Each question is worth 5 points. To receive full credit for a correct answer please, show your work and processes in solving each question. Total points = 125. 1- Simply the expression: (5X3); x y 2- Find the domain and range of the function f( x ) = x + 4 For questions 3 10, Solve for each equation, inequality, or system of equations: 3- 2(x5)+10=3(x+2) 4- - 2+5x7=o 5- 3x2-7x+2=0 6- x"'- 6x1" + s = o For questions 11 13, Graph each equation, function or system. If two functions are indicated, graph both together. 11- 100x2+ y2 = 25 12- 2xyz4 x<2 13- f(x)'=x'4x5 14- Letf(x)= x22x+1andg(x)=x-1. a) Findf(|(2)). b) Findg(f(-1). c) Findf(2)g(3). d) Find f(1)-g(2). e) Findf(x),lg(x). 15- 16- 17- 18- 19- It is a known fact that the snout of a Florida crocodile that is seen above water varies directly with its body length. If you see a crocodile's snout above water that is 2 feet long, the normal body length is really 7 feet. Therefore, if you see a snout above water while canoeing that is approximately 3 '/2 feet, how long to you estimate the crocodile's length? (Bonus question: What do you do next if you are in the canoe?) Write the linear function in slope intercept format of the line passing through A(1,3)andB(4,4). Find the distance between the points A ( 2, 3 ) and B (-1, 1 ). ' Write an equation for the line that passes through (3, 3 ) and is parallel to the line with the equation 4x + y - 5 = 0. (10 points) Take the indicated quadratic inequality xz 13x + 36 5 0. a) Determine the type of graph. b) Determine whether it goes up, or down. c) Determine its size. d) Graph the equation. e) Dene the vertex. f) Dene the x - intercept. 9) Dene the y intercept. h) Dene the focal point. Find the average rate of change of the function f( x ) = 3x2 + x from x1= 5 to X2 = 6. 21- Find the accumulated value of an investment of $680 @ 4% compounded monthly for 17 years. Hint: Use the formula on page 407. 22- A paper manufacturing company recycles paper. cans and other sheet metal. The prot on the paper is $500 and the prot on the cans is $350 per pound. a) Write the objective function that models the daily prot. .1 ' b) The manufacture is bound by the following constraints, namely: 0 Equipment in the factory allows at most a total of 360 lbs of paper and cans a day at the same time. 0 Equipment in the factory allows for recycling at most 200 lbs of paper in a day. 0 Equipment in the factory requires at least 10 lbs of paper and 80 lbs of cans to operate the recycling machinery. 0 Equipment in the factory allows for recycling at most 300 lbs of cans per day c) Graph the inequalities from the equations in b. d) Evaluate the objective function at each of the vertices and determine the mix of cans and paper for maximum prot. 23- Suppose we select without looking, one marble from a bag containing 6 red marbles and 8 blue marbles. a) What is the probability of selecting a blue marble on the rst try? b) If you return the marble to the bag, what is the probability of selecting a red marble? c) If you rst selected a red marble and you don't return that marble to the bag, what is the probability of the second draw being a red marble? d) If you rst selected a red marble and you don't return that marble to the bag, what is the probability of the second draw being a blue marble? e) Starting over with all the marbles in the bag, how many draws would you need to guarantee blue marble? 24- You and your wife are planning on having three kids. a) How many different combinations of kids can you have? b) Diagram all the various permutations. c) What is the probability of having all three as girls? d) What is the probability of having only one boy? 6 2 Q1) ( ) xy 4 x y 1 2 y5 3 x ( ) = y5 x3 2 ( ) 1 10 = (y ) 6 x y 10 [ = x6 Q2) at f(x) = 0 0= x+4 x 21 0=x+4 X = -4 F(x) 4 Q3) -2(x-5) + 10 = 3(x+2) -2x + 10 + 10 = 3x + 6 -2x + 20 = 3x + 6 5x = 14 X = 2.8 Q4) -x2 + 5x -7 = 0 Divide all terms by negative X2 - 5x + 7 = 0 + 2 X =-b b 4 ac 2a + 2 X =5 5 417 21 + =5 3 2 i= 1 + =5 1 3 2 + =5 i 3 2 X= 5+ i 3 2 OR X= 5i 3 2 Q5) 3x2 - 7x + 2 = 0 3x2 - 6x -x + 2 = 0 3x(x-2) -1(x-2) = 0 (3x-1)(x-2) = 0 X = 1/3 oR x=2 Q6) x1/2 - 6x1/4 + 8 = 0 Multiply each power by 4 X2 - 6x + 4096 = 0 + 2 X =-b b 4 ac 2a + 2 X =6 6 414096 21 + =6 16348 2 i= 1 + =5 1 16348 2 + =5 i 16348 2 X= 5+ i 16348 2 OR X= 5i 16348 2 Q7) 2 x +4 -x+3-1=0 Squaring each term 2x + 4 - x2 + 4 = 0 -x2 + 2x + 8 = 0 X2 - 2x - 8 = 0 X(x-4) + 2(x-4) = 0 (x+ 2)(x-4) = 0 X = -2 OR X=4 Q8) 6x2 - 6 < 5x 6x2 - 5x - 6 < 0 6x2 - 9x + 4x - 6 < 0 3x(2x - 3) + 2(2x-3) < 0 (3x+2)(2x-3) < 0 X< 2 3 OR X< 3 2 Q9) 4x2 + 3y2 = 48 3x2 + 2y2 = 35 Solving by elimination (4x2 + 3y2 = 48)3 (3x2 + 2y2 = 35)4 12x2 + 9y2 = 144 12x2 + 8y2 = 140 Finding the difference Y2 = 4 + Y= 2 Substituting to 4x2 + 3y2 = 48 4x2 + 3* 4 = 48 4x2 = 36 X2 = 9 X= 3 Q10) X2 - x - y = -1 -x + y = 1 Solving by substitution Y=1+x X2 - x - (1+x) = -1 X2 - 2x = 0 X2 = 2x X=2 Substitute in -x + y = 1 -2 + y = 1 Y=3 Q11) 100x2 + y2 = 25 Y2 = 25 - 100x2 X2 Y2 0 25 1 -75 4 -375 9 -875 16 -1575 12 10 8 y2 6 4 2 0 0 f(x) = 2 4 6 8 10 12 14 16 18 x2 Q12) 2x - y Y 4 2 x4 x y -2 -8 -2.5 -2 f(x) = 2x -1.5- 4 -1 -6 -1 0 -4 -0.5 0 -1 0 -2 -3 y -4 -5 -6 -7 -8 -9 x 1 -2 0.5 1 2 0 1.5 2 2.5 Q13) f(x) = x2 - 4x - 5 x y -4 27 -2 -7 0 -5 2 -9 4 -5 30 25 20 15 f(x) = - 3.3x + 0.2 10 f(x) Y-Values Linear (Y-Values) 5 -5 -4 -3 -2 -1 0 -5 0 -10 -15 y Q14a) f(x) = x2 - 2x + 1 g(x) = x - 1 f(g(2)) = f(2-1) = f(1) = 12 - 2*1 + 1 =1-2+1 =0 b) g(f(-1)) = g(12 - 2* -1 + 1 = g(4) g(4) = 4 -1 =3 c) f(2) - g(3) = (22 - 2* + 1) - (3-1) 1 2 3 4 5 = 1- 2 = -1 d) f(-1) * g(-2) = (-1)2 - 2*-1 * -2 -1 = 4 * -3 = -12 g ( x) e) f(x) 2 = x 2 x +1 x 1 = x 2xx +1 x1 = x ( x1)1( x +1) x1 ( x1)1(x1) = x1 =x-1 Q15) 2ft long = 7ft length 3.5ft long = ? length = 3.57 2 = 12.25 ft When you are in the canoe you will drive away along 12.25 ft before driving forward Q16) A(1,3) and B(4,-1) G= y 31 = x 14 4 = 3 Linear equation 4 3 = y 3 x1 3(y-3) = -4(x-1) 3y - 9 = -4x + 4 3y = -4x + 4 = 9 Y= 4 13 x+ 3 3 Q17) B A AB = 3 unit 4 2+3 2 = 5 unit Q18) parallel line gradient are equal 4x + y - 5 = 0 Y = -4x + 5 Gradient = -4 4 1 = y 3 x3 (y-3) = -4(x-3) y - 3 = -4x + 12 y = -4x + 12- 3 Y= 4 x +15 Q19a) x2 - 13x + 36 Parabolic graph b) it goes up 0 4 unit 0 c) x2 - 13x + 36 0 x2 - 4x - 9x + 36 x(x-40 - 9(x-4) = 0 (x-9)(x-4) = 0 X = 9 OR x = 4 d) y = x2 -13x + 36 x y -2 66 -1 50 0 36 1 24 2 14 70 60 50 40 y 30 20 10 -4 -2 0 0 2 4 6 x e) vortex h= b 13 = 2 a 21 = 6.5 k = 6.52 - 13*6.5 + 36 = -6.25 Vortex is at (6.5, -6.25) f) x - intercept x = 9 0r x = 4 8 10 12 14 16 g) y - intercept y = 38 h) diameter of the parabola = 17 unit radius = 8.5 unit square of radius = 72.25 unit square depth = 63 unit and multiply with 4 = 4*63 = 252 units divide = 72.25/252 = 0.2867 Q20) f(x) = 3x2 + x = f ( 6 )f (5) ( 6 )(5) (362 +6)(35 2+ 6) = ( 6 )(5) = 114 81 1 = 33 Q21) A = P(1+r/n)nt P = $680 R = 0.04 t= 17 years n = 12 A = $680(1 + 0.04/12)12*17 = $1340.72 Q22a) let x represent recycle papers and y = cans X + y = $500 + $350 $500X + $350y = p 360 lb b) x +y x 200 lb x+y 10+ 80 x+y 90 y 300 x 0 y 0 c) d) max profit is at vertex point (360,0) $500 * 360 + 0 = $180,000 Q23a) total number of marbles = 8 + 6 = 14 P(blue) = b) P(red) = 8 4 = 14 7 6 3 = 14 7 c) p(2nd red) = 6 13 d) p(2nd blue) = 8 13 e) drawing red marble = 6 drawing blue marble = 8 at least one of each you draw one more 6 + 8 + 1 = 15 Q24a) 3! = 3*2*1 = 6 b) sample space ggb, bbb, bbg, ggg, gbg, bgb c) P = 1/3 * 1/3 *1/3 = 1/27 d) P = 1/3 x +y x 360 lb 200 lb x+y 10+ 80 x+y 90 y 300 x 0 y 0 check out x +y 360 lb X Y 0 360 360 0 Plot the values During shading you pick to points on left example (0,0) and right example (400,0), and substitute You will find out that the point (0,0) is accepted as per in inequality since 0 is less than 300, so we shade the region we don't require which is the right hand region Same applies to x + y X Y 90 0 90 x 200 lb y 300 x 0 y 0 make the line pass through the respective values on x axis and y axis 90 0 shading the regions Note also, an inequality like x 200 lb means we are interested on values that are less than 200 lb so we will shed the region which we are not interested on that will be values greater than 200 lb, that's why I shaded on my right hand on that line, Same will apply on the other single inequalities \f6 2 Q1) ( ) xy 4 x y 1 2 y5 3 x ( ) = y5 x3 2 ( ) 1 10 = (y ) 6 x y 10 [ = x6 Q2) at f(x) = 0 0= x+4 x 21 0=x+4 X = -4 F(x) 4 Q3) -2(x-5) + 10 = 3(x+2) -2x + 10 + 10 = 3x + 6 -2x + 20 = 3x + 6 5x = 14 X = 2.8 Q4) -x2 + 5x -7 = 0 Divide all terms by negative X2 - 5x + 7 = 0 + 2 X =-b b 4 ac 2a + 2 X =5 5 417 21 + =5 3 2 i= 1 + =5 1 3 2 + =5 i 3 2 X= 5+ i 3 2 OR X= 5i 3 2 Q5) 3x2 - 7x + 2 = 0 3x2 - 6x -x + 2 = 0 3x(x-2) -1(x-2) = 0 (3x-1)(x-2) = 0 X = 1/3 oR x=2 Q6) x1/2 - 6x1/4 + 8 = 0 Multiply each power by 4 X2 - 6x + 4096 = 0 + 2 X =-b b 4 ac 2a + 2 X =6 6 414096 21 + =6 16348 2 i= 1 + =5 1 16348 2 + =5 i 16348 2 X= 5+ i 16348 2 OR X= 5i 16348 2 Q7) 2 x +4 -x+3-1=0 Squaring each term 2x + 4 - x2 + 4 = 0 -x2 + 2x + 8 = 0 X2 - 2x - 8 = 0 X(x-4) + 2(x-4) = 0 (x+ 2)(x-4) = 0 X = -2 OR X=4 Q8) 6x2 - 6 < 5x 6x2 - 5x - 6 < 0 6x2 - 9x + 4x - 6 < 0 3x(2x - 3) + 2(2x-3) < 0 (3x+2)(2x-3) < 0 X< 2 3 OR X< 3 2 Q9) 4x2 + 3y2 = 48 3x2 + 2y2 = 35 Solving by elimination (4x2 + 3y2 = 48)3 (3x2 + 2y2 = 35)4 12x2 + 9y2 = 144 12x2 + 8y2 = 140 Finding the difference Y2 = 4 + Y= 2 Substituting to 4x2 + 3y2 = 48 4x2 + 3* 4 = 48 4x2 = 36 X2 = 9 X= 3 Q10) X2 - x - y = -1 -x + y = 1 Solving by substitution Y=1+x X2 - x - (1+x) = -1 X2 - 2x = 0 X2 = 2x X=2 Substitute in -x + y = 1 -2 + y = 1 Y=3 Q11) 100x2 + y2 = 25 Y2 = 25 - 100x2 X2 Y2 0 25 1 -75 4 -375 9 -875 16 -1575 12 10 8 y2 6 4 2 0 0 f(x) = 2 4 6 8 10 12 14 16 18 x2 Q12) 2x - y Y 4 2 x4 x y -2 -8 -2.5 -2 f(x) = 2x -1.5- 4 -1 -6 -1 0 -4 -0.5 0 -1 0 -2 -3 y -4 -5 -6 -7 -8 -9 x 1 -2 0.5 1 2 0 1.5 2 2.5 Q13) f(x) = x2 - 4x - 5 x y -4 27 -2 -7 0 -5 2 -9 4 -5 30 25 20 15 f(x) = - 3.3x + 0.2 10 f(x) Y-Values Linear (Y-Values) 5 -5 -4 -3 -2 -1 0 -5 0 -10 -15 y Q14a) f(x) = x2 - 2x + 1 g(x) = x - 1 f(g(2)) = f(2-1) = f(1) = 12 - 2*1 + 1 =1-2+1 =0 b) g(f(-1)) = g(12 - 2* -1 + 1 = g(4) g(4) = 4 -1 =3 c) f(2) - g(3) = (22 - 2* + 1) - (3-1) 1 2 3 4 5 = 1- 2 = -1 d) f(-1) * g(-2) = (-1)2 - 2*-1 * -2 -1 = 4 * -3 = -12 g ( x) e) f(x) 2 = x 2 x +1 x 1 = x 2xx +1 x1 = x ( x1)1( x +1) x1 ( x1)1(x1) = x1 =x-1 Q15) 2ft long = 7ft length 3.5ft long = ? length = 3.57 2 = 12.25 ft When you are in the canoe you will drive away along 12.25 ft before driving forward Q16) A(1,3) and B(4,-1) G= y 31 = x 14 4 = 3 Linear equation 4 3 = y 3 x1 3(y-3) = -4(x-1) 3y - 9 = -4x + 4 3y = -4x + 4 = 9 Y= 4 13 x+ 3 3 Q17) B A AB = 3 unit 4 2+3 2 = 5 unit Q18) parallel line gradient are equal 4x + y - 5 = 0 Y = -4x + 5 Gradient = -4 4 1 = y 3 x3 (y-3) = -4(x-3) y - 3 = -4x + 12 y = -4x + 12- 3 Y= 4 x +15 Q19a) x2 - 13x + 36 Parabolic graph b) it goes up 0 4 unit 0 c) x2 - 13x + 36 0 x2 - 4x - 9x + 36 x(x-40 - 9(x-4) = 0 (x-9)(x-4) = 0 X = 9 OR x = 4 d) y = x2 -13x + 36 x y -2 66 -1 50 0 36 1 24 2 14 70 60 50 40 y 30 20 10 -4 -2 0 0 2 4 6 x e) vortex h= b 13 = 2 a 21 = 6.5 k = 6.52 - 13*6.5 + 36 = -6.25 Vortex is at (6.5, -6.25) f) x - intercept x = 9 0r x = 4 8 10 12 14 16 g) y - intercept y = 38 h) diameter of the parabola = 17 unit radius = 8.5 unit square of radius = 72.25 unit square depth = 63 unit and multiply with 4 = 4*63 = 252 units divide = 72.25/252 = 0.2867 Q20) f(x) = 3x2 + x = f ( 6 )f (5) ( 6 )(5) (362 +6)(35 2+ 6) = ( 6 )(5) = 114 81 1 = 33 Q21) A = P(1+r/n)nt P = $680 R = 0.04 t= 17 years n = 12 A = $680(1 + 0.04/12)12*17 = $1340.72 Q22a) let x represent recycle papers and y = cans X + y = $500 + $350 $500X + $350y = p 360 lb b) x +y x 200 lb x+y 10+ 80 x+y 90 y 300 x 0 y 0 c) d) max profit is at vertex point (360,0) $500 * 360 + 0 = $180,000 Q23a) total number of marbles = 8 + 6 = 14 P(blue) = b) P(red) = 8 4 = 14 7 6 3 = 14 7 c) p(2nd red) = 6 13 d) p(2nd blue) = 8 13 e) drawing red marble = 6 drawing blue marble = 8 at least one of each you draw one more 6 + 8 + 1 = 15 Q24a) 3! = 3*2*1 = 6 b) sample space ggb, bbb, bbg, ggg, gbg, bgb c) P = 1/3 * 1/3 *1/3 = 1/27 d) P = 1/3 x +y x 360 lb 200 lb x+y 10+ 80 x+y 90 y 300 x 0 y 0 check out x +y 360 lb X Y 0 360 360 0 Plot the values During shading you pick to points on left example (0,0) and right example (400,0), and substitute You will find out that the point (0,0) is accepted as per in inequality since 0 is less than 300, so we shade the region we don't require which is the right hand region Same applies to x + y X Y 90 0 90 x 200 lb y 300 x 0 y 0 make the line pass through the respective values on x axis and y axis 90 0 shading the regions Note also, an inequality like x 200 lb means we are interested on values that are less than 200 lb so we will shed the region which we are not interested on that will be values greater than 200 lb, that's why I shaded on my right hand on that line, Same will apply on the other single inequalities \f