MAT 1033 Test 2 Review (chapters 6,7,11) Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Multipl both sides of each equation b a common denominator to eliminate the fractions. Then solve the sstem. 1) 7 3 + 4 = 4 1) 6-2 = 21 (-6, 8) (6, 8) (6, -8) (-6, -8) Solve the problem b writing and solving a suitable sstem of equations. 2) Carole's car averages 16.0 miles per gallon in cit driving and 23.7 miles per gallon in highwa driving. If she drove a total of 87.2 miles on 29 gallons of gas, how man of the gallons were used for cit driving? 16 gallons 13 gallons 1 gallons 21 gallons 2) Write an augmented matri for the sstem of equations. 3) 4 + 7 + 2z = 4 2 + 7 + z = 76 2 + 9 + z = 86 4 2 7 4 76 7 2 86 9 2 4 2 2 4 7 7 9 76 2 86 4 7 2 2 7 2 9 4 7 2 4 2 7 76 2 9 86 3) Obtain an equivalent sstem b performing the stated elementar operation on the sstem. 4) Replace the third equation b the sum of itself and -1 times the second equation. - 2-7z = 17 - + 9 + 6z = -9 3 + 4 - z = -4-2 - 7z = 17 - + 9 + 6z = -9 8 - - 7z = - 2-7z = 17 - + 9 + 6z = -9-2 + 13 + z = -13-2 - 7z = 17 8 - - 7z = 3 + 4 - z = -4-2 - 7z = 17 - + 9 + 6z = -9-8 + + 7z = 4) 1
) Replace the fourth equation b the sum of itself and 3 times the second equation ) - 2 + z - 6w = 4 3 - z - 4w = - 3-4z + 2w = -3 4-3z - 4w = 8-2 + z - 6w = 4 9-3z - 12w = -1 3-4z + 2w = -3 4-3z - 4w = 8-2 + z - 6w = 4 3 - z - 4w = - 3-4z + 2w = -3 13-6z - 16w = -7-2 + z - 6w = 4 3 - z - 4w = - 3-4z + 2w = -3 - + 6z + 8w = 23-2 + z - 6w = 4 3 - z - 4w = - 3-4z + 2w = -3 9 + 3z - 16w = -7 The reduced row echelon form of the augmented matri of a sstem of equations is given. Find the solutions of the sstem. 6) 6) 1 0 0 0 0 3 0 1 0 0 0 8 0 0 1 0 0 0 0 0 1 0-0 0 0 0 0 1 (3, 8,, -, w) for an real number w (3, 8,, -) (3, 8,, -, 1) No solution Solve the sstem of equations. If the sstem is dependent, epress solutions in terms of the parameter z. 7) 2 + - z = 2-3 + 2z = 1 7-7 + 4z = 7 7) (2,, 7) (1, 0, 0) No solution 7 + z 7, z, z for an real number z 7 8) - + z = 2 + + z = -2 + - z = 0 (1,-2,-1) No solution (1,-1,-2) (-1, 1, -2) 8) 2
Solve the problem b writing and solving a suitable sstem of equations. 9) Julia is preparing a meal b combining three ingredients. One unit of each ingredient provides the following quantities (in grams) of carbohdrates, fat, and protein. 9) Protein(g) Carbohdrates(g) Fat (g) Ingredient A 3 3 1 Ingredient B 2 4 2 Ingredient C 4 1 Ideall the meal should contain 28 grams of protein, 3 grams of carbohdrates, and 11 grams of fat. How man units of each ingredient should Julia use? 4 grams of ingredient A, 2 grams of ingredient B, 3 grams of ingredient C 4 grams of ingredient A, 3 grams of ingredient B, 2 grams of ingredient C 3 grams of ingredient A, 2 grams of ingredient B, 4 grams of ingredient C 2 grams of ingredient A, 4 grams of ingredient B, 3 grams of ingredient C Perform the indicated operation where possible. 9 2 9 10) -6 6 9 4 - - 3-0 1-1 -8-6 2 11 6-3 9-7 4-2 3 2-16 12-3 9-4 -2 3 6 6-6 -7-9 7-6 -14-1 -2 16 6 7 9-7 6 14 1 2-16 10) Perform the indicated operation. 1 11) Let C = -3 and D = 2-4 12-8 -1 3-2. Find C - 3D. 4-6 4 4-12 8-2 6-4 11) Write a matri to displa the information. 12) When the owner of a picture framing store took inventor of her unused frames, she found that she had 40 oak-8", 22 oak-12", 47 oak-18", 6 walnut-8", and 9 walnut-18". Write this information as a 2 3 matri. 40 6 22 0 47 9 40 22 47 6 0 9 40 22 47 6 0 9 40 22 47 9 0 6 12) Given the matrices A and B, find the matri product AB. 13) A = 3-1, B = 0-1 4 0 2 6 Find AB. -4 0 30-2 -2-9 0-4 -9-2 -4 0 0 1 8 0 13) 3
14) A = 0-2 2 3 0 2-2 -2 3 7 0-2 2 3-2 7, B = -1 3 2 0-1 1 Find AB. AB is not defined. 0-6 -8 0-3 3 14) 1) A = 3-2 1 0 4-3 1-10 -6 12-8, B = 0-2 2 Find AB. AB is not defined. 1) 1-6 -10 12-8 1 0 0 8 4
Graph the feasible region for the sstem of inequalities. 16) + 2 2 + 0 16) - - - - - - - - - -
Find the value(s) of the function on the given feasible region. 17) Find the maimum and minimum of z = + 7. 17) 4, 20 0, 28 30, 3-20, -4 Find the value(s) of the function, subject to the sstem of inequalities. 18) Find the maimum and minimum of Z = 16 + 4 subject to: 0 10, 0, 3 + 2 6. 180, 160 180, 12 20, 12 160, 12 18) Use a calculator to estimate the limit. 19) lim 1 4-1 - 1 19) 4 2 Does not eist 0 Use the properties of limits to evaluate the limit if it eists. 20) lim -3 2-2 - 1 + 3 0 Does not eist -8 20) Find the average rate of change for the function over the given interval. 21) = 2 between = 2 and = 8 21) 2 1 3 7-3 10 Solve the problem. 22) Compute the instantaneous rate of change of the function at at = a. f() = -72 -, a = -3. 42 41-39 39 22) 6
Find the point from those given that has the given propert. 23) The point where the slope of the tangent is greatest 10 23) -10 10-10 (0,0) (2,-6) (-2,2) Find the equation of the tangent line to the curve when has the given value. 24) f() = 4 ; = 24) = 13-16 = 20 + 1 = - 4 2 + 8 = -39-80 Find f'() for the function, then find f'() for the given. 2) g() = 3 +, g'(1) g'() = 32 + ; g'(1) = 8 g'() = 3 + ; g'(1) = 8 g'() = 32 + ; g'(1) = 8 g'() = 3; g'(1) = 3 2) Find the derivative. 26) f() = 24 + 3 + 2 43 + 32 43 + 32-7 83 + 12 83 + 12-7 26) 27) f() = 7 ; ( > 0) 6(6 ) - 8 7-8/7 8 7 8/7 1 7-6/7 27) 28) = 2-4 28) ' = 1 + 4 2 ' = 1-4 2 ' = 1 + 4 ' = + 4 2 Find the following. 29) f'(4) if f() = 7-29) - 3 16-11 16 11 16 3 16 7
Find the derivative of the function. 30) f() = (2-3 + 2)(3-2 + ) f'() = 24-643 + 392 + 6-1 f'() = 4-603 + 392 + 6-1 f'() = 4-643 + 392 + 6-1 f'() = 24-603 + 392 + 6-1 30) Use the quotient rule to find the derivative. 3 31) = - 1 31) ' = -2 3 + 32 ( - 1)2 ' = -2 3-32 ( - 1)2 ' = 2 3 + 32 ( - 1)2 ' = 2 3-32 ( - 1)2 32) = 2 + 8 + 3 32) ' = 2 + 8 ' = 3 2 + 8-3 23/2 ' = 3 2 + 8-3 ' = 2 + 8 23/2 Find the derivative of the function. 33) = 4 + 2 2 ' = 4 + 2 ' = 1 4 + 2 ' = 4 4 + 2 ' = 8 4 + 2 33) Find the derivative. 34) = 4e2 8e2 8e 2 8e 42 8e 34) Find the derivative of the function. 3) = ln 92 2 2 + 9 1 2 + 9 2 18 3) Find the derivative. 36) f() = e ln, > 0 e (ln + ) e ( ln + 1) e ln e 36) 37) f() = ln e 37) 1 e 1 - ln e ln e 1 - ln e 8
38) f() = e + ln 38) e + eln e( + 1) e(1 + 1 ) e + ln 9
Answer Ke Testname: 1033FALL2013FINALEXAMREVIEWB 1) C ID: MWA10L 6.1.3-2) B ID: MWA10L 6.1.4-3 3) D ID: MWA10L 6.2.3-3 4) A ID: MWA10L 6.2.1-4 ) C ID: MWA10L 6.2.1-6) D ID: MWA10L 6.2.6-2 7) D ID: MWA10L 6.2.8-6 8) A ID: MWA10L 6.2.8-4 9) A ID: MWA10L 6.2.10-2 10) D ID: MWA10L 6.4.2-7 11) C ID: MWA10L 6.4.3-12) C ID: MWA10L 6.4.4-3 13) B ID: MWA10L 6..2-3 14) A ID: MWA10L 6..2-6 1) B ID: MWA10L 6..2-16) A ID: MWA10L 7.1.2-7 17) B ID: MWA10L 7.2.1-4 18) B ID: MWA10L 7.2.3-4 19) A ID: MWA10L 11.1.2-4 20) D ID: MWA10L 11.1.4-9 21) B ID: MWA10L 11.3.1-3 22) B ID: MWA10L 11.3.4-23) C ID: MWA10L 11.4.2-3 10
Answer Ke Testname: 1033FALL2013FINALEXAMREVIEWB 24) C ID: MWA10L 11.4.1-3 2) A ID: MWA10L 11.4.4-4 26) C ID: MWA10L 11..1-2 27) D ID: MWA10L 11..1-6 28) A ID: MWA10L 11..1-9 29) B ID: MWA10L 11..2-6 30) A ID: MWA10L 11.6.1-3 31) D ID: MWA10L 11.6.2-4 32) B ID: MWA10L 11.6.2-8 33) A ID: MWA10L 11.7.4-2 34) D ID: MWA10L 11.8.1-4 3) C ID: MWA10L 11.8.2-9 36) B ID: MWA10L 11.8.3-2 37) B ID: MWA10L 11.8.3-38) B ID: MWA10L 11.8.3-4 11