n\[ = \frac{\sqrt{3}}{2} \]\n\nNow, we can evaluate the inverse function as we did earlier.\n\n\[ {\sin }^{-1}\left... | Yes |
Find a simplified expression for \( \cos \left( {{\sin }^{-1}\left( \frac{x}{3}\right) }\right) \) for \( - 3 \leq x \leq 3 \) . | Solution We know there is an angle \( \theta \) such that \( \sin \theta = \frac{x}{3} \). \n\n\( {\sin }^{2}\theta + {\cos }^{2}\theta = 1\; \) Use the Pythagorean Theorem. \n\n\[ \begin{aligned} {\left( \frac{x}{3}\right) }^{2} + {\cos }^{2}\theta & = 1 & & \text{ Solve for cosine. } \\ {\cos }^{2}\theta & = 1 - \fra... | Yes |
Verify the following equivalency using the even-odd identities:\n\n\[ \left( {1 + \sin x}\right) \left\lbrack {1 + \sin \left( {-x}\right) }\right\rbrack = {\cos }^{2}x \] | Solution Working on the left side of the equation, we have\n\n\[ \left( {1 + \sin x}\right) \left\lbrack {1 + \sin \left( {-x}\right) }\right\rbrack = \left( {1 + \sin x}\right) \left( {1 - \sin x}\right) \;\text{ Since }\sin \left( {-x}\right) = - \sin x \]\n\n\[ = 1 - {\sin }^{2}x \] Difference of squares\n\n\[ = {\c... | Yes |
Verify the identity \( \frac{{\sec }^{2}\theta - 1}{{\sec }^{2}\theta } = {\sin }^{2}\theta \) | Solution As the left side is more complicated, let's begin there.\n\n\[ \frac{{\sec }^{2}\theta - 1}{{\sec }^{2}\theta } = \frac{\left( {{\tan }^{2}\theta + 1}\right) - 1}{{\sec }^{2}\theta }\;{\sec }^{2}\theta = {\tan }^{2}\theta + 1 \]\n\n\[ = \frac{{\tan }^{2}\theta }{{\sec }^{2}\theta } \]\n\n\[ = {\tan }^{2}\theta... | Yes |
Verify the identity:\n\[ \frac{{\sin }^{2}\left( {-\theta }\right) - {\cos }^{2}\left( {-\theta }\right) }{\sin \left( {-\theta }\right) - \cos \left( {-\theta }\right) } = \cos \theta - \sin \theta \] | Solution Let's start with the left side and simplify:\n\n\[ \frac{{\sin }^{2}\left( {-\theta }\right) - {\cos }^{2}\left( {-\theta }\right) }{\sin \left( {-\theta }\right) - \cos \left( {-\theta }\right) } = \frac{{\left\lbrack \sin \left( -\theta \right) \right\rbrack }^{2} - {\left\lbrack \cos \left( -\theta \right) ... | Yes |
Using the formula for the cosine of the difference of two angles, find the exact value of \( \cos \left( {\frac{5\pi }{4} - \frac{\pi }{6}}\right) \) . | Solution Use the formula for the cosine of the difference of two angles. We have\n\n\[ \cos \left( {\alpha - \beta }\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]\n\n\[ \cos \left( {\frac{5\pi }{4} - \frac{\pi }{6}}\right) = \cos \left( \frac{5\pi }{4}\right) \cos \left( \frac{\pi }{6}\right) + \sin \left... | Yes |
Find the exact value of \( \cos \left( {75}^{ \circ }\right) \) . | As \( {75}^{ \circ } = {45}^{ \circ } + {30}^{ \circ } \), we can evaluate \( \cos \left( {75}^{ \circ }\right) \) as \( \cos \left( {{45}^{ \circ } + {30}^{ \circ }}\right) \) . Thus,\n\n\[ \cos \left( {{45}^{ \circ } + {30}^{ \circ }}\right) = \cos \left( {45}^{ \circ }\right) \cos \left( {30}^{ \circ }\right) - \sin... | Yes |
Find the exact value of \( \tan \left( {\frac{\pi }{6} + \frac{\pi }{4}}\right) \). | Solution Let’s first write the sum formula for tangent and substitute the given angles into the formula.\n\n\[ \tan \left( {\alpha + \beta }\right) = \frac{\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta } \]\n\n\[ \tan \left( {\frac{\pi }{6} + \frac{\pi }{4}}\right) = \frac{\tan \left( \frac{\pi }{6}\right) + \t... | Yes |
Write tan \( \frac{\pi }{9} \) in terms of its cofunction. | Solution The cofunction of \( \tan \theta = \cot \left( {\frac{\pi }{2} - \theta }\right) \) . Thus,\n\n\[ \tan \left( \frac{\pi }{9}\right) = \cot \left( {\frac{\pi }{2} - \frac{\pi }{9}}\right) \]\n\n\[ = \cot \left( {\frac{9\pi }{18} - \frac{2\pi }{18}}\right) \]\n\n\[ = \cot \left( \frac{7\pi }{18}\right) \] | Yes |
Let \( {L}_{1} \) and \( {L}_{2} \) denote two non-vertical intersecting lines, and let \( \theta \) denote the acute angle between \( {L}_{1} \) and \( {L}_{2} \) . Show that \[ \tan \theta = \frac{{m}_{2} - {m}_{1}}{1 + {m}_{1}{m}_{2}} \] where \( {m}_{1} \) and \( {m}_{2} \) are the slopes of \( {L}_{1} \) and \( {L... | Solution Using the difference formula for tangent, this problem does not seem as daunting as it might. \[ \tan \theta = \tan \left( {{\theta }_{2} - {\theta }_{1}}\right) \] \[ = \frac{\tan {\theta }_{2} - \tan {\theta }_{1}}{1 + \tan {\theta }_{1}\tan {\theta }_{2}} \] \[ = \frac{{m}_{2} - {m}_{1}}{1 + {m}_{1}{m}_{2}}... | Yes |
Use the double-angle formula for cosine to write \( \cos \left( {6x}\right) \) in terms of \( \cos \left( {3x}\right) \) . | \[ \cos \left( {6x}\right) = \cos \left( {2\left( {3x}\right) }\right) \]\n\[ = {\cos }^{2}{3x} - {\sin }^{2}{3x} \]\n\[ = 2{\cos }^{2}{3x} - 1 \] | Yes |
Verify the identity:\n\[ \tan \left( {2\theta }\right) = \frac{2}{\cot \theta - \tan \theta } \] | Solution In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.\n\n\[ \tan \left( {2\theta }\right) = \frac{2\tan \theta }{1 - {\tan }^{2}\theta } \]\nDouble-angle formula\n\n\[ = \frac{2\tan \theta \left( \frac{1}{\tan \theta }\right) }{\le... | Yes |
Find \( \sin \left( {15}^{ \circ }\right) \) using a half-angle formula. | Solution Since \( {15}^{ \circ } = \frac{{30}^{ \circ }}{2} \), we use the half-angle formula for sine:\n\n\[ \sin \frac{{30}^{ \circ }}{2} = \sqrt{\frac{1 - \cos {30}^{ \circ }}{2}} \]\n\n\[ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \]\n\n\[ = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} \]\n\n\[ = \sqrt{\frac{2 - \sqrt{3}... | Yes |
Write the following product of cosines as a sum: \( 2\cos \left( \frac{7x}{2}\right) \cos \frac{3x}{2} \) . | Solution We begin by writing the formula for the product of cosines:\n\n\[ \cos \alpha \cos \beta = \frac{1}{2}\left\lbrack {\cos \left( {\alpha - \beta }\right) + \cos \left( {\alpha + \beta }\right) }\right\rbrack \]\n\nWe can then substitute the given angles into the formula and simplify.\n\n\[ 2\cos \left( \frac{7x... | Yes |
Write the following difference of sines expression as a product: \( \sin \left( {4\theta }\right) - \sin \left( {2\theta }\right) \) . | Solution We begin by writing the formula for the difference of sines.\n\n\[ \sin \alpha - \sin \beta = 2\sin \left( \frac{\alpha - \beta }{2}\right) \cos \left( \frac{\alpha + \beta }{2}\right) \]\n\nSubstitute the values into the formula, and simplify.\n\n\[ \sin \left( {4\theta }\right) - \sin \left( {2\theta }\right... | Yes |
Evaluate \( \cos \left( {15}^{ \circ }\right) - \cos \left( {75}^{ \circ }\right) \). | Solution We begin by writing the formula for the difference of cosines.\n\n\[ \cos \alpha - \cos \beta = - 2\sin \left( \frac{\alpha + \beta }{2}\right) \sin \left( \frac{\alpha - \beta }{2}\right) \]\n\nThen we substitute the given angles and simplify.\n\n\[ \cos \left( {15}^{ \circ }\right) - \cos \left( {75}^{ \circ... | Yes |
Identify all exact solutions to the equation \( 2\left( {\tan x + 3}\right) = 5 + \tan x,0 \leq x < {2\pi } \) . | Solution We can solve this equation using only algebra. Isolate the expression \( \tan x \) on the left side of the equals sign.\n\n\[ 2\left( {\tan x}\right) + 2\left( 3\right) = 5 + \tan x \]\n\n\[ 2\tan x + 6 = 5 + \tan x \]\n\n\[ 2\tan x - \tan x = 5 - 6 \]\n\n\[ \tan x = - 1 \]\n\nThere are two angles on the unit ... | Yes |
Solve the equation exactly: \( 2{\sin }^{2}\theta - 5\sin \theta + 3 = 0,0 \leq \theta \leq {2\pi } \) . | Solution Using grouping, this quadratic can be factored. Either make the real substitution, \( \sin \theta = u \), or imagine it, as we factor: \[ 2{\sin }^{2}\theta - 5\sin \theta + 3 = 0 \] \[ \left( {2\sin \theta - 3}\right) \left( {\sin \theta - 1}\right) = 0 \] Now set each factor equal to zero. \[ 2\sin \theta - ... | Yes |
Solve exactly:\n\n\[ 2{\sin }^{2}\theta + \sin \theta = 0;0 \leq \theta < {2\pi } \] | Solution This problem should appear familiar as it is similar to a quadratic. Let \( \sin \theta = x \) . The equation becomes \( 2{x}^{2} + x = 0 \) . We begin by factoring:\n\n\[ 2{x}^{2} + x = 0 \]\n\n\[ x\left( {{2x} + 1}\right) = 0 \]\n\nSet each factor equal to zero.\n\n\[ x = 0 \]\n\n\[ \left( {{2x} + 1}\right) ... | Yes |
Solve exactly: \( \cos \left( {2x}\right) = \frac{1}{2} \) on \( \lbrack 0,{2\pi }) \) . | Solution We can see that this equation is the standard equation with a multiple of an angle. If \( \cos \left( \alpha \right) = \frac{1}{2} \), we know \( \alpha \) is in quadrants I and IV. While \( \theta = {\cos }^{-1}\frac{1}{2} \) will only yield solutions in quadrants I and II, we recognize that the solutions to ... | Yes |
Graph the function \( y = - 4\cos \left( {\pi x}\right) \) using amplitude, period, and key points. | Solution The amplitude is \( \left| {-4}\right| = 4 \) . The period is \( \frac{2\pi }{\omega } = \frac{2\pi }{\pi } = 2 \) . (Recall that we sometimes refer to \( B \) as \( \omega \) .) One cycle of the graph can be drawn over the interval \( \left\lbrack {0,2}\right\rbrack \) . To find the key points, we divide the ... | Yes |
Example 8 Finding the Displacement, Period, and Frequency, and Graphing a Function\n\nFor the given functions,\n\n1. Find the maximum displacement of an object.\n\n2. Find the period or the time required for one vibration.\n\n3. Find the frequency.\n\n4. Sketch the graph. | Solution\n\na. \( y = 5\sin \left( {3t}\right) \)\n\n\n\nFigure 12 \n\nFigure 13\n\n1. The maximum displacement ... | Yes |
In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth. | Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle \( \gamma = {85}^{ \circ } \), and its corresponding side \( c = {12} \), and we know side \( b = 9 \) . We will use this proportion to solve for \( \beta \) .\n\n\[ \n\frac{\sin \left( {... | Yes |
Plot the point \( \left( {-2,\frac{\pi }{6}}\right) \) on the polar grid. | Solution We know that \( \frac{\pi }{6} \) is located in the first quadrant. However, \( r = - 2 \) . We can approach plotting a point with a negative \( r \) in two ways:\n\n1. Plot the point \( \left( {2,\frac{\pi }{6}}\right) \) by moving \( \frac{\pi }{6} \) in the counterclockwise direction and extending a directe... | Yes |
Write the polar coordinates \( \\left( {-2,0}\\right) \) as rectangular coordinates. | Solution See Figure 7. Writing the polar coordinates as rectangular, we have\n\n\[ x = r\\cos \\theta \]\n\n\[ x = - 2\\cos \\left( 0\\right) = - 2 \]\n\n\[ y = r\\sin \\theta \]\n\n\[ y = - 2\\sin \\left( 0\\right) = 0 \]\n\nThe rectangular coordinates are also \( \\left( {-2,0}\\right) \) . | Yes |
Using the equation in Example 1, find the zeros and maximum \( \left| r\right| \) and, if necessary, the polar axis intercepts of \( r = 2\sin \theta \) . | Solution To find the zeros, set \( r \) equal to zero and solve for \( \theta \).\n\n\[ 2\sin \theta = 0 \]\n\n\[ \sin \theta = 0 \]\n\n\[ \theta = {\sin }^{-1}0 \]\n\n\[ \theta = {n\pi }\;\text{where}n\text{is an integer} \]\n\nSubstitute any one of the \( \theta \) values into the equation. We will use 0 .\n\n\[ r = ... | Yes |
Sketch the graph of \( {r}^{2} = 4\cos {2\theta } \) . | The equation exhibits symmetry with respect to the line \( \theta = \frac{\pi }{2} \), the polar axis, and the pole.\n\nLet's find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution \( u = {2\theta } \).\n\n\[ 0 = 4\cos {2\theta } \]\n\[ 0 = 4\cos ... | Yes |
Sketch the graph of \( r = 2\sin \left( {5\theta }\right) \) . | The graph of the equation shows symmetry with respect to the line \( \theta = \frac{\pi }{2} \) . Next, find the zeros and maximum.\n\nWe will want to make the substitution \( u = {5\theta } \).\n\n\[ 0 = 2\sin \left( {5\theta }\right) \]\n\n\[ 0 = \sin u \]\n\n\[ {\sin }^{-1}0 = 0 \]\n\n\[ u = 0 \]\n\n\[ {5\theta } = ... | Yes |
Plot the complex number \( 2 - {3i} \) in the complex plane. | Solution From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See Figure 1. | Yes |
Find the absolute value of \( z = \sqrt{5} - i \) . | Solution Using the formula, we have\n\n\[ \left| z\right| = \sqrt{{x}^{2} + {y}^{2}} \]\n\n\[ \left| z\right| = \sqrt{\sqrt{{5}^{2}} + {\left( -1\right) }^{2}} \]\n\n\[ \left| z\right| = \sqrt{5 + 1} \]\n\n\[ \left| z\right| = \sqrt{6} \] | Yes |
Given \( z = 3 - {4i} \), find \( \left| z\right| \) . | Solution Using the formula, we have\n\n\[ \left| z\right| = \sqrt{{x}^{2} + {y}^{2}} \]\n\n\[ \left| z\right| = \sqrt{{\left( 3\right) }^{2} + {\left( -4\right) }^{2}} \]\n\n\[ \left| z\right| = \sqrt{9 + {16}} \]\n\n\[ \left| z\right| = \sqrt{25} \]\n\n\[ \left| z\right| = 5 \] | Yes |
Find the rectangular form of the complex number given \( r = {13} \) and \( \tan \theta = \frac{5}{12} \) . | Solution If \( \tan \theta = \frac{5}{12} \), and \( \tan \theta = \frac{y}{x} \), we first determine \( r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{12}^{2} + {5}^{2}} = {13} \) . We then find \( \cos \theta = \frac{x}{r} \) and \( \sin \theta = \frac{y}{r} \) .\n\[ z = {13}\left( {\cos \theta + i\sin \theta }\right) \]\n\[ ... | Yes |
Find the product of \( {z}_{1}{z}_{2} \), given \( {z}_{1} = 4\left( {\cos \left( {80}^{ \circ }\right) + i\sin \left( {80}^{ \circ }\right) }\right) \) and \( {z}_{2} = 2\left( {\cos \left( {145}^{ \circ }\right) + i\sin \left( {145}^{ \circ }\right) }\right) \) . | Solution Follow the formula\n\n\[ \n{z}_{1}{z}_{2} = 4 \cdot 2\left\lbrack {\cos \left( {{80}^{ \circ } + {145}^{ \circ }}\right) + i\sin \left( {{80}^{ \circ } + {145}^{ \circ }}\right) }\right\rbrack \n\]\n\n\[ \n{z}_{1}{z}_{2} = 8\left\lbrack {\cos \left( {225}^{ \circ }\right) + i\sin \left( {225}^{ \circ }\right) ... | Yes |
Find the quotient of \( {z}_{1} = 2\left( {\cos \left( {213}^{ \circ }\right) + i\sin \left( {213}^{ \circ }\right) }\right) \) and \( {z}_{2} = 4\left( {\cos \left( {33}^{ \circ }\right) + i\sin \left( {33}^{ \circ }\right) }\right) \) . | Solution Using the formula, we have\n\n\[ \frac{{z}_{1}}{{z}_{2}} = \frac{2}{4}\left\lbrack {\cos \left( {{213}^{ \circ } - {33}^{ \circ }}\right) + \mathrm{i}\sin \left( {{213}^{ \circ } - {33}^{ \circ }}\right) }\right\rbrack \]\n\n\[ \frac{{z}_{1}}{{z}_{2}} = \frac{1}{2}\left\lbrack {\cos \left( {180}^{ \circ }\righ... | Yes |
Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. | Solution\n\nMethod 1. First, let’s solve the \( x \) equation for \( t \) . Then we can substitute the result into the \( y \) equation.\n\n\[ x = {3t} - 2 \]\n\n\[ x + 2 = {3t} \]\n\n\[ \frac{x + 2}{3} = t \]\n\nNow substitute the expression for \( t \) into the \( y \) equation.\n\n\[ y = t + 1 \]\n\n\[ y = \left( \f... | Yes |
Sketch the graph of the parametric equations \( x\left( t\right) = {t}^{2} + 1, y\left( t\right) = 2 + t \) . | Solution Construct a table of values for \( t, x\left( t\right) \), and \( y\left( t\right) \), as in Table 1, and plot the points in a plane.\n\n<table><thead><tr><th>\( t \)</th><th>\( x\left( t\right) = {t}^{2} + 1 \)</th><th>\( y\left( t\right) = 2 + t \)</th></tr></thead><tr><td>\( - 5 \)</td><td>26</td><td>\( - 3... | Yes |
Construct a table of values for the given parametric equations and sketch the graph:\n\n\[ x = 2\cos t \]\n\n\[ y = 4\sin t \] | Solution Construct a table like that in Table 2 using angle measure in radians as inputs for \( t \), and evaluating \( x \) and \( y \) . Using angles with known sine and cosine values for t makes calculations easier.\n\n<table><thead><tr><th>\( t \)</th><th>\( x = 2\cos t \)</th><th>\( y = 4\sin t \)</th></tr></thead... | Yes |
Given vector \( v = \langle 3,1\rangle \), find \( {3v},\frac{1}{2}v \), and \( - v \) . | Solution See Figure 11 for a geometric interpretation. If \( v = \langle 3,1\rangle \), then\n\n\[ {3v} = \langle 3 \cdot 3,3 \cdot 1\rangle \]\n\n\[ = \langle 9,3\rangle \]\n\n\[ \frac{1}{2}v = \left\langle {\frac{1}{2} \cdot 3,\frac{1}{2} \cdot 1}\right\rangle \]\n\n\[ = \left\langle {\frac{3}{2},\frac{1}{2}}\right\r... | Yes |
Given \( u = \langle 3, - 2\rangle \) and \( v = \langle - 1,4\rangle \), find a new vector \( w = {3u} + {2v} \) . | Solution First, we must multiply each vector by the scalar.\n\n\[ \n{3u} = 3\langle 3, - 2\rangle \n\]\n\n\[ \n= \langle 9, - 6\rangle \n\]\n\n\[ \n{2v} = 2\langle - 1,4\rangle \n\]\n\n\[ \n= \langle - 2,8\rangle \n\]\n\nThen, add the two together.\n\n\[ \nw = {3u} + {2v} \n\]\n\n\[ \n= \langle 9, - 6\rangle + \langle ... | Yes |
Determine whether the ordered pair \( \left( {5,1}\right) \) is a solution to the given system of equations. | Solution Substitute the ordered pair \( \left( {5,1}\right) \) into both equations.\n\n\[ \left( 5\right) + 3\left( 1\right) = 8 \]\n\n\[ 8 = 8\;\text{True} \]\n\n\[ 2\left( 5\right) - 9 = \left( 1\right) \]\n\n\[ 1 = 1 \]\n\nThe ordered pair \( \left( {5,1}\right) \) satisfies both equations, so it is the solution to ... | Yes |
Solve the given system of equations in two variables by addition.\n\n\[{2x} + {3y} = - {16}\]\n\n\[{5x} - {10y} = {30}\] | Solution One equation has \( {2x} \) and the other has \( {5x} \) . The least common multiple is \( {10x} \) so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate \( x \) by multiplying the first equation by -5 and the second equation by 2 .\n\n\[- 5\left( {{2x} + ... | Yes |
Solve the given system of equations in two variables by addition. | Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.\n\n\[6\left( {\frac{x}{3} + \frac{y}{6}}\right) = 6\left( 3\right) \]\n\n\[{2x} + y = {18}\]\n\n\[4\left( {\frac{x}{2} - \frac{y}{4}}\right) = 4\left( 1\right) \]\n\n\[{2x} - y = 4\]\n\nNow multipl... | Yes |
The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? | Solution Let \( c = \) the number of children and \( a = \) the number of adults in attendance.\n\nThe total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day.\n\n\[ c + a = 2,000 \]\n\nThe revenue from all children can be found by multiplying $25.00 by the ... | Yes |
In the problem posed at the beginning of the section, John invested his inheritance of \$12,000 in three different funds: part in a money-market fund paying \( 3\% \) interest annually; part in municipal bonds paying \( 4\% \) annually; and the rest in mutual funds paying 7% annually. John invested \$4,000 more in mutu... | Solution To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:\n\n\( x = \) amount invested in money-market fund\n\n\[ y = \text{amount invested in municipal bonds} \]\n\n\( z = \) amount invested in mutual funds\n\nTh... | Yes |
Solve the system of equations.\n\n\\[ \nx - y = - 1 \n\\]\n\n\\[ \ny = {x}^{2} + 1 \n\\] | Solution Solve the first equation for \( x \) and then substitute the resulting expression into the second equation.\n\n\\[ \nx - y = - 1 \n\\]\n\n\\[ \nx = y - 1\\;\\text{ Solve for }x. \n\\]\n\n\\[ \ny = {x}^{2} + 1 \n\\]\n\n\\[ \ny = {\\left( y - 1\\right) }^{2} + 1\\; \\text{ Substitute expression for } x. \n\\]\n\... | Yes |
Solve the system of nonlinear equations. | Solution Let's begin by multiplying equation (1) by -3, and adding it to equation (2).\n\n\[ \left( {-3}\right) \left( {{x}^{2} + {y}^{2}}\right) = \left( {-3}\right) \left( {26}\right) \]\n\n\[ - 3{x}^{2} - 3{y}^{2} = - {78} \]\n\n\[ \begin{aligned} 3{x}^{2} + {25}{y}^{2} & = {100} \\ {22}{y}^{2} & = {22} \end{aligned... | Yes |
Find a partial fraction decomposition of the given expression. | \[ \frac{8{x}^{2} + {12x} - {20}}{\left( {x + 3}\right) \left( {{x}^{2} + x + 2}\right) } = \frac{A}{\left( x + 3\right) } + \frac{{Bx} + C}{\left( {x}^{2} + x + 2}\right) } \]\n\nWe follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denomin... | Yes |
Given matrix \( A \) :\n\na. What are the dimensions of matrix \( A \) ?\n\nb. What are the entries at \( {a}_{31} \) and \( {a}_{22} \) ?\n\n\[ A = \left\lbrack \begin{array}{lll} 2 & 1 & 0 \\ 2 & 4 & 7 \\ 3 & 1 & - 2 \end{array}\right\rbrack \] | Solution\n\na. The dimensions are \( 3 \times 3 \) because there are three rows and three columns.\n\nb. Entry \( {a}_{31} \) is the number at row 3, column 1, which is 3 . The entry \( {a}_{22} \) is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column. | Yes |
Find the sum of \( A \) and \( B \) . | \[ A = \left\lbrack \begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right\rbrack \;\text{ and }B = \left\lbrack \begin{array}{ll} 5 & 9 \\ 0 & 7 \end{array}\right\rbrack \] \n\nSolution Add corresponding entries. Add the entry in row 1, column 1, \( {a}_{11} \), of matrix \( A \) to the entry in row 1, column \( 1,{b}_{11... | Yes |
Find the difference of \( A \) and \( B \) . | \[ A = \left\lbrack \begin{array}{rr} - 2 & 3 \\ 0 & 1 \end{array}\right\rbrack \;\text{ and }B = \left\lbrack \begin{array}{ll} 8 & 1 \\ 5 & 4 \end{array}\right\rbrack \] Solution We subtract the corresponding entries of each matrix. \[ A - B = \left\lbrack \begin{array}{rr} - 2 & 3 \\ 0 & 1 \end{array}\right\rbrack -... | Yes |
Example 5 Finding the Sum and Difference of Two \( 3 \times 3 \) Matrices\n\nGiven \( A \) and \( B \) :\n\na. Find the sum.\n\nb. Find the difference.\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & - {10} & - 2 \\ {14} & {12} & {10} \\ 4 & - 2 & 2 \end{array}\right\rbrack \text{ and }B = \left\lbrack \begin{array}{rrr}... | Solution\n\na. Add the corresponding entries.\n\n\[ A + B = \left\lbrack \begin{array}{rrr} 2 & - {10} & - 2 \\ {14} & {12} & {10} \\ 4 & - 2 & 2 \end{array}\right\rbrack + \left\lbrack \begin{array}{rrr} 6 & {10} & - 2 \\ 0 & - {12} & - 4 \\ - 5 & 2 & - 2 \end{array}\right\rbrack \]\n\n\[ = \left\lbrack \begin{matrix}... | Yes |
Multiply matrix \( A \) by the scalar 3 . | Solution Multiply each entry in \( A \) by the scalar 3 .\n\n\[ {3A} = 3\left\lbrack \begin{array}{ll} 8 & 1 \\ 5 & 4 \end{array}\right\rbrack \]\n\n\[ = \left\lbrack \begin{array}{ll} 3 \cdot 8 & 3 \cdot 1 \\ 3 \cdot 5 & 3 \cdot 4 \end{array}\right\rbrack \]\n\n\[ = \left\lbrack \begin{array}{rr} {24} & 3 \\ {15} & {1... | Yes |
Find the sum \( {3A} + {2B} \) . | \[ A = \left\lbrack \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & - 1 & 2 \\ 4 & 3 & - 6 \end{array}\right\rbrack \;\text{ and }B = \left\lbrack \begin{array}{rrr} - 1 & 2 & 1 \\ 0 & - 3 & 2 \\ 0 & 1 & - 4 \end{array}\right\rbrack \] Solution First, find \( {3A} \), then \( {2B} \). \[ {3A} = \left\lbrack \begin{array}{lll} 3 ... | Yes |
Example 9 Multiplying Two Matrices\n\nGiven \( A \) and \( B \) :\n\na. Find \( {AB} \) . b. Find \( {BA} \) .\n\n\[ A = \left\lbrack \begin{array}{rrr} - 1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right\rbrack \;\text{ and }B = \left\lbrack \begin{array}{rr} 5 & - 1 \\ - 4 & 0 \\ 2 & 3 \end{array}\right\rbrack \] | ## Solution\n\na. As the dimensions of \( A \) are \( 2 \times 3 \) and the dimensions of \( B \) are \( 3 \times 2 \), these matrices can be multiplied together because the number of columns in \( A \) matches the number of rows in \( B \) . The resulting product will be a \( 2 \times 2 \) matrix, the number of rows i... | Yes |
Find the system of equations from the augmented matrix. | \[ \left\lbrack \begin{array}{rrrr} 1 & - 3 & - 5 & - 2 \\ 2 & - 5 & - 4 & 5 \\ - 3 & 5 & 4 & 6 \end{array}\right\rbrack \rightarrow \begin{aligned} x - {3y} - {5z} & = - 2 \\ {2x} - {5y} - {4z} & = 5 \\ - {3x} + {5y} + {4z} & = 6 \end{aligned} \] | Yes |
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