MONROE COMMUNITY COLLEGE ROCHESTER, NY MTH 104 INTERMEDIATE ALGEBRA DETAILED SOLUTIONS. 4 th edition SPRING 2010

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MONROE COMMUNITY COLLEGE ROCHESTER, NY * MTH 04 INTERMEDIATE ALGEBRA FINAL EXAM REVIEW DETAILED SOLUTIONS * 4 th edition SPRING 00 Solutions Prepred by Prof. Jn (Yhn) Wirnowski

Evlute the function in problems: 3. R(x) - x 4x + 8; R() - () 4() + 8-4 8 + 8-4; *R() - 4* f(x) x 3x + 7; f(-) (-) 3(-) + 7 4 + 6 + 7 7; ff(-) 7 3 g(x) x 3 5x; g(-) (-) 3 5(-) -6 + 0-6 *g(-) - 6* Simplify problems: 4. Assume ll vribles re positive. 4 5 5 3 3 3 5 3 3 8 0 (7 x y ) (7) ( x ) ( y ) 3 3 3 ( 7 ) x y 6 (3 ) (3 ) (3 )(3 ) 3 9 8 0 9x y 7 8 3 3 3 4 4 ( 6y 4 ) ( 6 4 ) ( ) ( ) ( ) 4 y 6 y EXTRA: 3 3 ( y ) 3 3 3 8 y9 8 y 4 3 4 3 3 b 3 3 7 7 3 b b b -3 7 b4 Note: -3 moved to the numertor becomes 3 ; b 3 moved to the denomintor becomes b -3 ; when the bse is moved from the denomintor to numertor or vice vers, the sign of the exponent chnges to the opposite. 0 5 5 75 b 5i3 b 5 b 7 6 3 8 9 5 3 3 8x y z 5 3 4 5 ( ) ( x ) ( y ) ( z ) ( 5) 3 3 3 9 5 ( 8) ( x ) ( y ) ( z ) 8 9 ( ) ( ) ( ) ( 5) 3 4x y z 5 6 8 0 3 4 4 4 4 4 4 b 8 8 4 8 b b b 4 8 b 3 5 7 3 5 7 4 50x y 5i x y 8 b 8-5 xy x y -5 xy 3 x y Note bout problem # 0: Divide the rdicl s index (index is not mrked over the squre roots) into the exponents under the rdicl. The quotients resulting from tht division become exponents over nd b (7 nd 6, respectively) outside the rdicl nd the reminder becomes the non-written exponent over, under the rdicl. Problem # is hndled in similr wy. 98 98 49i 7i i i Fctor completely problems: 3 9. 3 xz x yz y x z y z 3x -y 5z + (Fctored by grouping) 5 + 6 5 3 (5 + ) (5 + ) ( )( ) 4 4b + 4b -b ( ) Perfect squre trinomil Squre of binomil 8/7/0 jw

5 ( )( ) 5x 4x 8 5x 4 x + Tril nd Error AC Method (By Grouping): 5 8 40; 40-0 4 5x 0x 4x + 8 5x (x ) 4 (x ) ((5x 4) (x )) 6 00 y 49 (0y) (7) ((0 y + 7) (0 y 7)) Difference of two squres yields product of the sum nd difference of two terms 7 36 + 4b +b 6[6 + 7b +b ] *6 (3 + b) ( + b)) 8 5 3 + 8 (5) 3 + () 3 [(5) + ()] [(5) (5)() + () ] ((5 + ) (5 0 +4)) 9 64m 4 7mn 3 m[(4m) 3 (3n) 3 ] m[(4m) (3n)] [(4m) + (4m)(3n) + (3n) ] *m(4m 3n) (6m +mn + 9n )) Perform the indicted opertions in problems: 0 8. Assume ll vribles re positive rel numbers. 0 (x+3) (x 3 x + ) x 4 x +x + 3x 3 6x +3 xx 4 +3x 3 x 5x +33 (3 + ) (3) +(3)() + () 99 + + 44 (4y 3) (4y + 3) (4y) (3) 6y 99 Product of the sum nd difference of two terms Difference of two squres 3 3 4 3 3 4 + i x + x + x + x x + x x + x 5 5 5 3 ( 5)( 3) 3x 9 4 3x + 5 + ( x + 5)( x 3) ( x + 5)( x 3) ( x + 5)( x 3) 3 3 ( x + 5) ( x + 5) ( 3) x -3 x 4 4 3x + 4 4 x 5 3x + 4 x 4x 0 3x 6x + 4x 8 i i x x 5 x x 5 x 5 x ( x )( x 5) ( x )( x 5) 4x 0 3x x 8 (4x 0) (3x x 8) 4x 0 3x + x + 8 ( x )( x 5) ( x )( x 5) ( x )( x 5) ( x )( x 5) -3x +6x- -3(x -x +4) (x-)(x-5) ( x-)(x-5) 5 3 3 3 5 3 3 5 4i3 x + x x x x + x x 3x 3x 3x 3x + 0x 3x 6 5 3 7 5 6i 36i 5i 4 6 0 6 4 8/7/0 jw

3 5x 3 3 3 3 3 7 40x 5x 3 8i5x 5x 5x 5x 8 + 3 8 + 4 + 3 8 + 0 + 4 ( + 6)( 3) x x x + x + x x x + x x + x + x x x x x x x x x x 0 4 4 ( 3) ( 4) ( x + 6) ( x 3) ( x 3) ( x 4) x +6 - x-4 ( x + 4) ( x + 6) ( x + 6) ( x 4) Divide using long division: 9 3. 9 (x -x +3) (x 4) x 8 x 4 x x + 3 x 4x 8x + 3 8x + 3 R 0 *Ans.: x 8* 30 3 x x x + 5 6 x Ans.: x + 4x + 3 x + 4x + 3 3 x x + x 5x 6 x x 3 4x 5 4x 8 x x 3x 6 3x 6 R 0 3 (x +0x -5) (x +) 6x + x x x + + 0 5 x + 6 x 4x 5 4x + R 7 7 Ans.: 6x + - x + 3 Simplify the frction: 3 34. 8 8 x + (Rtionlize x x x + 8 x +6 8( x + ) denomintor) x -4 ( x ) () 33 Symplify: 9 9 y y y 9 y 3 3 + + y y + 3y y y (3 + y) (3 y) y ( y + 3) 3- y y 34 4 4 y y 3 y y 3 y (y 3) (y 3) 4y y 6 4y 3 3 + + y (y 3) (y 3) + 3y y 3 + 3y y y 3 y y 3 -y-6 y +6 OR - 4y-3 4y-3 8/7/0 jw 3

MTH 04 Finl Exm Review Detiled Solutions MCC 35 37 Solve the eqution. For the specified vrible : 35 37.? v t + v0 v v0 t + v0 v0 v v t r? 0 v -v 0 t S S - r - r - r 36 mv + mp bv v(m - b) - mp mv + mp - mp bv - mp v? mv bv - mp mv - bv - mp - r S S - - S r - S S m p v - m - b m p b- m S - S r S r S - Solve the eqution. Include ny complex solutions: 38 43. 38 ( ) ( ) 5x + 4 5x + 4 5x + 6 5x 5 x 3 Must check: 5(3) + 4(?) 6 4 4 4 Solution set 3 { } 39 5+ 4x+ 8 4x+ 8 6 ( x ) ( ) 4 + 8 6 4x+ 8 36 4x 8 x 7 Must check: 5+ 4(7) + 8 (?) 5+ 36 5+ 6 Solution set 7 { } 40 4 Multiply both sides of the eqution by LCD to eliminte the denomintors. 3 8 3 8 + + LCD (x + 5)(x 5) x 5 x + 5 x 5 (x + 5)(x 5) x + 5 x 5 3 8 + i (x + 5)(x 5) 3 + (x 5) 8(x + 5) + x 8x + 40 (x + 5)(x 5) x + 5 x 5 4 x 6; D/C 7x 4 x (Denomintor check): (-6)+5 0, (-6)-5 0; 7 Solution set -6 { } Multiply both sides of the eqution by LCD to eliminte the denomintors. 6 4 i( 6) 6 4( 6) 6 4 4 6 3 4 3 8 6 6 8 6 D/C : Solution set (Denomintor check) (6) 6 0 { } 3 Since the denomintor 0, nd since division by zero yields undefined result, there is no solution to the eqution. 8/7/0 jw 4

MTH 04 Finl Exm Review Detiled Solutions MCC 4 x + 96 0 x 96 x ± 96 x ± i 96 x ± 4i Solution set -4i,4i { } 43 3x + 3 3x + 4 3x + 4 or 3x + 4 Solution set, x or x ; 3 3 Solve ech inequlity, express the solution using set nottion, intervl nottion, nd grph the solution on the rel number line (44 5). < 3x + 7 < 4 7 7 7 44 ( ) 9< 3x < 3 (3) 3< x < Intervl nottion : -3,- -3,- -3,- -3,- 45 AND Inequlity Tip is in, expression in between x 3 5 5 x 3 5 x 8 [ -,8 ] Intervl nottion : Set nottion: { x -3 < x -} -3 - - 8 Set nottion: { x - x 8}} OR Inequlity Tip is out, expression left nd right. x OR x > 7 46 5 x 5 x or 5 x x 7 or x 3 Multiply by - nd reverse the inequlities. x 7 or x 3,3 ] [ 7, Intervl nottion : ] ] - [7, 47 Solution: 7 Set nottion: {{ x x } 3 7 Intervl nottion: [, ) Set nottion: { x x 3 or x 7 }} 48 x < 3 AND x 7 7 Solution: 3 Set nottion: {{ x x < 3 }} Intervl nottion: (-, 3) 3 8/7/0 jw 5 49 Solution: x 4 OR x < 7 4 7 All rel numbers Set nottion: { x x is rel number }} Intervl nottion: ((-, )

x < 3 AND x > 0 x < 0 OR x 6 3 0 50 0 5 6 Solution: No solution Solution: 0 6 Set nottion: { } Ø Set nottion: {{ x x < 0 or x 6 }} Intervl nottion: (( -, 0 ) [ 6, ) 5 Solution: x 0 AND x > - - - 0-0 Set nottion: { x - < x 0 }} Intervl nottion: ((-, 0]] Solve the qudrtic eqution by completing the squre. Express the solution in set nottion: (53 54). 53 ( ) x 6x 6 0 x 6x 6 6 6 x 6x 6 + + x 6x + 3 6 + 3 (x 3) 6 + 9 (x 3) 5 (x 3) 5 ± (x 3) ± 5 x 3 5 or x 3 5 Solution set -,8 x or x 8 { } 54 0 0 y + 0y + 0 y + 0y y + 0y + + ( ) (y 5) 3 (y 5) 3 y 5 3 or y 5 3 { } y + 0y + 5 + (5) (y + 5) + 5 (y + 5) 3 + ± + ± + + y 5 3 or y 5 + 3 Solution set -5-3,-5 + 3 8/7/0 jw 6

Solve the qudrtic eqution by using the qudrtic formul. Express the solution in set nottion. (55 57). Use Qudrtic Formul The Qudrtic Formul ± b b 4c x is used in the next three problems: 55, 56, 57 Discriminnt b - 4c ( Expression under the rdicl) is used in problems 58 nd 59. 55 4x 3 8x 4x 8x 3 0 + (8) (8) 4(4)( 3) ± x (4) 8 ± 64 + 48 8 ± 8 ± 6i7 8 8 8 i 4 ±i 4 7 ± 7 i 4 - - 7 - + 7 Solution set, Use Qudrtic Formul 0 5 4 4 0 + 5 0 Use Qudrtic Formul x x 0 + + ± (4) ( 0) ( 0) 4(4)(5) () () 4()() ± x () 56 0 ± 400 400 0 ± 0 0 5 8 8 8 5 Solution set 57 ± 4 8 ± 4 ± i 4 i ± i i ± i ± i i Solution set - -i, - + i { } Use the discriminnt to determine whether the qudrtic eqution hs one rel number solution, two rel number solutions, or two complex number solutions: (58 59). 58 9x + 30x + 5 0 Discriminnt b 4c (30) 4(9)(5) 900 900 0 The Qudrtic Formul will yield 59 3t t + 0 Discriminnt b 4c ( ) 4(3)() 4 3 The Qudrtic Formul will yield only one rel solution. two complex solutions. 8/7/0 jw 7

Perform the indicted opertions nd simplify. Write the result in + bi form: (60 63). 60 5i (3 i) 5i 0i 5i + 0 i0 + 5ii 6 (3+i) (3) + (3)(i) + (i) 9 + i + 4i 9 + i 4 i5 + ii 6 3 + i 3 + i 7 + 6i + 8i + 4i + i i 7 6i 7 6i 7 + 6i (7) (6i) 9 3 + 3i 9 + 3i 9 + 3i + i 85 85 63 49 36i 49 + 36 85 4 4 3 + 7i + 8i i 3 7i 3 7i 3 + 7i (3) (7i) 8 + 8i + 8i + 8i + 58 58 9 49i 9 + 49 58 6 4 + i 9 9 64 Use clcultor to evlute to four deciml plces: ccos 80 0.7366 65 Use clcultor to evlute to four deciml plces: ttn 4. 0.5 66 (i) (ii) Solve using right tringle ABC with C 90 : problems 66 70. Use exct vlues in your nswers (leve rdicls, no decimls), unless otherwise indicted. c 3cm, b cm,? c Find the exct vlue of side B 5 c + b c b c b 3 Find the exct vlue of : sin A, cos A, tn A 5 5 b sin A sin A ; cos A cos A ; tn A tn A c 3 c 3 b (iii) Find, to the nerest tenth : A nd B 5 OR 3 3 From clcultor : A cos 48. A sin 48. From the tringle : since A + B 90 B 90 A B 90 48. 4.8 A b C A 48. o B 4.8 o nd 8/7/0 jw 8

67 A 8 b 5 Find vlues of sides:? nd c? (To the nerest tenth.) tna b tna 5 tn8 5 b (0.537).7 b b 5 5 cos A c c c c 5.7 c cos A cos 8 0.889 A c b B C 68, b 5; Find: sin A? cos A? tn A? (exct vlues, no decimls) c 9 c + b c () + (5) c 4 + 5 c 9 9 sin A i c 9 9 9 9 b 5 9 5 9 cos A i c 9 9 9 tn A tn A 5 b 5 9 sin A 9 5 9 cos A 9 A c b B C Hypotenuse c, oneleg : 7; Find the three trig functions of the smller ngle. sin A? cos A? tn A? (To four deciml plces.) 69 c + b b c b c b () (7) b 6 b 7 b 36i 7 sin A 0.6364 sin A 0.6364 c 6 b cos A 0.774 cos A 0.774 c 7 0.850 tn A 0.850 tn A 6 b A c b B C B 4, 9; Find ll other ngles nd sides; round to one deciml plce: B 70 A 90 4 48 Finding side c 9 9 9. sin A cos B sin 48 c c c sin 48 0.743 b b 8. tn B tn 4 b 9i tn 4 9 i(0.9004) 9 A NSWERS : A 48 o, b 8. c. A, c b C 8/7/0 jw 9

Find the eqution of the line tht stisfies the informtion: problems 7 73. Write your nswers in slope-intercept form. Eqution of the line through the point (4,-), with slope of ¾; in slope-intercept form. 7 3 3 3 y y m(x x ) y ( ) (x 4) y + (x 4) y + x 3 4 4 4 3 y x-4 4 7 Eqution of the line through the points (-, ) nd (, 3); in slope-intercept form. y y 3 m m m m 3 x x ( ) + y y m(x x ) y 3 (x ) y 3 x y x + 3 3 3 3 3 3 8 y x + 3 3 Eqution of the line through the point (3, ) nd perpendiculr to the line x + y - 3; in slopeintercept form. l x + y 3 y x 3 m l m m m m ( ) l 3 3 y y m(x x ) y (x 3) y x y x + y x- 73 Eqution of the line through the point (3, ) nd prllel to the line x + y - 3; in slope-intercept form. l x + y 3 y x 3 m l m m m l y y m(x x ) y (x 3) y x + 6 y x + 6+ y -x +7 8/7/0 jw 0

74. Grph the solution to the inequlity *x + y > 6* FOUR STEPS Y (0,3). Write the corresponding eqution, in order to drw the boundry line: x + y 6; find the intercepts (0,3) nd (6,0) nd plot the points.. Drw dshed/dotted boundry line. (0,0) (6,0) X 3. Check if the point (0,0) stisfies the inequlity: (0) + (0) > 6? It does not. 4. Hence, shde the solution to the inequlity on the other side of the boundry line thn the point (0,0) is found. 75. Grph the solution to the inequlity *3x - 5y < 5* FOUR STEPS Y (0,0) (5,0) X. Write the corresponding eqution, in order to drw the boundry line: 3x - 5y 5; find the intercepts (0,-3) nd (5,0) nd plot the points.. Drw dshed/dotted boundry line. 3. Check if the point (0,0) stisfies the inequlity: (0,-3) 3(0) - 5(0) < 5? It does. 4. Hence, shde the solution to the inequlity on the sme side of the boundry line tht the point (0,0) is found. 76. Given the eqution y x + 6x +55, do the following: Find the vertex nd the eqution of the xis of symmetry 4 Grph the qudrtic eqution on the coordinte xes below nd lbel the points with the coordintes. Stte the y-intercept. 5 NOTE: Since > 0, prbol opens up. 3 Stte the x-intercepts, if ny. SOLUTION:. VERTEX (VTX) nd the Eqution of Symmetry When the prbol opens upwrd, the vertex is the lowest point on the curve. 8/7/0 jw

Find the vertex coordintes: b (6) x 3 ; *x - 3 is both the x-coordinte of the VERTEX nd the eqution of the xis () of symmetry of the prbol. To find the y-coordinte, substitute x -3 in the eqution: y (-3) (-3) + 6 (-3) + 5-4 Hence, the VTX coordintes re: (-3, - 4).. Y-INTERCEPT nd the Y-INTERCEPT S COUSIN POINT To find the y-intercept, it is necessry to set x 0 in the eqution. y x + 6x +5 y (0) (0) + 6 (0) +5 5 Hence, the y-intercept is (0, 5). On the other side of the xis of symmetry, eqully distnced from tht xis is the "cousin point whose coordintes re ( 6, 5); this my lso be determined by inspection. 3. X-INTERCEPTS To find the x-intercepts, it is necessry to set y 0 in the eqution. 0 x + 6x +5 (x + ) (x + 5) 0 So x -, x -5 Hence, the x-intercepts re: (-, 0) nd (-5, 0). IMPORTANT: X-intercepts should only be considered if the eqution is esy to fctor. Otherwise, plotting dditionl points, s shown in step 4 will suffice. 4. PLOTTING ADDITIONAL POINTS x -3 y y x + 6x + 5- (-6,5) (0,5) To plot dditionl points, it is convenient to use tble: X Y (-5,0) (-,0) x - -3-4 -3 ANSWERS: (-4,-3) VTX (-3,-4) (-,-3) Hence, the dditionl points re: (-,-3) nd (-4, -3). These points re lso symmetriclly locted ( cousins ) nd when we know one, we cn, by inspection, determine the other; s it ws done in Step. This completes the grph. Vertex: (-3, -4); xis of symm. :x -3 3 X-intercepts....(-, 0) nd (-5, 0) Y-intercept... (0, 5) 4 The grph is shown bove, in Step #4. 8/7/0 jw

77. Given the eqution *y -x + 4x 3, do the following: Find the vertex nd the eqution of the xis of symmetry 4 Grph the qudrtic eqution on the coordinte xes below nd lbel the points with the coordintes. Stte the y-intercept. 5 NOTE: Since < 0, prbol opens downwrd. 3 Stte the x-intercepts, if ny. SOLUTION:. VERTEX (VTX) nd the Eqution of Symmetry When the prbol opens downwrd, the vertex is the highest point on the curve. Find the vertex coordintes: b (4) x ; x is both the x-coordinte of the VERTEX nd the eqution of the xis of ( ) symmetry of the prbol. To find the y-coordinte, substitute x in the eqution: y () -() + 4 () -3 Hence, the VTX coordintes re: (, ).. Y-INTERCEPT nd the Y-INTERCEPT S COUSIN POINT To find the y-intercept, it is necessry to set x 0 in the eqution. y -x + 4x - 3 y (0) -(0) + 4(0) -3-3 Hence, the y-intercept is (0,-3). On the other side of the xis of symmetry, eqully distnced from tht xis is the "cousin point whose coordintes re (4, -3); this my lso be determined by inspection. 3. X-INTERCEPTS To find the x-intercepts, it is necessry to set y 0 in the eqution. 0 y -x + 4x - 3 (-x +) (x -3) 0 So x, x 3 Hence, the x-intercepts re: (, 0) nd (3, 0). IMPORTANT: X-intercepts should only be considered if the eqution is esy to fctor. Otherwise, plotting dditionl points, s shown in step 4, will suffice. 8/7/0 jw 3

4. PLOTTING ADDITIONAL POINTS Y (0,-3) (,0) (,) VTX (3,0) (4,-3) (-,-8) (5,-8) X X y -x + 4x -33 To plot dditionl points, it is convenient to use tble: X Y - -8 5-8 Hence, the dditionl points re: (-,-8) nd (5, -8). These points re lso symmetriclly locted ( cousins ) nd when we know one, we cn, by inspection, determine the other; s it ws done in Step. This completes the grph. ANSWERS: Vertex: (, ); xis of symm.: x 3 X-intercepts: (, 0) nd (3, 0) Y-intercept... (0, -3) 4 The grph is shown bove, in Step #4. 78. Solve the system of equtions by grphing. y x + x + y - () () (3) (-,-3) (0,) (-,0) (-,-3) (0,-) (-,-) Using the eqution, obtin two ordered pirs, plot the two points nd drw line. Using the eqution, obtin two ordered pirs, plot the two points nd drw line. Solution is where the two lines intersect, t (-,-). The system is consistent. 8/7/0 jw 4

Solve the system of equtions by grphing. 79. x +y 6 y - (½) x+ 3 () () (3) (0,3) (0,3) (0,3) (6,0) (6,0) (6,0) Using the eqution, obtin two ordered pirs, plot the two points nd drw line. Using the eqution, obtin two ordered pirs, plot the two points nd drw line. The two lines intersect on their entire length, becuse they re superposed. Hence, there re n infinite number of solutions. The system is dependent. Solve the system of equtions lgebriclly: problems 80 83. (Use either substitution or ddition/elimintion method). 80 Addition/elimintion method x + 3y 5 x + () 3 x + y 3 x x + 3y 5 ( b) x 4y 6 SOLUTION Sol.set, y y { ( ) } Substitution method x + 3y 5 x + () 3 x x + y 3 x 3 y (3 y) + 3y 5 SOLUTION Sol.set, 6 4y + 3y 5 y { ( ) } Addition/elimintion method 8 x 6y 5 ( ) 4x + y 0 Contrdiction, the system is inconsistent. 4x y 5 4x y 5 The lines re prllel,hence do not intersect. 0 5 No solution. NOTE: Another wy to show tht the system is inconsistent (equtions represent prllel lines: no intersection point), would be to write them in 5 slope-intercept form: sme slopes, different 5 b s. y x - y x - Eqution : 3 6 Eqution : 3 8/7/0 jw 5

8 83 (i) x + y z x y + z (c) x + y 3z 7 (ii) x + y z x y + z ( + b) 3x 3 x - (iii) ( ) + y z (c) ( ) + y 3z 7 Simplify, renme (A, C): (A) y z (C) y 3z 6 (i) 3x y + 4z 5x 3y + 5z (c) 6x y + 3z 5 (ii) ( i ) 3x + y 4z (c) 6x y + 3z 5 (A) 3x z 6 We hve now one eqution with two vribles, x & z; nother is needed. (iii) ( b) 0x + 6y 0z 4 (3c) 8x 6y + 9z 5 (B) 8x z Superpose equtions (A, B): (A) 3x z 6 (B) 8x z (iv) Solve for: z ( A) y + z (C) y 3z 6 z 4 z 4 (v) Sobstitute to nd solve for: ( ) + y (4) y 5 y 3 Solution set -,3,4 { ( ) } Check if the solution stisfies ll three equtions. (iv) Solve for: x ( ia) 3x + z 6 (B) 8x z 5x 5 x (v) Sobstitute x to (A) nd solve for: z (A) 3() z 6 3 z 6 z -3 (v) Sobstitute x & z 3 to, or (c);find y 3() y + 4( 3) y -4 y 8 Solution set,-4,-3 y { ( ) } Check if the solution stisfies ll three equtions. 8/7/0 jw 6

For the following word problems, identify the vrible(s) used, set up eqution(s) nd solve lgebriclly. 84. A 30-foot ldder, lening ginst the side of building, mkes 50 ngle with the ground. How fr up the building does the top of the ldder rech? Express your nswer to the nerest tenth of foot. VARIABLE(S): x distnce from the top of the ldder to the bse of the house. L 30 ft. ( Length of the ldder) L x 50 x sin50 x Lisin50 L x (30) i(0.7660).98 3.0ft. x 3 ft. ANSWER: The top of the ldder reches 3.0 feet up the side of the building. 85. A 70 foot rope is ttched to the top of one of the verticl poles used to hold up circus tent. The other end of the rope is nchored to the ground 40 feet from the bottom of the pole. Wht is the height of the tent nd wht ngle does the rope mke with the tent? Express both nswers to the nerest tenth (remember bout the units). VARIABLE(S): Ө ngle of the rope with the tent h height of the tent 40 4 4 70 7 7 sin θ sin θ sin 34.8 70 ft rope Ө T E N T h 40 ft ANSWERS: The height of the tent is 57.4 ft. The ngle Ө 86. Find three consecutive odd integers such tht seven times the sum of the first two integers is three more thn nine times the third integer. VARIABLES: x the first odd integer; x + the second odd integer; x + 4 the third odd integer 7 (x + (x + )) 9 (x + 4) + 3 7 (x + ) 9 x + 36 + 3 4x + 4 9 x + 39 5x 5 x 5 ANSWER: The integers re: 5, 7 nd 9 8/7/0 jw 7

87. A chemist must mix 8 L of 40% cid solution with some 70% cid solution to get 50% cid solution. How much of the 70% solution should be used? ID RAG (I D) (Rte) (Agent) (Gin) Solution Type Acid % Decimls Solution liters Pure cid liters VARIABLE(S): x liters of 70% sol. 8 + x liters of 50% sol. 40% solution 0.40 8 (0.40) (8) 3. 70% solution 0.70 x 0.70 x 50% solution 0.50 (8 + x) 0.50 (8 + x) ANSWER: Needed re 4 liters of 70% cid solution. THE AMOUNT OF PURE ACID BEFORE MIXING THE AMOUNT OF PURE ACID AFTER MIXING (0.40) (8) + 0.70 x 0.50 (8+ x) 3.0 + 0.70 x 4.00 + 0.50 x 0.0 x 0.80 0.80 8 x 4 0.0 88. A privte irplne leves n irport nd flies due est t 80 mph. Two hours lter, jet leves the sme irport nd flies due est t 900 mph. How long will it tke for the jet to overtke the privte plne? ID RAG (I D) (Rte) (Agent) (Gin) Type of Plne Speed mph Time hrs Distnce mi Privte Airplne 80 t + 80 (t + ) The Jet 900 t 900 t VARIABLE(S): t jet s ctch up time. t + privte irplne trvel time. ANSWER: t ½ hour of ctch up time for the jet. DISTANCE TRAVELED BY PRIVATE AIRPLANE DISTANCE TRAVELED BY JET 360 8 0( t + ) 9 0 0t 80 t + 360 900 t 70 t 360 t t hour. 70 8/7/0 jw 8

89. Jonthn invests $ 7,500 @ 0.4 % simple interest. How much dditionl money must he invest t simple interest rte of 4 % so tht the totl interest erned is % of the totl investment? ID RAG (I D) (Rte) (Agent) (Gin) Type of Account Rte Amount Invested Interest 0.4 % 0.04 7,500 0.04 7,500 4 % 0.4 x 0.4 x % 0. 7,500 + x 0. (7,500 + x ) VARIABLE(S): x $$ invested t 4 % 7,500 + x $$ of totl investment. ANSWER: Jonthon must invest $6,000 in dditionl money. 0. 0 4 7,5 0 0 + 0. 4 x 0. 7,5 0 0 + x 7 8 0 + 0. 4 x 9 0 0 + 0. x ( ) ( ) ( ) ( ) 0 0.0 x 0 x x 6,0 0 0 0.0 90. Flying with the wind, Rob flew 800 miles between Pittsburgh nd Atlnt in 4 hours. The return trip ginst the wind took 5 hours. Find the rte of the plne in clm ir nd the rte of the wind. ID RAG (I D) (Rte) (Agent) (Gin) Direction To Atlnt with the wind From Atlnt ginst the wind Speed mph Time hrs Distnce mi r + w 4 800 r - w 5 800 VARIABLE(S): r the rte of plne. w the rte of wind. ANSWER: 80 mph the rte of plne in clm ir. 0 mph the rte of wind. (TRAVEL RATE) (TRAVEL TIME) TRAVEL DISTANCE r + w 4 8 0 0 D iv id e b o th s id e s b y 4 r + w 0 0 ( ) ( ) ( ) A d d b o th s id e s b r - w 5 8 0 0 D iv id e b o th s id e s b y 5 b r - w 6 0 ( ) ( ) ( ) r 3 6 0 r 8 0 w 0 0-8 0 0 8/7/0 jw 9

9. A member of the City Volunteer Corp. cn mow nd clen up lrge lwn in 9 hours. With two members of the City Volunteer Corp. working, the sme job cn be done in 6 hours. How long would it tke the second member of the tem, working lone, to do the job? ID RAG (I D) (Rte) (Agent) (Gin) Member Defined s: Member s Work Rte Time Working Together Ech Member s Work Contribution First member / 9 6 ( / 9) 6 Second member /t 6 ( / t ) 6 VARIABLE(S): t the time needed by the second member to complete the work lone. ANSWER: 8 hrs. The time needed by the second member to complete the work lone. INDIVIDUAL WORK CONTRIBUTIONS ADDED ENTIRE JOB DONE 6 6 6 8 3 8 9 + 3 + t + t t t t LCD 3t 9. How mny pounds of gourmet cndy selling for $.80 per pound should be mixed with 3 pounds of cndy selling for $.60 pound to obtin mixture selling for $.04 per pound? ID RAG (I D) (Rte) (Agent) (Gin) Type of Cndy Price $ per lb. Quntity lbs. Cost $ Gourmet cndy.80 x.80 x Other cndy.60 3.60 3 Mixture.04 x + 3.04 (x + 3) VARIABLE(S): x lbs of Gourmet cndy x + 3 lbs of mixture. ANSWER: 7 lbs of Gourmet cndy should be mixed. COST OF THE CANDY BEFORE MIXING COST AFTER MIXING.80 x +.60 3.04 (x + 3).80 x + 7.80.04 x + 6. 0.4 x.68.68 x x 7 0.4 8/7/0 jw 0

93. A fenced re is 300 squre feet. If the width is 5 feet less thn the length, find the length nd the width of the fenced re. () A rectngle s re. L rectngle s length. W rectngle s width. (6) L 5L 300 0 () A L W (7) (L 0) (L + 5) 0 (3) W L 5 (8) L 0, L 5 Reject 5 (No negtive dimension) (4) (L 5) L 300 (9) W L 5 0 5 5 (5) L 5L 300 Rectngle L ANSWER The length 0 ft. The height 5 ft. W 94. At business meeting t Pner Bred, the bill for two cppuccinos nd three house lttes ws $4,55. At nother tble, the bill for one cppuccino nd two house lttes ws $8.77. How much did ech type of beverge cost? x + 3y 4.55 x + (.99) 8.77 x + y 8.77 x + 5.98 8.77 ------------------------------- x 40 x + 3y 4.55 x.79 (-b) -x 4y 7.54 ------------------------------- y.99 y.99 VARIABLE(S): x cost of one cppuccino. y cost of one house ltte. ANSWER: One cppuccino costs $.79. One house ltte costs $.99. 8/7/0 jw