Solutions to the Practice Questions for the Final Exam MA 153 1. 15512=312=31⋅21= 6 D 2. 16x2− 4y8= 4(4x2− y8) = 42x()2− y4()2[]= 42x + y4()2x − y4() C 3. 4a4b8c−2⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ −12= 4a4b8c2()−12=14a4b8c2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 12=14a4b8c2=12a2b4c A 4. 3x3x +1−xx − 2=3x(x− 2)(3x +1)( x − 2)−x(3x+1)(3x +1)(x − 2)=3x2− 6x − 3x2− x(3x + 1)( x − 2)=−7x(3x +1)(x − 2) C 1 5. x− 2(x +1)(x − 3)÷(x− 2)(x+ 1)(x + 3)( x −3)=x − 2(x +1)( x −3)⋅(x + 3)(x − 3)(x − 2)(x +1)=x + 3(x +1)2 B6. let t = time second person working alone16+1t=14 or 16(4) +1t(4)=112t16+1t⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =14⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 12t 3t23+4t⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1[]3t2t +12 = 3t 2t + 12 = 3tt = 12 t=12 E 7. xy−1(x + y)−1=xy1(x + y)=xy⋅(x + y)1=x(x + y)y A 8. 32 + 3⋅(2 − 3)(2 − 3)=3 2 − 3()2 + 3()2 − 3()=23− 3()22()2− 3()2=23− 34 − 3= 23− 3 C 9. let x = first positive integer x<y()y = x +1y2− x2=145⎧ ⎨ ⎪ ⎩ ⎪ ⇒ substitute eq. (1) into eq.(2) for y:x +1()2− x2= 145x2+ 2x +1− x2= 1452x + 1 =1452x = 144x = 72 B 210. A =P1+ rt()A = P + Pr tA − P = Pr tA − PPr= t E 11. let t = # hours truck has been traveling 40t = 55 t −1()40t = 55t − 5555 =15tt =5515=113 hours, so distance is 40113⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =4403 miles 55(t – 1) 40t t - 1 t 55 40 car truck distance time rate A 12. Let A=area of circleArea of circle ⇒ A(r) =π r2Diameter (d) of circle ⇒ x2+ x2= d2 2x2= d2 d =± 2x2 d = x 2Radius (r) of circle ⇒ r =x 22So, A(x) =πx 22⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 2=πx2(2)4⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ =π x22 or π2x2 A xx 313. 42 p − 3+104 p2− 9=12p + 3 Domain : p ≠±32⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 p + 3()2p − 3()42 p − 3+102p + 3()2 p − 3()⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ =12 p + 3⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2p + 3()2 p − 3()42p +3()+ 10 = 2 p − 38p + 12 +10 = 2p − 36 p =−25p =−256 D 14. let x= # ml of the 50% solutionlet y = total # of mlx + 40 = yx .50()+ 40 .20()= y .25()⎧ ⎨ ⎪ ⎩ ⎪ x .50()+ 8 = x + 40().25().50 x + 8 = .25 x + 10.25 x = 2x = 8 ml B 15. x = 14 + 5x Check :x()2= 14 + 5x()2 If x = 7:x2=14 + 5x 7 = 14 + 35x2− 5x − 14 = 0 7 = 49⇒ yesx − 7()x + 2()= 0 If x =−2:x = 7, x =−2 − 2 = 14 −10 − 2 = 4 ⇒ no E x = 7[] 416 m4− m2− 6 = 0m2− 3()m2+ 2()= 0m2− 3 = 0, m2+ 2 = 0m2= 3, m2=−2m =± 3, m =± −2 m =± i 2 D 17. 3x− 2 > 6x+1−3x > 3x <−1−∞, −1() A 18. 6 − 2x ≤ 3−3 ≤ 6− 2x ≤ 3−9 ≤−2x ≤−392≥ x ≥3232≤ x ≤92 C 19. 2x2− 4x + k=0x =4 ± 16 − 4(2)(k)2(2)=4 ± 16 − 8k4Need : 16− 8k ≥ 0 − 8k ≥−16 k≤ 2 D 520. Let h = height of original triangle then base of original triange = h+3 New: h+3 h Area of new triangle =14 in212h + 3()h + 6()= 14h + 3()h + 6()= 28h2+ 3h + 6h +18 = 28h2+ 9h −10 = 0h +10()h −1()= 0h =−10, h = 1Original height = 1 in.Original base = 1 + 3 = 4 in. A h+3 h+3+3 21. 2x2+ y2=1x − y = 1 ⇒ y = x −1⎧ ⎨ ⎪ ⎩ ⎪ 2x2+ x − 1()2=12x2+ x2− 2x +1 = 13x2− 2x = 0x 3x − 2()= 0x = 0, x =23 B 622. A 1,−2(), Midpoint M 2,3(), Bx, y()Midpoint ⇒ x1+ x22,y1+ y22⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1+ x2,−2 + y2⎛ ⎜ ⎝ ⎞ ⎠ ⎟ ⇒ 2,3()1 + x2Midpoint of AB⇒ 7 = 2, −2 + y2= 3 1+ x = 4, − 2 + y = 6 x = 3, y = 8 so B 3,8() C 23. slope of line ⇒ m =−13slope of line perpendicular ⇒ m = 3 D 24. m =xykz 3 =(4)(2)k6 18 = 8k k =188=94 B 25. 2x− 3y = 7−3y =−2x + 7y =23x −73slope m =23slope of parallel line m =23 point is 4,−1() ; m =23y = mx + b−1 =23⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (4) + b−1 =83+ bb =−113 C so y =23x −11326. Center ⇒ 0, 2() (x−h)2+y−k()2=r2radius = 2 x − 0()2+ y − 2()2= 22 x2+ y− 2()2= 4 x2+ y2− 4y + 4 = 4 x2+ y2− 4y = 0 B 27. f (x) = 1− x, g(x) =1xg o f()(x) = gf(x)[]= g 1 − x()=11 − x D 28. f (x) =xx2+11f (3)=13(3)2+1=1310=103 D 29. Vertex ⇒ V(0,2) y=a(x−h)2+kpoint on parabola ⇒ (1,0) y = a(x − 0)2+ 2 y = ax2+ 2 0 = a(1)2+ 2 a =−2 y =−2x2+ 2 B 89 30. A 31. logby3+ logby2− logby4=logb(y3y2)−logby4= logby5− logby4= logby5y4⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = logby B 32. f(x) = logax if a >1example : if a = 2, then f(x) = log2x, Graph of y = log2x ⇒ 2y= x D y A. y xE. y B. C. y x xD. y x xf is increasing,fdoes not have a as an x - intercept (the x- int. is (1,0)), f does not havea y - intercept, the domain of f is 0,∞(). y x33. log432.095()72.13()⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = log432.095()1272.1()13⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = log432− log .095()1272.1()13⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = log432− log .095()12+ log 72.1()13⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = log432−12log.095 −13log72.1 B 34. 3x− 5= 4log3x− 5= log4x − 5()log3 = log4x − 5 =log4log3x =log4log3+ 5 C 35. log32x + 3 = 232= 2x + 32x +3 = 92x + 3()2= (9)2 Check : 2(39) + 3 = 92x + 3 = 81 9 = 92x = 78 Check : log32(39) + 3 = 2x = 39 32= 81 C 1036. log3m = 8 log3mnp3⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = log3mn()12− log3p3log3n =10 ⇒ = log3m12n12⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ − log3p3log3p = 6 = log3m12+ log3n12− log3p3 =12log3m +12log3n − 3log3p =12(8) +12(10) − 3(6 ) = 4 + 5− 18 =−9 A 37. D y = 2 + 2xWhen x = 0, y = 2 + 20 y = 2 +1= 3 38. y = x2x − 1()x +1()2x - intercepts: x2x −1()x …
View Full Document
Unlocking...