Chapter : Linear Equations and Functions Eercise.. 7 8+ 7+ 7 8 8+ + 7 8. + 8 8( + ) + 8 8+ 8 8 8 8 7 0 0. 8( ) ( ) 8 8 8 8 + 0 8 8 0 7. ( ) 8. 8. ( 7) ( + ) + + +. 7 7 7 7 7 7 ( ). 8 8 0 0 7. + + + 8 + 8. ( ) ( ) 0 0 Check: () 0 is the solution. 7. ( + )( ) ( )( ) + + ( + ) 7( ) + 7 7 7 7 Check: 7 7 ( ) + 7 7 is the solution.
Eercise.. + Multipl each term b ( ). + ( ) ( ) + + Check: ( ) + + + is the solution.. + ( + ) Multipl each term b ( + ). ( 8+ 0) 8 0 or ( ) Check: ( 0 ) + ( ) + 0 0 + 0 0 and 0 is the solution. 7.. 8. 8. 8( 8. +. 7). 8. 8. 77 8. 8. +. 77 8. 8 8. 8. 0 0. 0 0. 0 0. 7. 0. 0. +. 8. (. 0. 00). 8 0. +. 8. 0. 000. 8 0. + 0. 000.. 8. 8 0. 000 7. 7. 7. 0. 000. + +. + ( ) 8+ 8+ +. I Prt I Prt rt rt I P rt + + 8 + 7 + FE + FE 7. 0 FE + 8 0 FE FE A is the lowest grade that can be earned on the final.. a. w + 0 ( 0) w + 0 0 w w b. ( 0) + 0 ( h 0) 80 + 0 + 0 h h 80 h 7. inches. p 7.0 0.708t a. p 7.0 0.708().% b. 0 7. 0 0. 708t 7. 0 t 0. 7 ears or in 08. 0. 708. $ billion $ million. + t t. 78. 7 t ears or in 88 + 0. total served active + dropouts + 8, + 8, ouths served 7. a. 7n T T 7n+ 7n T + 7n T b. 8 chirps in seconds means chirps in minute (or 0 seconds). 7( ) t Temperature is.. A P+ Prt P+ P( 0. )( ). P P $. A r 0. t
Chapter : Linear Equations and Functions. amount in safe fund 0, amount in risk fund Yield: 0.0 + 0.(0, ), 0. 0+, 00 0., 0. 0 00 0, $0, in % fund 0, 0 $0, in % fund.. Reduced salar: 0.0() $800 Increased salar: 800 + 0.0(800) $0 0 R% of R 0 8 00 $0 is an 8% increase.. C + MU SP (MU Profit) + 0. 80. 80 00 0. 80 0. 8 80 00 Profit is $80 on. Loss is $0 on. The collector lost $0 on the transaction. 7. cost + markup selling price. 0 + 0.. 0. 0. 7. 07 07. The selling price is $07.. Wholesaler. 8 + 0. W W 0. W. 8 W $ 7. 0 Retailer 7. 0 + 0. R R 0. 7R 7. 0 R $. 00 Supplementar Eercises. Suppose we want to solve +. a. Should we remove parentheses or clear the equation of fractions first? b. Answer True or False for each Step : I. 0 0 0 + ( ) II. 8 + III. +. In a surve of 00 colleges it was found that enrollments fell into the following ranges: Number of Colleges Range of Enrollment 0, to, 0,00 to,,00 to 0, Which of the following could be the average enrollment for these 00 colleges? I. 8, II., III., a. I onl b. II onl c. I and II onl d. II and III onl e. I, II, III. Item A costs the store $8 and the markup is % of the cost. Item B costs the store $0 and the markup is % of the selling price. What is the total selling price of A and B? a. $80 b. $8 c. $87 d. $ e. none of these Solutions to Supplementar Eercises. a. Remove parentheses first. b. + (We removed parentheses.) 0 0 0 + (We multiplied both sides b 0.) Thus, I is False, II is True, and III is True.. Minimum average ( 0, ) + 0(, 00) + (, 00) + 0 +, 7 Maimum average (, ) + 0(, ) + ( 0, ) + 0 +, 70 I and II are the onl possibilities. Answer: c. Cost + Markup Selling price A: 8 + 8 0 ( ) B: 0 + 0 or 80 Total selling price $0 + 80 $8 Answer: b
Eercise. 7 Eercise.. is a function of.. is not a function of. If, for eample, there are two possible values for.. is a function of since for each there is onl one. D {,,, 8, }, R {,, } 7. R() 8 0 a. R(0) 8(0) 0 0 b. R() 8() 0 c. R( ) 8( ) 0 d. R(.) 8(.) 0.8. C ( ) a. C( 0) ( 0) b. C( ) ( ) c. C( ) ( ) d. C. f( ) a. f + 8 8 8 b. f ( ) 8 c. f ( ) ( ) 8+. f( ) + + a. f( + ) f( ) + + f() + f() 7 + 0 f() + f() f( + ) b. f( + h) + ( + h) + ( + h) c. No. This is equivalent to a. d. f( )+ h + + + h No. f( + h) f() + h e. f( + h) + ( + h) + ( + h) + + h+ + h+ h f( ) + + f ( + h) f ( ) h + h + h h( + + h) f( + h) f( ) + + h h. f( ) a. f ( + h) ( + h) ( + h) h h + + h b. f( + h) f( ) ( + h) ( + h) ( ) + h h h + h h h f( + h) f( ) h h h h h h 7. The vertical line test shows that graph (a) is a function of, and that graph (b) is not a function of.
8 Chapter : Linear Equations and Functions. Since (, 0) and (, ) are points on the graph, (a) f() 0 and (b) f().. a. The ordered pair (a, b) satisfies the equation. Thus b a a. b. The coordinates of Q (, ). Since the point is on the curve, the coordinates satisf the equation. c. The coordinates of R (, ). The satisf the equation. d. The values are 0 and. These values are also solutions of 0.. f() g ( ) a. ( f + g)( ) + b. ( f g)( ) c. ( f g)( ) d. f ( ) g. f( ) g ( ) a. ( f + g)( ) + b. ( f g)( ) c. ( f g)( ) d. f ( ) g 7. f( ) ( ) g ( ) a. ( f o g)( ) f( ) ( ) 8 b. ( go f )( ) g (( ) ) ( ) c. f( f( )) f(( ) ) [( ) ] d. ( f f)( ) ( ) ( ) ( ) [(f f)() f(f())]. f( ) g ( ) + a. ( f o g)( ) f( + ) + b. ( go f)( ) g( ) ( ) + + c. f( f( )) f( ) d. ( f f)( ) [(f f)() f(f())]. + There is no division b zero or square roots. Domain is all the reals, i.e., { : Reals}. Since 0, +, the range is reals or { : }.. + There is no division b zero. To get a real number, we must have + 0 or. Domain:. The square root is alwas nonnegative. Thus, the range is { : Reals, 0}.. D: { :, } 7. D: { : 7 7}. a. f(0) 0, means it will take 0 ears to pa off a debt of $0, (at $800 per month and 7.% compounded monthl.) b. f( + ) f(0),; f() + f() 80,; No.. a. f(0). means that in 0 there were. workers supporting each person receiving Social Securit benefits. b. f(0). c. The points based on known data must be the same and those based on projections might be the same. d. Domain: 0 t 00 Range:. n.. a. Since the wind speed cannot be negative, s 0. b. f ( 0). +. 7( 0). 0. At a temperature of F and a wind speed of 0 mph, the temperature feels like. F. c. f(0)., but f(0) should equal the air temperature, F. t. E 0. + t a. E is a function of t. b. The domain of the function is reals ecept t. c. Because negative times should not be considered, the domain is t 0. 7. C F 0 a. C is a function of F. b. Mathematicall, the domain is all reals.
Eercise. c. Domain: {F : F } Range: {C : 0 C 00} d. C( 0) ( ) 0 0 00 0 0. C 700p. Cp ( ) 00 p a. Domain: {p : 0 p < 00} 700( ) 8, 00 b. C( ) $ 7. 7 00 700( 0) 7, c. C( 0) $, 700 00 0 0 700( ) 7, 700 d. C( ) $ 7, 700 00 700(. ) e. C(. ) 00. 77, 080 0. $, 87, 700 In each case above, to remove p% of the particulate pollution would cost C(p).. a. A is a function of. b. A() (0 ) sq ft A(0) 0(0 0) 00 sq ft c. For the problem to have meaning we have 0 < < 0.. a. Pqt ( ( )) P( + 0t) ( + 0t) 80( + 0t) 00 00, 800 + 00t t b. q() + 0() 0 P(q()) $,7. fab (, ) a + b f (, ) + f(, f(, )) f(, ) +. length width L + 00 or 00 L + 00 + 00 7. Revenue (no. of people)(price per person) Eample: R 0 0 R. 80 R. 0 Solution: R (0 + )(0 0.0) Supplementar Eercises. Suppose that f( ) +. What is the domain of f()?. If fab (, ) a + b, then f(, )?. If fab (, ) a + b, then f(, f(, ))?. If f( ), then a. a + a+ a b. a + a+ a c. a + a+ a + d. a + a+ a + e. f( + a)? f( ) + f( a) Solutions to Supplementar Eercises. f( ) + Rule out division b zero: must be 0: Domain: {:, }. fab (, ) a + b f (, ) +. f( + a) ( + a) + a+ a a + a+ ; f( ) ; f( a) a f( + a) a + a+ a + a+ f( ) + f( a) + ( a ) a (Note: The purpose of this question is to show that f( + a) f() + f(a).) Answer: b
0 Chapter : Linear Equations and Functions Eercise.. + -intercept: 0 then. -intercept: 0 then. +. -intercept: 0 then. -intercept: 0 then 7. + or +, m, b.. m, b + - - -. m, b + + - - -. + 0 -intercept: 0 then 0 Likewise, -intercept is 0. - - + 0 7. (, ) and (, ) m ( ). (, ) and (, ) m 0 0 ( ) For problems 7, use m + b to obtain slope and -intercept. 7., m 7, b. or 0 +, m 0, b. 8 Slope is undefined. There is no -intercept. - -. P( 0, ), m m( ) + - - - - -. P(, ), m ( ( )) + - - - - - - +
Eercise. 7. P(, ), m is undefined - - - - -. P (, ), P (, ) m ( ). P ( 7, ), P (, ) m 7 ( 7) 7 + + or +. The -coordinates are the same. The line is horizontal. 0. + + Lines are perpendicular since. 7. or Parallel. Lines are the same.. If +, then +. So, m. A line parallel will have the same slope. Thus, m and P (, 7) gives ( 7) ( ( )) which simplifies to.. If, then. Slope of the perpendicular line is. Thus m and P (, ) gives ( ) which simplifies to +.. a. 0, 80, 0, 00 0 0 b. 0 0, 00 0 0 months 00 In 0 months, the building will be completel depreciated. c. (0, 70,) means that after 0 months the value of the building will be $70,.
Chapter : Linear Equations and Functions. a. b 00 00 00 00 00 00 00 00 00 00 00 R FS R SP. R FS 8.7 + 88.8 R SP 7.7t + 0.8 7 c. If t 0, then R F $ 88. 8. If t 0, then R SP $ 0.. The tables are eact. The equations are a best fit for the data. The equations are not eact. d. The equations are based on past performance. No one knows the results of future investments. 7. 0..0 a. m 0. b.0 b. 0 means that there was a negative amount of transactions. Restrictions are > 0 and 0. c. With an increase of (thousand) terminals, the amount of transactions increases b $0. (billion).. 0.088 +. Both units are in dollars.. a. (m, f) is the reference. Slope 88 0. 88. f 7,00 0.88(m,00) or f 0.88m 87. b. f 0.88(0,) 87. $,7.0. (, p) is the reference. ((0, 8) is one point. m 700 700 p 8, 700( 0) or p 700 + 8,. (t, R) is the ordered pair. P 7,, P (, ) 8 m. 7 R.( t ) or R. t. + or R.t 0. t t 7. P ( 00, ) P ( 0, ) m 08. 0 00 0 0. 8( 00) or 0. 8 7 Supplementar Eercises. What is the slope of the line with equation +? a. b. c. d. e.. In problem, what is the -intercept? a. b. c. d. 0 e.. The cost of making widgets is $ 0 for widgets $ for widgets. If the cost is a linear $8 for 0 widgets function, what are the fied costs?. A firm determines that the cost and revenue functions for a product are C + 0 and R 7, respectivel. If the firm will make a profit on the sale of 00 units, how much profit will the firm make on the sale of the0st unit? Solutions to Supplementar Eercises. + + + Answer: c. -intercept means 0. Answer: c. We need the equation for the cost. 0 m 7 C 0 7( ) or C 7 + Fied costs are $.
Eercise.. We need the marginal profit or the slope of the profit function. P R C 7 ( + 0) 0 Thus, m. The firm will make a profit of $ on the sale of the 0st unit. Eercise.. Not linear 0. 0 - -0. 0. 0 + 0. 7+ 0-0 0-0 a. 0.0 + 0. 7 + 0. Not linear + 0-00 80 - -0 0 b. 0-0 -0 0. Not linear 0-0 0 +. + 00 a. -0 0.0-0 7. Not linear + 0-0 0-00 00-0.0 b. Standard Window 0. Not linear -0 0-0 0-0 + -0 0-0 7. a. The equation is linear, so the graph will be a line. Use the intercepts to determine a window. b. Window: -min -min 0.0 -ma -ma 0.0 c. 0.00 0.0 0.00 0.0 0.0 - -0.0
Chapter : Linear Equations and Functions and. Complete graphs can be seen with different windows. A hint is to look at the equation and tr to determine the ma and/or min of. Also, find the -intercepts.. 0. ( 0. ) + 0 There is no min. Ma value of 0. -intercepts: 0 0. ( 0. ) + 0 0 ( 0. ). 0. 0. ±. 8 0. ± 8 or. or 8. Suggested: -min -ma -min -ma as one choice. - - Use this window to graph or use tet answers for graph. + 80. If 0,. 0 Starting point: -min -ma -min 0 -ma 0 Use tet answers for graph. + 80 0 7-7. + + + - -. f( ) + f () () + 0 f. + 8 Use our graphing calculator, and evaluate the function at these two points. If either of our answers differ, can ou eplain the difference?. As gets large, approaches. When 0,, intercepts at ±. Suggested: -min -min -ma -ma - -. + + ( )( + ) ( + )( + ) What happens to as approaches?? 0-7. + - -. + + + - -0 Note: For problems 7, find the zeros and find the -intercepts are equivalent statements. In part (a) we use a graphing calculator s TRACE OR ZERO function and in part (b) we find the eact solutions.. a. Graphing calculator approimation to decimal places:.08, 8.08 7 b. ± b the quadratic formula. 7. a. The graphing calculator gives eact answers of, using the ZERO function. b. When this numerator is zero the function is zero or when or. - -0
Eercise. 8,. -intercept: When R 0, 0, 008. R-intercept: When 0, R 8, 0, R 8, 0.08 0 00, 0. a. c. The coordinates of the point mean that the cost of obtaining stream water with % of the current pollution levels would cost $8,0. d. The p-intercept means that the cost of stream water with 00% of the current pollution levels would cost $0. 00. a. 0 F 0.88M. 0 800 0 0 0 b. 0.0 + 7.08 00 0.0 + 7.08 b. The coordinates mean that when a male s salar is $0 thousand, a female s salar is $0. thousand.. a. E 0,p 00p 00, -0 0-0, 0 800 0 0.8 + 0.8 0.8 + 0.8 00 b. E 0 when 0 p 00.. a. R 0. 0t + 0. 77t+ 0. 7 R 0.0t + 0.77t + 0.7 8 0 0 b. R 0. 0t + 0. 77t+ 0. 7 R 0.0t + 0.77t + 0.7 0 0 800 0 c. Using the first equation as the better fit, it predicts $ (nearest dollar) while the actual is $0. No Supplementar Eercises Eercise.., - - - - 0 0 7. a. -0 c. The revenue reaches a maimum during but zeros out in 00. The model could not be valid until 00. 0, 8, C p 80. No solution. - - - -/ -0 0 00 0 b. Near p 0, cost grows without bound.
Chapter : Linear Equations and Functions. Multipl first row b. + + Add the two equations. 8 Solve for the variable. 8 Substitute for this variable in ( ) either original equation and solve for the other variable. 8 The solution of the sstem is and. 7. Multipl first row b. 8 8 + + Add the two equations. Solve for the variable. Substitute for this variable in ( ) either original equation and solve for the other variable. 8 The solution of the sstem is and.. 8 Solve for. 8 Substitute for this variable in () first equation and solve for the other variable. + 0 0 0 The solution of the sstem is and.. + Multipl first equation b. + Multipl second equation b. Add the two equations. Substitute for this variable in either original equation and solve for the other variable. 8 + () 8 The solution of the sstem is and.. 7 Multipl first equation b. 8 + Multipl second equation b 7. + 77 Add the two equations. 7 7 Substitute for this variable in either original equation and solve for the other variable. 8 () + The solution of the sstem is and or (, ).. 0. u 0. v 0. Multipl first row b. 0. u 0. v. 0. u 0. v 0. 0. u 0. v 0. Subtract the two equations. 0. There is no solution. The sstem is inconsistent. Note: If ou have 0 a non zero number the sstem is alwas inconsistent.
Eercise. 7 7. 0. 0. 00. 0... Multipl nd row b 0.. 0. 0. 0. Subtract the two equations. 0.. Solve for the variable. 7 Substitute, solve for. 8 7 8 The solution of the sstem is and. 7 7. + + + Multipl second row b. Add the two equations: 0 0 There are infinitel man solutions. The sstem is dependent. Solve for one of the variables in terms of the remaining variable:. Then a general solution is c c, where an value of c will give a particular solution. +. Before we can solve this sstem + 8 we must arrange each equation in a + b c form. 7 7 Add and solve for. Substitute in an original equation and solve for remaining variable. 7 The solution is and 7 or (, 7). Use the standard window and graph each equation. Use the TRACE or INTERSECT feature to find the solution.. 8-8 - - +8 Solution: (, ). + + 7-8 - + - -7 - Solution: (, )
8 Chapter : Linear Equations and Functions 7. Eq. + + z Steps,, and of the sstematic Eq. + z 8 procedure are completed. Eq. z 0 Step : z From Eq. + () 8or 7 From Eq. + 7 () + or 7 The solution is 7, 7, z.. Eq. 8z 0 Steps and of the sstematic Eq. + z 8 procedure are completed. Eq. + z Step : ( ) Eq added to Eq. gives z or z. From Eq. + ( ) 8or From Eq. 8( ) 0 or The solution is,, z or (,, ).. Eq. + z Step is completed. Eq. Eq. + + z + 8z Step : + z Eq. Eq. + z ( ) Eq. added to Eq. Eq. z ( ) Eq. added to Eq. Step is also completed. Step : z from Eq.. From Eq. + or From Eq. + ( ) or The solution is,, z or,,.. amount of safe investment. amount of risk investment. +,00 Total amount invested 0. + 0.8 0, Income from investments The solution is the solution of the above sstem of equations. +, 00 +. 8 00, ( 0) second equation 0. 8, 00 8, Subtract equations Solve for or amount of risk investment. Substituting 8, into one of the original equations we have + 8,,00 or $77,00. Solution: Put $77,00 in a safe investment and $8, in a risk investment.. amount invested at 0%. amount invested at %. +,00 Total amount invested 0.0 + 0. 0 Investment income Solve the sstem of equation: +, 00 +., 00 ( 0) second equation 0. Subtract equations 0, Solve for Substituting into the first equation we have + 0,,00 or,00. Thus, $,00 is invested at 0% and $0, is invested at %.
Eercise. 7. A ounces of substance A. B ounces of substance B. Required ratio A B gives A B 0. Required nutrition is %A + %B 00%. This gives A + B 00. The % notation can be trouble. Be careful! Now we can solve the sstem. A B 0 A+ B 00 B 00 Subtract first equation from second. 00 0 B Substituting into the original equation gives A 0 0 A The solution is ounces of substance A and ounces of substance B.. population of species A. population of species B. + 0,00 units of first nutrient +,0 units of second nutrient 8+, 00 ( ) first equation +, 0, 70 Subtract 0 Solve for Substituting 0 into an original equation we have (0) + 0,00. So, 00. Solution is 0 of species A and 00 of species B.. amount of 0% concentration. amount of % concentration. + 0 amount of solution 0.0 + 0.0 0.(0) concentration of medicine Solving this sstem of equations: + 0 + 0. 7. 7 ( ) second equation 07.. Subtract equations Solve for Substituting into the first equation we have + 0 or 7. The solution is cc of % concentration and 7 cc of 0% concentration.. number of $0 tickets. number of $0 tickets. +, total number of tickets 0 + 0 80, total revenue To solve the sstem of equations multipl Eq. b 0. 0+ 0 80, 0+ 0 80, 0 00, or 0, Substituting into the first equation gives. Sell 0, tickets for $0 each and tickets for $0 each.. amount of 0% solution to be added. 0.0 concentration of nutrient in 0% solution. 0.0(00) is the concentration of nutrient in % solution. 00. + 00. ( + 00) 00. + 00. + 0 0. 8or 80cc of 0% solution is needed.
0 Chapter : Linear Equations and Functions 7. ounces of substance A, ounces of substance B, and z ounces of substance C. + + z 00 Nutrition requirements z Digestive restrictions z Digestive restrictions Since both and are in terms of z, we can substitute in the first equation and solve for z. So, z + z + z 00 or 0z 00. So, z. Now, since z, we have. Since z, we have. The solution is ounces of substance A, ounce of substance B, and ounces of substance C.. A number of A tpe clients. B number of B tpe clients. C number of C tpe clients. A + B + C 00 Total clients 00A + 00B + 00C 0, Counseling costs 00A + 00B + 00C 00, Food and shelter To find the solution we must solve the sstem of equations. Eq. Eq. A+ B+ C 00 A+ B+ C 00 Original equation divided b 00 Eq. A+ B+ C Original equation divided b 00 A+ B+ C 00 Eq. Eq. B+ C 00 ( ) Eq. added to Eq. Eq. B C 00 ( ) Eq. added to Eq. A+ B+ C 00 Eq. B+ C 00 Eq. C Eq. added to Eq. C 00 Substituting C 00 into Eq. gives B + 00 00 or B 00. So, B 00. Substituting C 00 and B 00 into Eq. gives A + 00 + 00 00. So, A 00. Thus, the solution is 00 tpe A clients, 00 tpe B clients, and 00 tpe C clients. Supplementar Eercises 0. If +, then? 7 a. b. 0 c. d. e. 7. If (a, b) is the solution to the sstem of equations 0. + 0, then a + b? 0. + 0. 8 a. 0 b. 8 c. 80 d. 0 e. none of these
Eercise.. A bank loaned $8, for the development of two products, A and B. The amount loaned for product B was $, less than twice the amount loaned for product A. A + B 8, is one equation needed to find the amounts loaned for A and B. The other equation is a. B A, b. A B, c. A B, d. B A, e. A, B Solutions to Supplementar Eercises. + 0 7 + 0 Answer: d. 0. + 0 0. + 0. 8 0. + 0. + 0. 0 0 0 0.( 0)+ 0 0 a + b 80 Answer: c. amount loaned for B was, less than twice the amount loaned for A B A, This simplifies to A B,. Eercise.. a. Total cost Variable cost + Fied cost C() 7 + 00 b. C(00) 7(00) + 00 $800. a. R() b. R(00) (00) $0,00. a. P() R() C() (7 + 00) 7 00 b. P(00) 7(00) 00 $700 7. a. P ( ) R ( ) C ( ) 80 ( + 80) 7 80 b. P( 0) 7( 0) 80 $ 70 The total costs are more than the revenue. c. P ( ) 0 or 7 80 0 80 So, 0 units is the break-even 7 point.. C() + 0 a. m, C-intercept: 0 b. MC means that each additional unit produced costs $. c. Slope marginal cost. C-intercept fied costs. d. $, $ ( MC at ever point). R 7 a. m 7 b. 7; each additional unit sold ields $7 in revenue. c. In each case, one more unit ields $7.. R() 7, C() + 0 a. P ( ) 7 ( + 0) 0 b. m c. Marginal profit is. d. Each additional unit sold gives a profit of $. To maimize profit sell all that ou can produce. Note that this is not alwas true.. (, P) is the correct form. P ( 00, 00) P ( 0, ) 00 m 8 0 00 P 00 8( 00) or P 8 800 The marginal profit is 8. 7. a. The revenue function is the graph that passes through the origin. b. At a production of zero the fied costs are $. c. From the graph, the break-even point is 00 units and $ in revenue or costs. d. Marginal cost 00 0 0 Marginal revenue 00 0. 7.
Chapter : Linear Equations and Functions. R() C() 8 + 0 or 0 0 or. Thus, necklaces must be sold to break even.. a. R(), C() 8 + 00 b. R() C() if 8 + 00 or 00 or 00. It takes 00 units to break even.. a. P ( ) R ( ) C ( ) ( 8+ 00) 00 b. B setting P() 0 we get 00. Same as (b).. a. R().0 b. P (, 0) P ( 800, 0), 0 0, 7 880 m,. 0 800 00 0,. 0( ) or. 0+ 0, 00 C( ) c. From.0.0 + 0,00 we have 0 units to break even. 7. If price increases, then the demand for the product decreases.. a. If p $00, then q 00 (approimatel). b. If p $00, then q 00. c. There is a shortage since more is demanded.. Demand: p+ q 00 ( 0) + q 00 q 80 q Suppl: p q 0 0 q 0 q 0 q There will be a surplus of units at a price of $0.00.. Remember that (q, p) is the correct form. P ( 0, 00) P (, 80) 80 00 0 m 0 7 Note: m < 0 for demand equations. p 00 ( q 0) or p q+ 00. (q, p) is the correct form. P ( 0,. 0) P (,. 00). 0 00 m. 0. 0 Note: m > 0 for suppl equations. p 0.(q ) or p 0.q + 0. 7. a. The decreasing function is the demand curve. The increasing function is the suppl curve. b. Reading the graph, we have equilibrium at q 0 and p.. a. Reading the graph, at p 0 we have 0 units supplied. b. Reading the graph, at p 0 we have 0 units demanded. c. At p 0 there is a shortage of 0 units.. B observing the graph in the figure, we see that a price below the equilibrium price results in a shortage.. q+ 8 q+ q+ 8 q+ 8 Required condition. Multipl both sides b to simplif. q 00 q 0 Substituting into one of the original equations gives p 0 + 8 8 Thus, the equilibrium point is (q, p) (0, 8).. q+ 0 q+ 0 Required condition. 0 q q 0 Solve for q. Substituting q 0 into one of the original equations gives p 80. Thus, the equilibrium point is (q, p) (0, 80).
Eercise. 0 00 7. Demand: (80, 0) and (0, 00) are two points. m 80 0 p p m( q q) or p 00 ( q 0) or p q+ 0 80 70 Suppl: (0, 80) and (0, 70) are two points. m 0 0 8 p p m( q q) or p 80 ( q 0) or p q+. 8 8 Now, set these two equations for p equal to each other and solve for q. 8 q+. q+ 0 Required for equilibrium. q+ 700 0q+ 00 Multipl both sides b 8 to simplif. q 00 q 00 Substituting q 00 into one of the original equations gives p. Thus, the equilibrium point is (q, p) (00, ).. a. Reading the graph, we have that the ta is $. b. From the graph, the original equilibrium was (00, 00). c. From the graph, the new equilibrium is (0, 0). d. The supplier suffers because the increased price reduces the demand.. New suppl price: p q + 0 + 8 q + 8 q+ 8 q+ 0 Required condition q q 8 Substituting q 8 into one of the original equations gives p 88. Thus, the new equilibrium point is (q, p) (8, 88). q q. New suppl price: p + + + 0 0 0 q q 0 + 0 + Required condition q+ 00 q+ 00 q q 00 Thus, p 00 0 + 0. The new equilibrium point is (00, 0). q. Demand: p +00 q Suppl: p + 0 0 0 q q q New suppl: p + 0 + + 0 0 + + 00 0 0 0 0 q+ 00 q+ 00 0 0 Required condition q+ 00 q+ 00 Multipl both sides b 0 q 00 q 00 Thus, p 00+ 00 0. The new equilibrium quantit is 00. The new equilibrium price is $.
Chapter : Linear Equations and Functions Supplementar Eercises. A portable TV sells for $8. The production of TVs has a fied cost of $7 and a variable cost of $ for each TV produced. How man TVs must be sold to break even? a. b. c. 8 d. 7 e.. For the graph given below, what quantit is demanded at a price of $0? 80 70 0 0 0 0 0 0 a. 0 b. 0 c. 0 d. 0 e. 70 (0, ) 0 0 0 0 0 0 70 80. For the graph in problem, what quantit is supplied at a price of $0? a. 0 b. 0 c. 0 d. 0 e. 70 Solutions to Supplementar Eercises. Break even means profit is zero. P R C P 8 ( + 7) 7 P 0 at. Answer: b. The demand curve is a falling curve. At p 0, q 0. Answer: c. The suppl curve is a rising curve. At p 0, q 70. Answer: e Review Eercises For this set of eercises we will not give reasons for an steps or list an formulas.. + 7 7 7. 8. 8 + + ( ). 8 + 8 ( ) ( + ) 0( ) 8+ 0 0 8 8. + + + +. 0. + 00. 0. 0. 0. 7. + ( + ) ( ) + + 8 8. 0. + + + 7 ( + 7) ( + ) ( + 7) + ( ) + 0 + + 0 7 7 There is no solution since we have division b zero when 7.
Review Eercises 0. Yes.., is not a function of. If, then ±.. Yes.. Domain: 0 or or. Range: Positive square root means 0.. f( ) + + a. f ( ) ( ) + ( ) + + b. f ( ) ( ) + ( ) + + + 7. f( ) +, g( ) a. ( f + g) ( + ) + + + b. f g + or + + c. f( g( )) f( ) + d. ( f o f) f( + ) ( + ) + + 0. + 0 -intercept: If 0, -intercept: If 0, c. f + + + + + 0. g ( ) + a. g( ) ( ) + 0 b. g + + c. g( 0. ) ( 0. ) + 00. + 0 00. 0.. f( ) f( + h) ( + h) ( + h) + h h h f( ) f( + h) f( ) h h h h( h) f( + h) f( ) h h 7. is a function of. (Use vertical rule test.) 8. No, fails vertical line test.. f() 0. 0,. a. f() 7 b. f() if, c. 8 - - - - - - -. + -intercept: If 0, -intercept: If 0 or - - +. -intercept: There is no -intercept. - - -. P(, ); P(, ) ( ) m 7. (.8, 7.) and (.8,.) m 7.. 8. 8. ( 8. ) 0 Slope is undefined.
Chapter : Linear Equations and Functions 8. + 0 + m, b. + m, b 0. m, b, +. m, b, +. P (, ), m ( + ) or +. (, 7) and (, ) m 7 ( ) 8 7 ( ( )) or 8 7 + 8. P( 8, ); P(, ) The line is vertical since the -coordinates are the same. Equation:. Parallel to means m. ( ) or +. P(, ); to + or + m ( + ) or 0 + 7. + 0 + + 8. 0-0 0-0 0 + + -0 0. a. ( + )( )( ) b. -700 00 - ( + )( )( ) 0 ( + )( )( ) -0 0-0 c. The graph in (a) shows the complete graph. The graph in (b) shows a piece that rises toward the high point and a piece between the high and low points. 0. is a parabola opening upward. a. b. 0 - -0 0-0 0-0 c. (a) shows the complete graph. The -min is too large in (b) to get a complete graph. +. 0; + 0 or ; Domain: 0,. + Then, 8 + 8 8 ( ) Solution: (, ) -0 7 +
Review Eercises 7. + Then, + 8 0 0 0 ( ) + Solution: (0, ). + Then, + () + Solution: (, ). + + + ( + ) + No solution.. 8 0 ( 0) + + 7 70 7 0 Solution: (0, 7) 7. + + z Steps and : Nothing to be done. + z Step : + + z + z + z z Step : z + ( ) + ( ) + ( ) Solution is,, z. 8. + z Thus z z 7 7 7 + 7z 0 0 or 0 + z + 0 z 7 z Solution: (, 0, ). Student has total points of + 8 + 88 + 0 + +. Total of possible points is 00 + 0 + 00 0. To earn an A students need at least 0.(0) 8 points. Student must earn 8 0 points on the final. This is the same as %. 0. Diesel: C 0. + 8, 0. +, 0. + 8, Gas: C 0. +, 0. 0 0, Costs are equal at 0, miles. A truck is used more than 0, miles in ears. Bu the diesel.
8 Chapter : Linear Equations and Functions. a. Yes b. No c. f(00). a. f(80). b. The monthl pament on a $70, loan is $.7. c. f( 0) f( 0) (. 07) $ 88. Two $0, loans are the same as a $0, loan.. P ( ) 80 00 00 q() t + 0t a. ( Po q) t P( + 0t) ( + 0t) 80( + 0t) 00 00 b. q( ) + 0( ) 0 units produced ( 0) P( 0) 80( 0) 00 $, 7 00 ( t ). WL ( ) kl, Lt ( ) 0 0, 0 t 0 0 ( t ) ( t ) ( W L)( t) W. 0 0 o 0 00 0 0 0. d a. t 8. d d t.8. t 8 0 b. (., ) means that the thunderstorm is two miles awa if flashes are. seconds apart. H c 00 80 0 H c 0 T a 0 0 0 0 0 80 00 T a 7. (, P) is the required form. P ( 00, 00), P ( 0, ) 00 00 m 8 0 00 0 P 00 8( 00) or P ( ) 8 800
Review Eercises 8. Use (C, F) to get equation. P( 0, ); P( 00, ) 80 m 00 0 00 F ( C 0) or F C+. Also C ( F ).. a. 700 0 0 0 8-00 b. Algebraicall, 0 if 0 0 0 ( ) 0. Answer: 0 0. v 0( h+ 0) v h + 0 0 v h 0 0 00 v h 0 0 0 00-0. amount of safer investment and amount of other investment. + 0 0. 0+ 0. Solving the sstem: 0. + 0. 00 0. 0+ 0. 0. 0 00 00 Then 0. Thus, invest $00, at.% and $0, at %.. liters of 0% solution liters of 70% solution + 0. + 07.. + +. 7.. +. 8. Answer:.8 liters of 0%,. of 70%.. S: p q+, D: p q+ 8. a. S: q+ D: q+ 8 q 8 q 8 q q b. Demand is greater. There is a shortfall. c. Price is likel to increase. 0 0 00 0 80 0 0 p Suppl p q + 0 0 0 0 0 0 0 (0, 0) Market Equilibrium p + q 0 Demand q. C() 8.80 + 00, R().0 a. Marginal cost is $8.80. b. Marginal revenue is $.0. c. Marginal profit is $.0 8.80 $.0. d.. 0 8. 80+ 00. 0 00 00 units to break even.. FC $00, VC - $ per unit, R $ per unit a. C() + 00 b. R() c. P R C 0 00 d. MC e MR f. MP 0 g. Break even means 0 00 0 or 0.
0 Chapter : Linear Equations and Functions 00 00 00 0 7. Suppl: m Demand: m 00 00 00 00 p 00 ( q 00) p 0 ( q 00) p q p q+ 00 So, q q+ 00 or q 00. The equilibrium price is p 00 0 ( ) $. q q 8. New suppl equation: p + 8+ + 0 0 0 q q Demand: p + 00 + 0 0 0 q q + + 0 0 0 0 q 0 or q 700 0 700 p + 0 80 0 Solution: (700, 80) Chapter Test. + 8 + 7 8 8 7... + + ( + ) + ( + ) ( ) + + + 7 7 7 7 ( ) ( ) 7 0 8 f( ) 7+ f( + h) 7+ ( + h) ( + h) 7+ + h h h f( ) 7+ f ( + h) f ( ) h h h f( + h) f( ) h h. 0 -intercept: -intercept: - - 8 - -. 7 + -intercept: -intercept: 0 - - - - 7. f( ) + 7 + a. + 0 Domain: ; Range: 0 For range, note square root is positive. b. f () + 7 c. f () 0+
Etended Applications 8. (, ) and (, ) m ( ) ( ( )) +. + + m, b 0. Point (, ) a. Undefined slope means vertical line. b. to + means m. Thus, + ( + ) or.. a. is not a function since for each there are two s. b. is a function since for each there is onl one. c. is not a function for same reason as (a).. + + + 8 8 + 7 + ( ) Solution: (, ). f( ), g( ) + a. ( fg)( ) ( )( + ) b. gg ( ( )) g ( + ) ( + ) + + c. ( f o g)( ) f( + ) ( + ) ( + ) + 0+ + 7+. R() 8, C() 0 + 00 a. MC $0 b. P ( ) 8 ( 0+ 00) 8 00 c. Break-even means P() 0. 8 00 or 0 units b. C( 00) 0( 00) + 8 $, It costs $, to make 00 units. c. 0 0+ 8 0 8 0 units. S : p q + 00, D : p q + 00 q+ 00 q+ 00 8q 00 or q 00 p(00) (00) + 00 $00 7. 0 00 a. b 0, The original value is $0,. b. m 00. The building is depreciating $00 each month. c. 00, 00, 00, 00, 0, 00 0 00 0 00 0 8. number of reservations 0. 0 0 00 Accept 00 reservations.. amount invested at % amount invested at % + 0 Amount 0.0 + 0.0 0 Interest 0. 0+ 0. 0 800 0. 0+ 0. 0 0 0. 0 0 $ 8 Invest $8 at % and $ at %. Etended Applications I. Hospital Administration. Revenue per case $ Annual fied costs $80, + 70, $0, Annual variable costs ($ 80 + + 0 ) $ 00, where is the number of operations per ear. d. MP $8. Each unit sold makes a profit of $8.. a. R() 0
Chapter : Linear Equations and Functions. Break-even occurs when Revenue Total Costs 0, + 00 00 0, 70 The hospital must perform 70 operations per ear to break even.. We have (70 operations/month)( months/ear) gives 80 operations/ear with a savings of (80 operations)($0 savings) $, on supplies. However, leasing the machine would cost $0,. Thus adding the machine would reduce the hospital s profits b $8 a ear at the current level of operations. (Note that operations must be performed each ear to cover the cost of the machine: [($0)00) $0,].). Profit Revenue Cost P ( ) ( 0, + 00) 00 0, At current level of operations, the annual profit is: P( 80) 00( 80) 0, 0, 0, $, With (0 new operations/month)( months/ear) 80 new operations/ear, the new level of operations is 80 + 80 0. The advertising costs are ($0,/month)( months/ear) $0, per ear. At the new level of operations, the profit would be: P( 0) 00( 0) 0, 0, 7, 70, $, The increase in profit is $,, $8,.. Each etra operation adds $ 00 $00 of profit. If the ad campaign costs $0, per month it must generate $ 0, per month operations/month to $ 00 per operation cover its cost.. Recall that the break-even point for leasing the machine is operations per ear. If the ad campaign meets its projections, 0 operations per ear will be performed, with a savings of (0)($0) $, on medical supplies b leasing the machine. The should reconsider their decision. (Note that this eample illustrates that if the assumptions on which a decision was made change, it ma be time to take another look at the decision.) II. Fund Raising Answers will var.