Practice Math 102 Final Exam

1 Practice Math 102 Final Exam 1. 2. Sketch the graph of f(x) = r%x%% . What is the domain of f? Use transformations to sketch the graph of g(x) = r%%–%x%% – 2 . 1 a. Given f(x) = ))))) 3x – 2 [0,4 ) What is the domain of g? 1 and g(x) = )) x What are the domains of f and g? Find fBg(x) and the domain of fBg . 3. 4a. 4b. 5. Find the formula for f –1(x) given x f(x) = ))))) . 3x – 2 x– 6 Analyze the function given by f(x) = )))))) x+ 2 State the domain and range of f. and sketch the graph, labelling everything. x2 + 4 and sketch the graph, labelling everything. Analyze the function given by f(x) = ))))) x2 – 4 x Solve )))))) < 2 . Express the solution set using interval notation. x+4 6. f(x+h) – f(x) For f(x) = 2x2 – 5x + 3, find & simplify completely the difference quotient ))))))))))) h 7a. Find the center and radius of the circle with equation 7b. Identify the curve and show any vertices, foci, asymptotes or axes: x 2 + y 2 – 4x – 6y – 3 = 0 x 2 ! 4y 2 + 8y = 8 . 6 8. a,b. Write an equation that would have the graph shown ± 9. Sketch the graph of P(x) = –x3 + 4x2 – 4x . Label everything! 10. For P(x) = 2x3 + 8x2 + 3x – 10 , a. List all the possible rational zeros of P(x) b. Show that – 2 is a zero of P(x). c. Find all the zeroes of P. a -7 b 11. Find a fourth degree polynomial P(x) with 1+2i a zero, 1 a zero of multiplicity 2, and a constant term of 10. Express your answer in the form P(x) = ax3 + bx3 + cx2 + dx + e, where a,b,c,d,e, are integers. 12. Simplify completely: 4 log4(2x) – log4 x5 + log4(4x) . 13. Sketch the graph of f(x) = ex— 2 and the graph of g(x) = lnx + 2. Describe the relationship between these graphs, and the relationship between f & g. 14. Solve for x: 15. Solve for x (exact answer.): 16. Find the maximum for P(x,y) = 2x + 3y, subject to conditions: x $ 0 ; y $ 0 ; y $ x + 1 ; x + y # 12 ; y # x+ 6 . Complete justification required. 17. Find all points on the line x ! y = 7 that lie exactly 5 units from the origin. (Hint: the points of the plane that lie 5 units from (0,0) form a conic section known as.... ) log 6 (x–3) + log 6(x+2) = 1 ex — 2 = 3x Practice Math 102 Final Exam 18. $2000 is invested in an account that pays 3% (annual percentage rate) compounded continuously. How long will it take for the balance to reach $3000? 19. A colony contained 2000 bacteria at 6AM, 3000 at 8AM. How many will there be at noon? When will there be 10,000 bacteria? 20. Find the quadratic function passing through the points (! 1,8) & (1,! 4) & (2,! 1). Hint: f(x) = ax2 + bx + c . 21. Find the dimensions of a rectangle whose perimeter is 105 meters and area is 441 square meters. Additional word problems are recommended:, for instance: Text §4.1 page 308 #69 Text §6.3 page 512 #73 Text §7.1 page 545 # 59, 61 Text §7.2 page 562 #77 (at least set it up) Text Chapter 7 Review page 628 #100, 102 2 3 Practice Math 102 Final Exam Practice Math 102 Final Exam PRACTICE MECHANICS– THESE ARE NOT TYPICAL FINAL QUESTIONS These problems are included as a reference for those who need a brush-up on such details. A. Simplify completely. Express your answer without negative exponents. (2x4y–4/5)5(16y–4)–1/4 First we give ou rselve s a little sp ace to thin k in, then d istribute th e expo nents, 5 and –1/4: (2 x4 y–4/5 )5 ( 16 y –4 )–1/4 = 25 x20 y–4 ( 16 y –4 )–1/4 = 25 x20 y–4 16–1/4 y –4 (–1/4) = 25 x20 y–4 16–1/4 y = 25 16–1/4 x20 y–4 y = 25 (24)–1/4 x20 y–4 y = 25 2–1 x20 y–4 y = 24 x20 y–3 24 x20 ))))))) y3 = B. x x 3/2 (x )))) 1/2 Regro up to facil itate com bining li ke factor s... This s implific ation req uires on ly one o r two step s, ... which were drawn ou t to reveal all details x3/2 + 6x1/2 + 9x–1/2 Factor completely. 1/2 e.g. (y– 4/5 )5 = y(– 4/5)5 = y–4 1/2 + 6x –1/2 + 9x Observation: Multiplication by x 1/2 would make that a lot nicer! ) (x3/2+1/2 + 6x1/2+1/2 + 9x–1/2+1/2 ) x ))))))))))))))))))))))))))))))))) 1/2 (x2 + 6x1 + 9 ) x )))))))))))))))) 1/2 (x + 3) 2 x ))))))))) 1/2 C. Multiply as indicated. Express the result in the form “a + bi”. 6i + 9 )))))) 1 – 3i = (6i + 9)(1 + 3i) 9 @1 — 6 @3 + 9 @3i +6i ))))))))))))) )) = )))))))))))))))))))) (1 – 3i)(1 + 3i) 1+9 = – 0.9 + 3.3i D. Multiply: (x ! (2 +3i))(x ! (2! 3i)) = xx ! (2! 3i)x ! (2+3i)x + (2+3i)(2! 3i) = x2 ! 2x + 3ix ! 2x ! 3ix + 4 ! 9i2 = x2 ! 2x ! 2x + 4 ! 9(! 1)2 = x2 ! 4x +13 E. Divide x7 ! 128 by x ! 2 and express x7 ! 128 as a product.... 2 1 1 0 2 0 4 0 8 0 16 0 32 0 64 2 4 8 16 32 64 x7 ! 128 = ( x ! 2)( x6 + 2 x5 + 4 x4 + 8x3 + 16 x2 + 32 x ) ! 128 128 0 4 Practice Math 102 Final Exam PRACTICE MECHANICS– THESE ARE NOT TYPICAL FINAL QUESTIONS These problems are included as a reference for those who need a brush-up on such details. F. Let A = (2,–5) and B = (–1,3) be points in the xy-plane. a. Find the length of segment AB. (simplify the answer) Length of AB = ))))))))))) / ()x)2 + ()y)2 = ( 3 2 + 8 2 )1/2 = 731/2 b. Find the midpoint of segment AB. Midpoint of AB = (halfway between 2 & –1, halfway between –5 & 3) ( 2 + –1 ) ' 2 , ( –5 =( + 3) '2 ) = ( ½ , –1 ) G. Solve. Express your answer using interval notation. 2| (½)x + 3| + 3 < 15 I like to see the x with coefficient 1, so I go ahead and multiply 2| (½)x + 3| | x + 6 | + 3 < 15 |x + 6 | < 12 |x – –6 | < 12 Here I LOOK at it and solve intuitively: ... says “distance between x & –6" < 12 [ –6 – 12, –6 + 12] [ –18 , 6 ] ! 18 !6 6 But this works, too: |x + 6 | < 12 if and only if –12 < x + 6 –18 < H. x < 12 < 6 so x is in [ –18 , 6 ] Simple word problem: The manager of a cheap plastic furniture factory finds that it costs $2400 to produce 200 chairs in a day, and $6200 to produce 400 chairs in a day. Assuming that the relationship between cost (y) and the number of chairs produced (x) is linear, find an equation that relates x and y. Reduced to its essence, this problem asks us to find the equation of a line through (x,y) = (200,2400) and (400,6200). SLOPE = )y/)x = 3800/200 = 19 Here I use the point-slope form: the equation is y – y0 = 19 (x – x0) where (x 0,y 0) is any point on the line y – 2400 = 19 (x – 200) y = 19x – 1400 Here I use the slope-intercept form: to find b, I take advantage of the fact that when x = 200, y = 2400: ² THIS IS SUFFICIENT ² But you may also write this. y = 19x + b 2400 = 19(200) + b So b = 2400 - 19(200) = -1400. 5 Practice Math 102 Final Exam 1. Sketch the graph of f(x) = r%x%% . What is the domain of f? Use transformations to sketch the graph of g(x) = r%%–%x%% – 2 . [0,4 ) What is the domain of g? We used the intermediate function y = r%%–%x%% y = r%%–%x%% f(x) = x r %%% x g(x) = 2. –x r%%%%% –2 1 Given f(x) = ))))) 3x – 2 1 & g(x) = )) x a. What are the domains of f and g? b. Find fBg(x) and its domain. a. Domain f is (!4, b) c (b, 4 ) ...because 3x! 2 must not be 0, thus x may not be b. Domain g is all reals except 0 ...aka (!4, 0) c (0, 4 ) 1 1 1 b. fBg(x) = f(g (x)) = ))))) = ))))) ² Here we see that 3 x) ! 2 must not be 0. 1 3g (x)! 2 3)x ! 2 ...in addition to the restriction x not be 0. So the domain of fBg is all reals except for 0 and 3/2. 3. Find the formula for f –1(x) given f(x) below: x f(x) = ))))) 3x – 2 or x y = ))))) 3x ! 2 We could solve this for x in terms of y, but instead we switch x & y, then solve for y in terms of x. y x = )))))) 3y – 2 Keep in mind you are solving for y ! 3xy – 2x = y y! 3xy = y + 2x y! 3xy ! y = 2x y! y(3x ! 1) = 2x 2x y = ))))))) 3x ! 1 4a. ...so: 2x f –1(x) = ))))))) 3x ! 1 x– 6 Analyze the function given by f(x) = )))))) x+ 2 and sketch the graph, labelling everything. x = !2 A typical linear rational function. Vertical asymptote where denominator = 0 ... at x = ! 2. y=1 Horizontal asymptote at y= 1 because as x 64, f(x) 6 1. x-intercept when x ! 6 = 0 ... at (x=6,y=0) y-intercept f(0) = (0! 6)/(0+2) = ! 3 ...i.e. at (x=0, y=! 3) (6,0) (0,!3) 6 Practice Math 102 Final Exam 4b. x2 + 4 Analyze the function given by f(x) = ))))) x2 – 4 and sketch the graph, labelling everything. Notic e this fu nction is EV EN: f( !a) = f(a) for every a. (Graph should be symmetric about the y-axis.) Horizontal asymptote at y= 1 because y=1 x2 + 4 1 + 4 /x 2 1+0 f(x) = ))))) = ))))) 6 ))))) as x 6 4 x2 – 4 1 ! 4 /x 2 1!0 (0,!1) Vertical asymptotes where x2 – 4 = 0... x = ± 2 y-intercept: f(0) = ! 1 x-intercept: x2 + 4 is never 0. (None) 5. x=2 Solve. Express your answer in interval notation. x < 2 x+4 )))))) Resist the temptation to multiply both sides. compare to 0, not 2 !! and SIMPLIFY x x+4 – 2 < 0 x x+4 2(x+4) – )))))) (x+4) )))))) )))))) x – 2(x+4) x+4 –x–8 x+4 )))))) < 0 < 0 ))))))))))) < 0 — 8 + –4 — )))))) |)))))))) |))))) –x – 8 + – – x+4 – – + quotient – + – check endpoints: at – 8: 0 < 0 (inequality is true) at – 4: inequality is NOT true State your conclusion 6. x=!2 Statement is true for x in (–4 ,–8] U (– 4,4 ) For f(x) = 2x2 – 5x + 3, find & simplify completely: f(x+h) – f(x) ))))))))))) h = 2(x+h)2 – 5(x+h) + 3 — ( 2x2 – 5x + 3) h )))))))))))))))))))))))))))))))))))))) = 2( x2 + 2xh + h2 ) – 5(x+h) + 3 — ( 2x2 ! 5x + 3) ))))))))))))))))))))))))))))))))))))))))))))))) h = )))))))))))))))))))))))))))))))))))))))))))))) = = 2x2 + 4xh + 2h2 – 5x – 5h + 3 — 2x2 + 5x — 3 h Distribute that — !! Notice many terms “cancel” ! 4xh + 2h2 ! 5h h )))))))))))))))) 4x + 2h – 5 ... And there you have it, in all the excruciating detail.... 7 Practice Math 102 Final Exam 7a. Find the center and radius of the circle with equation ... This is most easily done by completing the square. x 2 + y 2 – 4x – 6y – 3 = 0 x 2 – 4x + y 2 – 6y ( x – what) 2 + ( y – what) 2 x 2 – 4x + 4 ( x – 2) 2 X2 + = 3 = + y 2 – 6y + 9 + ( y – 3) 2 Y2 = = 16 3 + 13 Preparing to complete the sq uare 3 + whatever I need.... = 3+4+9 Fill in the gaps... and we com pare to the bas ic equation of a c ircle ... So we see this is the equation of a circle with center at (2,3) and radius 4. 6 8. a. Write an equation that would have the graph shown ± a Make note of the major and minor axes of the ellipse! These tell us a lot... enough to write the equation... We see the center is at (! 1,3) & the distance to the major vertices, at (! 5,3 & (3,3), is a = 4. The minor axis ( from ((!1,0) to (!1,6) ) is 6 units long, so b=3. This provides sufficient information to write the equation: (x + 1)2 )))))) + 42 (y ! 3)2 )))))) 32 = 1 b. Assuming this is the graph of a parabola (in x), the equation and seeing the vertex at (2,0), the equation could be: (y ! 0) = A (x ! 2)2 y = A (x ! 2)2 But what is A? Well, we have not used all the information available in the illustration... for instance, the y-intercept— substituting x=0, y= ! 2: ! 2 = A (0 ! 2)2 A = !½ Thus: 9. (2,0) (0,! 2) y = ! (½) (x ! 2)2 Sketch the graph of P(x) = –x3 + 4x2 – 4x . Label everything! As x 64, P(x) 6 – 4 since as x grows larger, the leading term dominates; that is, as x64, – x3 6!4 faster than 4x2 – 4x 6 +4 , so their sum 6!4. Similarly, as x 6 – 4 , P(x) 6 + 4 . P(0) = 0, so that’s the y=intercept, at (0,0) x-intercepts: –x3 + 4x2 – 4x = 0 –x (x2 – 4x + 4) = 0 . –x (x – 2)(x – 2) = 0 ... we see x-intercepts at 0, 2, and 2. Repeated root at 2 6 brush with x-axis. (make allowances for the graphing...) 8 Practice Math 102 Final Exam 10. For P(x) = 2x3 + 8x2 + 3x – 10 , a. List all the possible rational zeros of P(x) Because the coefficients of P(x) are all integers, any rational zero of P must be of the form: p ± ) where p divides (is a factor of) the constant term ! 10, and q divides the leading coefficient 2. q That gives us: b. = ± 1, 2, 5, 10, 1 ) 2 5 , 2) Show that – 2 is a zero of P(x). !2 2 2 c. 10 or 5 or 2 or 1 ± ))))))))))))) 2 or 1 8 3 !4 !8 !5 4 ! 10 This synthetic division shows, with the zero 10 remainder, that ! 2 is a zero of P(x). 0 It also shows us that P(x) = (x +2)(2x2 + 4x – 5) Find all the zeros of P(x). Simplify your answers. The remaining roots of P must be found in (2x2 + 4x – 5) Using the quadratic formula to solve 2x2 + 4x – 5 = 0... we find: x = –4 ± q %1% 6% %+ % %4 %0 % )))))))))) 4 = –1 ± 14 2 q %% % % ))) The third root of this cubic polynomial is –2, the root we found above. 11. Find a fourth degree polynomial P(x) with 1+2i a zero, 1 a zero of multiplicity 2, and a constant term of 10. Express your answer in the form P(x) = ax3 + bx3 + cx2 + dx + e, where a,b,c,d,e, are integers. If a,b,c,d and e are all integers. they must be real numbers. Since P has only real coefficients, any complex roots must occur in conjugate pairs (that is, for every root a+bi there must also be a root a - bi). Thus we know the four roots of P must be 1+2i, 1–2i, 1 and 1. And P must be the product of these factors: P(x) = A (x – (1+2 i) ) (x – (1–2i) ) (x – 1) (x –1) where A is some non-zero constant. P(x) = A (x2 – 2x + 5) (x2 – 2x + 1) which has constant term 5A. Since they want the constant term to be 10, A had better be 2. (5A = 10) P(x) = 2 (x2 – 2x + 5) (x2 – 2x + 1) P(x) = 2x4 – 4x3 + 20x2 – 24x + 10 12. Simplify completely: 4 log4(2x) – log4 x5 + log4(4x) . One way: Using the properties of logarithm functions, 4 log4(2x) – log4 x5 + log4(4x) 5 4 (4x) we can combine these terms into one log x + log log4(2x) – log4 4 4 (2x) (4x) expression, and see if the result can be log4 ))))))))) x5 simplified. It can. log4 64 9 Practice Math 102 Final Exam Alternate method: Using the properties of logarithm functions, we can expand & separate these terms so the log4x terms are alone. 4 log4(2x) – log4 x5 + log4(4x) 4 log42+ 4log4 x – 5log4x + log44 + log4x log416 + log44 + (4 – 5 + 1) log4x + 0 2 + 1 13. Sketch the graph of f(x) = ex— 2 and the graph of g(x) = lnx + 2. Describe the relationship between these graphs, and the relationship between f & g. . . . . . . . . . . . . . . Working from the known graph of y = ex , . . . . . . . . . . . . . . . . . . . . . . . . . . . . f(x) = ex–2 is the same graph, . . . . . . . . . . . . . . shifted to the right two units . . . . . . . . . . . . . . . . . . . . . . . . . . . . Again, working from a known graph, y=0 . . . . . . . . . . . . . . we sketch y=lnx, then shift two units . . . . . . . . . . (2,–1) . . . up to obtain the graph of g(x) = lnx +2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . The relationship between the graphs . . . . . . . . . . . . . . x–2 f(x) = e and g(x) = lnx +2: . . . . . . . . . . . . . . They are reflections of each other across the line y=x . That is appropriate, since f & g are inverses of each other. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . log 6 (x–3) + log 6(x+2) = 1 14. Solve for x: log 6 (x–3) + log 6(x+2) = 1 log 6 ((x–3)(x+2)) = 1 ((x–3)(x+2)) = 61 x2 – x – 6 = 6 x2 – x – 12 = 0 (x – 4)(x+3) = 0 x = 4 or –3 ex – 2 = 3x ln(e x – 2) = ln(3 x ) (x – 2) ln(e) = x ln (3) = x ln 3 Better to have one log expression than two, Now we can “undo” the log 6 using exponential base 6. CHECK THOSE SOLUTIONS!!!! Only x = 4 is a valid solution. 15. Solve for x (exact answer.): x–2 . . . . . . . . . . . . ex — 2 = 3x Std. Operating Procedure: x inside an exponential function, use ln.... and use those logarithmic properties! ... and keep in mind that ln3 is just a number. x – xln3 = 2 ( 1 – ln 3) x = 2 x = 2/(1– ln3) That’s an exact answer. 10 Practice Math 102 Final Exam 16. Find the maximum for P(x,y) = 2x + 3y, subject to conditions: x $ 0 ; y $ 0 ; y $ x + 1 ; x + y # 12 ; y # x+ 6 . Complete justification required. We first graph the “region of feasibility”– that part of the plane where the conditions are met: Conditions x $ 0 & y $ 0 restrict us to the first quadrant. . . . . . . . . . . . . . . y $ x + 1 ON & ABOVE the line y = x+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . x + y # 12 ON & BELOW the line y = 12 ! x . . . . . . . . . . . . . . . . . . . . . . . . . . . . y # x+ 6 ON & BELOW the line y = x + 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . The vertices (corners) of this region are: . . . . . . . . . . . . . . . . . . . . . . . . . . . . (0,1) (0,6) (3,9) and (6,6) . . . . . . . . . . . . . . . . . . . . . . . . . . . . Then we are guaranteed that the linear . . . . . . . . . . . . . . function P(x,y) = 2x+3y has both a maximum . . . . . . . . . . . . . . value and a minimum value somewhere in . . . . . . . . . . . . . . this polygonal region, and that those extreme values must be seen at the vertices. So evaluating P at all the vertices will provide sufficient information to locate them. P(0,1) = 2@0 + 3@1 = 3 P(0,6) = 18 P(3,9) = 33 ² Maximum value P(6,6) = 30 . . . . . . . . . . . . . . 17. Find all points on the line x ! y = 7 that lie exactly 5 units from the origin. (Hint: the points of the plane that lie 5 units from (0,0) form a conic section known as.... ) A point (x,y) lies 5 units from the origin if, and only if, x2 + y2 = 52 . If (x,y) lies on the line given, then (x,y) must satisfy x ! y = 7 . We must solve this system. We solve x ! y = 7 for either x or y, then substitute into the statement: Let’s choose y: Substituting: y=x!7 x2 + (x! 7)2 = 52 x2 + x2 ! 14x + 49 = 25 2x2 ! 14x + 24 = 0 x2 ! 7 x + 12 = 0 (x ! 3) (x ! 4) = 0 x = 3 or 4 There are two points on the line x ! y = 7 that lie 5 units from (0,0): (3,! 4) and (4,! 3). x2 + y2 = 52 . . . . . . . . . . . . . . . 11 Practice Math 102 Final Exam 18. $2000 is invested in an account that pays 3% (annual percentage rate) compounded continuously. How long will it take for the balance to reach $3000? When amount P0 is invested at r % interest compounded N times per year, the value of the account at time t years is Nt P(t) = P0 ( 1 + r/N ) rt As N 64, this approaches P0 e ... the value achieved with continuous compounding. .03t So, the value of this account after t years is $2000 e and we want to solve for t so that this value will be $3000. .03t $2000 e e = $3000 .03t = 1.5 = ln 1.5 .03 t t = ln 1.5 years .03 ))))) ( Ñ 13.52 years) 19. A colony contained 2000 bacteria at 6AM, 3000 at 8AM. How many will there be at noon? When will there be 10,000 bacteria? Bacteria multiply just like continuous compounding, but usually a lot faster! rt So we can expect the amount (A) of bacteria at time t to be A(t) = A0 e . rC0 Since A(0) = A0 e = A0 & A(0) = 2000 (considering 6AM to be time 0), we say A0 = 2000. Next we find r: At 8AM (t=2) there are 3000 bacteria, so: Taking ln of both sides, we get: r2 A(2) = 2000 e = 3000. Solve for r: r2 e = 3000/2000 = 3/2 or 1.5 2r = ln 1.5 r = (½) ln 1.5 So we now have a complete model for the bacteria population: (½) t ln1.5 A(t) = 2000 e At noon (t = 6) there will be: (½) 6 ln1.5 A(6) = 2000 e = 2000 (1.5)3 = 6750 (½) t ln1.5 There will be 10000 bacteria when A(t) = 10000.... When 2000 e = 10000 (½) t ln1.5 =5 We solve: e by taking ln of both sides: (½) t ln1.5 = ln5 t = 2 ln5 )))) (Hours after 6am) ln1.5 12 Practice Math 102 Final Exam 20. Find the quadratic function passing through the points (! 1,8) & (1,! 4) & (2,! 1). Hint: f(x) = ax2 + bx + c . f(! 1) = a(! 1)2 + b(!1) + c must = 8: f(1) = a(1)2 + b(1) + c must = ! 4: f(2) = a(2)2 + b(2) + c must = ! 1: a !b+c = 8 a + b + c = !4 4a +2b + c = ! 1 This system is easily solved for a & b & c (they are 3 & ! 6 & ! 1), so f(x) = 3x2 ! 6x ! 1 . 21. Find the dimensions of a rectangle whose perimeter is 105 meters and area is 441 square meters. A rectangle’s perimeter and area are understood in terms of its length and width, so we will call those dimensions x and y. Perimeter = 105 meters 2 Area = 441 m Y 2x + 2y = 105 m Y xy = 441 m2 Note this is a nonlinear system. Solving the first equation for y in terms of x: y = 52.5 ! x Substituting in the second equation for y: x(52.5 ! x) = 441 which yields the quadratic equation: x2 ! 52.5x + 441 = 0 using either the quadratic formula on this or factoring the equivalent: 2 x2 ! 105x + 882 = 0 (2x ! 21)(x ! 42) = 0 we solve for x, obtaining x = 10.5 and x = 42 as candidates... If x = 10.5, then y = 42. If x = 42, then y =10.5. The dimensions of the rectangle are 10.5 m by 42 m. Additional word problems: Setup for #77: (Using S B A) Text §4.1 page 308 #69 Answer in text Text §6.3 page 512 #73 Answer in text Text §7.1 page 545 # 59, 61 protein: 30S +15B + 3A = 78 Carbs: 20S + 2B + 25A = 59 Vitamin: 2S +20B + 32A = 75 Answer in text Text §7.2 page 562 #77 (set-up at right) Answer in text 320 Text Chapter 7 Review page 628 #100, 102 #100 answer: 466.25 acres in corn, 533.75 acres in soybeans. #102 answer: x$ 0 ; y$ 0 ; 4x + 3y # 960 ; 2x + 3y # 576. 200_ 192 (192,64) 240 288