Chapter P Prerequisites

Transcription

1 Section P. Real Numbers Chapter P Prerequisites Section P. Real Numbers Quick Review P.. {,,,,, 6}. {,, 0,,,,,, 6}. {,, }. {,,, }. (a) 87.7 (b).7 6. (a) 0.6 (b) ( ) -( )+ ; (.) -(.) ( ) +( )() ,,,,,, ,,,,,, 6, 7, 8, 9, 0,, Section P. Eercises..6 (terminates). 0. (repeats). -.6 (repeats). 0. (repeats). 0 all real numbers less than or equal to (to the left of and including ) all real numbers between and, including and ecluding all real numbers less than 7 (to the left of 7) 8. 0 all real numbers between and, including both and 9. 0 all real numbers less than 0 (to the left of 0) all real numbers between and 6, including both and ; all numbers between and including and ecluding. - q 6, or ; all numbers less than or equal to. - q 6 6, or 6 ; all numbers less then. - 6 ; all numbers between and, including and ecluding ; all numbers between and, ecluding both and 6. 6 q, or Ú ; all numbers greater than or equal to 7. (-, q); all numbers greater than 8. ( 7, ); all numbers between 7 and, ecluding both 7 and 9. (, ); all numbers between and, ecluding both and 0. [-, q); all numbers greater than or equal to. (, ]; all numbers between and, ecluding and including. (0, q); all numbers greater than 0. The real numbers greater than and less than or equal to 9.. The real numbers greater than or equal to, or the real numbers which are at least.. The real numbers greater than or equal to, or the real numbers which are at least. 6. The real numbers between and 7, or the real numbers greater than and less than The real numbers greater than. 8. The real numbers between and 0 (inclusive), or greater than or equal to and less than or equal to ; endpoints and ; bounded; half-open ; endpoints and ; bounded; open. 6 ; endpoint ; unbounded; open. Ú -6; endpoint 6; unbounded; closed. His age must be greater than or equal to 9: Ú 9 or [9, q); Bill s age. The costs are between 0 and (inclusive): 0 or [0, ]; cost of an item. The prices are between $.099 and $.99 (inclusive): or [.099,.99]; $ per gallon of gasoline 6. The raises are between 0.0 and 0.06: or (0.0, 0.06); average percent of all salary raises 7. a( +b)a# +a# ba +ab 8. (y-z )cy# c-z # cyc-z c 9. a +d a# +d# (a+d) # 0. a z+a wa z+a wa (z+w). The opposite of 6-, or (6- ) The opposite of 7, or ( 7)7. In, the base is.. In ( ) 7, the base is. # Copyright 0 Pearson Education, Inc.

2 Chapter P Prerequisites. (a) Associative property of multiplication (b) Commutative property of multiplication (c) Addition inverse property (d) Addition identity property (e) Distributive property of multiplication over addition 6. (a) Multiplication inverse property (b) Multiplication identity property, or distributive property of multiplication over addition, followed by the multiplication identity property. Note that we also use the multiplicative commutative property to say that # u u # u. (c) Distributive property of multiplication over subtraction (d) Definition of subtraction; associative property of addition; definition of subtraction (e) Associative property of multiplication; multiplicative inverse; multiplicative identity 7. ( ) y 8. y ( ) y y 9 y y y ( - y ) -. (y 6 - ) y -8 - y - 8 y - y * * * * * 0 8 mi * 0-9 C ,000,000,000 6.,870,000,000,000 mi ( zeros between the decimal point and the ) y a b a - y b a y b y y 8 a a b b a ba b a b b a a b ba a b b (.)(.) * * 0. * 0 9 * 0 9. * 0-9. * 0-8 (.7)(.) * * 0-. * 0 7. * ( ) 6 * * 0-8 a a b 6 ab 6. (a) When n0, the equation a m a n a m+n becomes a m a 0 a m+0. That is, a m a 0 a m. Since a Z 0, we can divide both sides of the equation by a m. Hence a 0. (b) When n m, the equation a m a n a m+n becomes a m a m a m+( m). That is a m-m a 0. We know from (a) that a 0. Since a Z 0, we can divide both sides of the equation a m a m by a m. Hence a -m a m. 66. (a) Step Quotient Remainder (b) When the remainder is repeated, the quotients generated in the long division process will also repeat. (c) When any remainder is first repeated, the net quotient will be the same number as the quotient resulting after the first occurrence of the remainder, since the decimal representation does not terminate. 67. False. If the real number is negative, the additive inverse is positive. For eample, the additive inverse of is. 68. False. If the positive real number is less than, the reciprocal is greater than. For eample, the reciprocal of is. 69. [, ) corresponds to - 6. The answer is E. 70. ( ) ( )( )( )( )6. The answer is A. 7. In -7 -(7 ), the base is 7. The answer is B. 6 # 7. The answer is D.. 7. The whole numbers are 0,,,,, so the whole numbers with magnitude less than 7 are 0,,,,,, The natural numbers are,,,,, so the natural numbers with magnitude less than 7 are,,,,, The integers are,,, 0,,,, so the integers with magnitude less than 7 are 6,,,,,,0,,,,,, 6. Copyright 0 Pearson Education, Inc.

3 Section P. Cartesian Coordinate System Section P. Cartesian Coordinate System Quick Review P Distance: ƒ 7 - ƒ 7 - L Distance: or 0. B - - a - 9 b D y y C 0 0 A C ƒ -ƒ - 9. Since p L. 6, ƒp - ƒ - p. 0. Since L.6 and or., (- - ) + [- - (-)] (-8) ƒ0.6 - (-9.)ƒ ƒ ƒ 9.9 ƒ -7 - (-)ƒ ƒ -7 + ƒ ƒ -ƒ (- - ) + (- - ) (-) + (-) + 6 L 6.0 (0 - ) + (0 - ) (- - ) + [ - (-)] (-) L.80 (- - ) + (0-0) (-7) (0-0) + [-8 - (-)] 0 + (-7) An isosceles triangle y D A B (0 - (-)) + (- - ) + (-) Section P. Eercises. A(, 0); B(, ); C(, ); D(0, ). A(0, ); B(, ); C(, 0); D(, ). (a) First quadrant (b) On the y-ais, between quadrants I and II (c) Second quadrant (d) Third quadrant. (a) First quadrant (b) On the -ais, between quadrants II and III (c) Third quadrant (d) Third quadrant. + ƒ -ƒ ƒ -ƒ ƒ(-)ƒ ƒ -6ƒ 6 (0 - (-)) + (- - ) (-) + (-) (- - ) + ( - ) (-9) + (-) 8 Perimeter + 8 L.86 Since () + () (8), this is a right triangle. Area ()() A square y This is a square with sides of length. Perimeter6; Area6. Copyright 0 Pearson Education, Inc.

4 Chapter P Prerequisites. A parallelogram y 0. y World Motor Vehicle Production 80,000 70,000 7 Number (in thousands) 60,000 0,000 0,000 0,000 0,000 [ - (-)] + [- - (-)] + 0 This is a parallelogram with base 8 units and height units. Perimeter L.9; Area 8#. A rectangle y. 0,000 y Year U.S. Imports from Meico Value (billions of dollars) This is a rectangle with length 6 units and height units. Perimeter; Area (-9.) (-) a - + a P, -, + 9 b a, b (, 6), a - 9 b - b a 9, + b Q P, Q a -, - b. Value (billions of dollars) 0 y Year U.S. Agricultural Eports a + (-), y - + (-) b a, -6 b (, -) U.S. Motor Vehicle Production. y Year U.S. Agricultural Trade Surplus Number (in thousands),000 0, Year Surplus (billions of dollars) Year Copyright 0 Pearson Education, Inc.

5 Section P. Cartesian Coordinate System. Value (billions of dollars) y. (a) about $.0 (b) about $.80 (c) about $ (a) 00: about $6; 006: about $9 9-6 L An increase of about 77% (b) 009: about $; 0: about $6 6 - L 0. An increase of about % 7. The three side lengths (distances between pairs of points) are ( - ) + (7 - ) (8 - ) + ( - 7) + (-) (8 - ) + ( - ) U.S. Eports to Canada Year Since two sides of the triangle formed have the same length, the triangle is isosceles. 8. (a) Midpoint of diagonal from ( 7, ) to (, ) is a (-), b (-, -) Midpoint of diagonal from (, ) to (, 6) is - + (-) a, + (-6) b (-, -) Both diagonals have midpoint (, ), so they bisect each other. (b) Midpoint of diagonal from (, ) to (6, 7) is a - + 6, b (, ) Midpoint of diagonal from (0, ) to (, ) is a 0 +, + b (, ) Both diagonals have midpoint (, ), so they bisect each other. 9. (a) Vertical side: length6-( )8; horizontal side: length-( ); diagonal side: length [6 - (-)] + [ - (-)] (b) (89), so the Pythagorean Theorem implies the triangle is a right triangle. 0. (a) ( - 0) + (- - 0) + (-) ( - ) + (- - ) + (-7) 0 ( - 0) + ( - 0) + 8 (b) Since () + (8) (0), the triangle is a right triangle.. (-) +(y-), or (-) +(y-). [-( )] +(y-), or (+) +(y-). [-( )] +[y-( )],or (+) +(y+) 9. (-0) +(y-0) (), or +y. (-) +(y-) 6, so the center is (, ) and the radius is [-( )] +(y-), so the center is (, ) and the radius is. 7. (-0) +(y-0) (), so the center is (0, 0) and the radius is. 8. (-) +[(y-( 6)], so the center is (, 6) and the radius is. 9. ƒ - ƒ 0. ƒy - (-)ƒ Ú, or ƒy + ƒ Ú. ƒ - cƒ 6 d. The distance between y and c is greater than d, so ƒy - cƒ 7 d. + a + b. and +a8 +b8 a7 b6. Show that two sides have the same length, but not all three sides have the same length: [ - (-)] + ( - 0) [ - (-)] + ( - ) ( - ) + ( - 0) The midpoint of the hypotenuse is a + 0, b a The distances from this point to, 7 b (.,.). the vertices are: (. - 0) + (. - 0) (. - ) + (. - 0) (-.) (. - 0) + (. - 7). + (-.) Copyright 0 Pearson Education, Inc. 6. ƒ - ƒ 6 means the distance from to must be less than. So must be between and. That is,

6 6 Chapter P Prerequisites 7. ƒ + ƒ Ú means the distance from to must be or more. So can be or more, or can be 8 or less. That is, -8 or Ú True. An absolute value is always greater than or equal to zero. If a 7 0, then ƒaƒ a 7 0. If a 6 0, then ƒaƒ -a 7 0. If a 0, then ƒaƒ 0. length of AM 9. True. because M is the midpoint of AB. length of AB length of AM length of AM By similar triangles, length of AC length of AB, so M is the midpoint of AC , so and ƒ - ƒ -( - ) -. The answer is B. 6. For a segment with endpoints at a and b, the a + b midpoint lies at The answer is C. 6. (-) +(y+) corresponds to (-h) +(y-k) (), with h and k. So the center, (h, k), is (, ). The answer is A. 6. In the third quadrant, both coordinates are negative. The answer is E. 6. (a) (b) - + (7 - (-)) (0) - + ; (7 - (-)) - + (0) (c) a + (b - a) a + b - a a + b (d) + (8 - ) + (6) + ; + (6) + 6 a + b (a + b) ; a + (b - a) a + b - a a + b a + b (a + b) a () + 7, () + b a 9, b (, ); a + (7), + () b a, b (, 8) (e) a a + c, b + d b; a a + c, b + d b 6. If the legs have lengths a and b, and the hypotenuse is c units long, then without loss of generality, we can assume the vertices are (0, 0), (a, 0), and (0, b). Then the midpoint of the hypotenuse is a a + 0, b + 0 b a a The, b b. distance to the other vertices is B a a b + a b b a B + b c c. (a) Area of ^BPQ area of ABCD - area of ^BCP - area of ^BAQ - area of ^DPQ a - (a)a a b - (a)a a b - a a ba ab a - a - a 8 - a 6 7a 6 (b) Area of ^BPQ 7 # ( area of ABCD), which is 6 just under half the area of the square ABCD. Note that the result is the same if a 6 0, but the location of the points in the plane is different. For #67 69, note that since P(a, b) is in the first quadrant, then a and b are positive. Hence, a and b are negative. 67. Q(a, b) is in the fourth quadrant, and since P and Q both have first coordinate a, PQ is perpendicular to the -ais. 68. Q( a, b) is in the second quadrant, and since P and Q both have second coordinate b, PQ is perpendicular to the y-ais. 69. Q( a, b) is in the third quadrant, and the midpoint of PQ is a a + (-a), b + (-b) b (0, 0). 70. Let the points on the number line be (a, 0) and (b, 0). The distance between them is (a - b) + (0-0) (a - b) ƒa - bƒ. Section P. Linear Equations and Inequalities Quick Review P. y a B A a Q(, 0 ) Copyright 0 Pearson Education, Inc. C a P( a, ) D a y - + y + ( + - ) + (y + y) + (7 + ) + y z + y - + y - z - ( - ) + (y + y) + (-z - z) + ( - ) + 7y - z +. ( - y) + (y - ) + + y 6 - y + y y + y. ( + y - ) + (y - + ) y - + y y +. y + y y 6. y - + y - y - (y - )(y - ) + (y - ) (y - )(y - ) y - y - + y - (y - )(y - ) (y - )(y - )

7 Section P. Linear Equations and Inequalities y - y y + y - y y y + - y y ( + ) ( - ) Section P. Eercises. (a) and (c): ( ) +( )(9)- 8-, and a b + a b a b + Meanwhile, substituting gives rather than (a): and Or: Multiply both sides by 6: 6a b + 6a 6 b 6a b, so +. Subtract from both sides: +0. Subtract from both sides:.. (b): Meanwhile, substituting or gives , which is undefined.. (c): (0-) / 8 /. Meanwhile, substituting 6 gives rather than ; substituting 8 gives 6 / L.8 rather than.. Yes: No. There is no variable in the equation. 7. No. Subtracting from both sides gives, which is false and does not contain the variable. 8. No. The highest power of is, so the equation is quadratic and not linear. 9. No. The equation has a root in it, so it is not linear. 0. No. The equation has in it, so it is not linear t. t t t y y y y + 8 -y y + -y 8 -y y -8 y a b a 7 8 b 0. a. a + b a b b () a + b ().. 6. ( - ) a - b z - 0z - z - 7-8z - 9 z - 7-8z z - 8-9z -8 z 8 9 z - 9-8z - z - 7z - z - 7z z + z. a - + b () z. a t + - t - b a 8 b (t + ) - (t - ) 8 t + - t + 8-9t t - t 9 8. a t - + t + b a b (t - ) + (t + ) 6 t - + t + 6 7t + 6 7t - t Copyright 0 Pearson Education, Inc. 9. (a) The figure shows that - is a solution of the equation

8 8 Chapter P Prerequisites (b) The figure shows that is a solution of the equation (a) The figure shows that is not a solution of the equation (b) The figure shows that - is a solution of the equation (a) (0)-0- >7. Meanwhile, substituting gives 7 (which is not less than 7); substituting 6 gives 9.. (b) and (c): ()-9- Ú, and ()--8 Ú.. (b) and (c): ()-8-7 and - 6 7, and also ()-- and - 6. Meanwhile, substituting 0 gives (which is not greater than ).. (a), (b), and (c): -( )+ and - ; -(0)-0 and - ; - () - - and Ú Ú 6+8 Ú 6+9 Ú < <9 < Ú >- 9+>- 0+> 0> > 6. a + 7 b (-) a - b 7 (-) -> > >. () Ú a y - b Ú (-) Ú y- Ú 6 7 Ú y Ú 7 y Ú - Ú - y 7 6. () 7 a y - b 7 (-) > y- > > y > > y > <y< 7. 0 z+<8 z < - z < 8. 6<t-<0 < t < < t < 9. a b 6 (-) (-)+(-)< -+-8< -< < a b 6 6(-) (-)+(-)< < 6 7+< 6 7< Copyright 0 Pearson Education, Inc a y - + y - b 6 0(y - ) (y-)+(y-)<0y-0 0y-+6y-<0y-0 6y-7<0y-0 6y<0y+7 6y<7 y 6 7 6

9 Section P. Linear Equations and Inequalities 9. a - y - y - b Ú ( - y) 6. True includes the possibility that 6 and this is, (-y)-(y-) Ú 8-y the case. -6y-6y+9 Ú 8-y y+ Ú 8-y Subtracting from each side gives -. y Ú 7-y The answer is E. y Ú 7 y Ú <6 Dividing each side by and reversing the<gives >. The answer is C.. c ( - ) - d [( - )] -- 0(-) c ( + ) + ( - ) d< 6c ( - ) d (+)+(-)<(-) +9+-8<-6-9<-6 <+ < < for for 0, for,,, for 0,, 9. Multiply both sides of the first equation by. 60. Divide both sides of the first equation by. 6. (a) No: they have different solutions (b) Yes: the solution to both equations is (a) Yes: the solution to both equations is (b) No: they have different solutions False. 6>, but 6< because 6 lies to the left of on the number line. 67. (+)0 0 or +0 The answer is A Multiplying each side by gives The answer is B (c) (d) (e) If your calculator returns 0 when you enter + 6, you can conclude that the value stored in is not a solution of the inequality P(L+W) PL+W P-LW P - L W P-L 7. A h(b +b ) h(b +b )A A b +b h A b h - b V pr p V r V B p r r V B p C (F - ) 9 9 C F - 9 C + F F 9 C + Copyright 0 Pearson Education, Inc.

10 0 Chapter P Prerequisites Section P. Lines in the Plane Eploration. The graphs of ym+b and ym+c have the same slope but different y-intercepts.. [.7,.7] by [.,.] m The angle between the two lines appears to be y a () + yb +y y + y y7+(-y) +y7+-y y7-y y7 7 y 8. +y-y y-y- y- y - [.7,.7] by [.,.] m [.7,.7] by [.,.] m (-8) (-) -0-6 Section P. Eercises. m. m [.7,.7] by [.,.] m In each case, the two lines appear to be at right angles to one another. Quick Review P (-)+(-) (7+)(-) y y + y - [.7,.7] by [.,.] m... m m m m so - 6 -, 8. - y - so y + y -, 6 9. y + so y6 + y +, 7 0. so ,. y-(-). y - - ( + ). y+ (-). y-(+). Since m, we can choose A and B. Since 7, y solves -y+c0, C must equal : -y+0. Note that the coefficients can be multiplied by any nonzero number, e.g., another answer would be -y+00. This comment also applies to the following problems. Copyright 0 Pearson Education, Inc.

11 Section P. Lines in the Plane 6. Since m, we can choose A and B. Since, y 8 solves -y+c0, C must equal : -y-0. See comment in #. 7. Since m0, we can choose A0 and B. Since, y solves 0+y+C0. C must equal : 0+y+0, or y+0. See comment in #. 8. Since m, we can choose A and B. Since, y solves +y+c0, C must equal 6: +y+60. See comment in #. 9. The slope is m A/B, so we can choose A and B. Since, y solves -y+c0, C must equal : -y+0. See comment in #. 0. Since m is undefined we must have B0, and we can choose A. Since, y solves +0y+C0, C must equal : -0. See comment in #.. Begin with point-slope form: y- (-0),so y +.. Begin with point-slope form: y - ( - ), so y +.. m - so in point-slope form, y - - ( + ),, and therefore y m so in point-slope form, y - ( - ), and 7, 7 therefore. Solve for y: y y Solve for y: y Graph y9-8; window should include (6., 0) and (0, 9), for eample, [, 0]*[ 0, 60]. 9. Graph y(9-)/7; window should include (.88, 0) and (0, 6.9), for eample, [, ]*[ 0, 80]. [, ] by [ 0, 80] 0. Graph y(0-00)/9-7; window should include (.686, 0) and (0, 9), for eample, [, ]*[ 0, 0]. [, ] by [ 0, 0]. (a): The slope is., compared to in (b). 7. (b): The slopes are and, respectively.. Substitute and solve: replacing y with gives, and replacing with 8 gives y.. Substitute and solve: replacing y with gives, and replacing with 8 gives y -8.. Substitute and solve: replacing y with gives -0, and replacing with 8 gives y Substitute and solve: replacing y with gives, and replacing with 8 gives y Ymin -0, Yma 0, Yscl 8. Ymin -0, Yma 0, Yscl 9. Ymin -0/, Yma 0/, Yscl / 0. Ymin -., Yma., Yscl. In #, use the fact that parallel lines have the same slope, while the slopes of perpendicular lines multiply to give.. (a) Parallel: y - ( - ), or y -. (b) Perpendicular: y - - ( - ), or [, 0] by [ 0, 60] 8. Graph y-; window should include (7., 0) and (0, ), for eample, [, 0]*[ 0, 0]. y (a) Parallel: y - -( + ), or y - -. (b) Perpendicular: y - ( + ), or y +.. (a) Parallel: + y 9, or y - +. (b) Perpendicular: - y 7, or y - 7. [, 0] by [ 0, 0]. (a) Parallel: - y, or y -. (b) Perpendicular: + y, or y - +. Copyright 0 Pearson Education, Inc.

12 Chapter P Prerequisites. (a) m(67,00-,000)/887., the y-intercept is b,000 so V87.t+,000. (b) The house is worth about $7,00 after 9.7 years. (d) [98, 0] by [0, ]. (a) Slope of the line between the points (00, 8.) and [0, ] by [0000, 00000] (c) 87.t+,0007,000: t0.0. (d) t years. 6. (a) (b) I (8,000-) (c),000 dollars. (b) (00, 9.9) is m Using the point-slope form equation for the line, we have y - 8..( - 00), so y.( - 00) [000, 0] by [0, 00] (d) 800 dollars. 7. y where y is altitude and is horizontal distance. 8, The plane must travel,000 ft horizontally just over 6 miles. 8. (a) m [0, 8000] by [0, 00] 6 ft 00 ft (b) 66.6 ft, or about 0.79 mile. (c) 7.6 ft. 9. m so asphalt shingles are , acceptable. 0. We need to find the value of y when 00, 006, and 007 using the equation y 0.8( - 00) y 0.8(00-00) y 0.8(006-00) y 0.8(007-00) Americans income in 00, 006, and 007 was, respectively, 0., 0.9, and. trillion dollars.. (a) Slope of the line between the points (990,.8) and (00, 0.) is m Using the point-slope form equation for the line, we have y ( - 990), so y 0.( - 990) +.8. (b) Using y 0.( - 990) +.8 and 00, the model estimates Americans ependitures in 00 were $8 trillion. (c) Using y 0.( - 990) +.8 and 0, the model predicts Americans ependitures in 0 will be $.8 trillion. (c) Using y.( - 00) + 8. and 0, the model predicts U.S. imports from Meico in 0 will be approimately $9. billion.. (a) Slope of the line between the points (0, ) and (b) (0, 6707) is m Using the point-slope form equation for the line, we have y - 8.0( - 0), so y [, 0] by [0, 0,000] [, 0] by [0, 0,000] (c) The year 00 is represented by 0. Using y and 0, the model predicts the midyear world population in 00 will be 7.8 billion. Copyright 0 Pearson Education, Inc.

13 Section P. Lines in the Plane. (a) Using the point-slope form equation for the line, we have y ( - 00), so y.( - 00) [000, 0] by [00, 0] (b) Slope of the line between the points (00, 69.9) and (00, 9.) is m [000, 0] by [00, 0] (c) Using y.( - 00) and 0, the model predicts total U.S. eports to Canada in 0 will be 0.6 billion. 8-0 a a - (a-) 6a- 9a a a a 7. AD ß BC Q b ; AB ß DC Q a - Q a 6 8. BC ß AD Q b ; AB ß CD Q a Q a False. The slope of a vertical line is undefined. For eample, the vertical line through (, ) and (, 6) would 6 - have a slope of which is undefined. - 0, 6. True. If b0, then a Z 0 and the graph of c is a a vertical line. If b Z 0, then the graph of y - a is a line with slope - a and y-intercept b + c b b c c If and a0, y which is a horizontal line. b. b Z 0 b, An equation of the form a+byc is called linear for this reason. 6. With (, y )(, ) and m, the point-slope form equation y-y m(- ) becomes y- [-( )] or y-(+). The answer is A. 6. With m and b, the slope-intercept form equation ym+b becomes y+( ) or y-. The answer is B. 6. When a line has a slope of m, a perpendicular line must have a slope of m - The answer is E. m. 66. The line through (, y )(, ) and (, y )(, ) has a slope of m y - y (-) - -. The answer is C. 67. (a) (b) (c) [, ] by [, ] [, ] by [, ] 9. (a) No, it is not possible for two lines with positive slopes to be perpendicular, because if both slopes are positive, they cannot multiply to. (b) No, it is not possible for two lines with negative slopes to be perpendicular, because if both slopes are negative, they cannot multiply to. 60. (a) If b0, both lines are vertical; otherwise, both have slope m a/b, and are therefore parallel. If cd, the lines are coincident. (b) If either a or b equals 0, then one line is horizontal and the other is vertical. Otherwise, their slopes are a/b and b/a, respectively. In either case, they are perpendicular. [, ] by [, ] Copyright 0 Pearson Education, Inc. (d) From the graphs, it appears that a is the -intercept and b is the y-intercept when c. Proof: The -intercept is found by setting y0. When c, we have Hence so a + 0 b. a a. The y-intercept is found by setting 0. 0 y When c, we have Hence so a + y b. b, yb.

14 Chapter P Prerequisites (e) 70. The line from the origin to (, ) has slope so the, tangent line has slope - and in point-slope form,, [ 0, 0] by [ 0, 0] [ 0, 0] by [ 0, 0] the equation is y - - ( - ). 7. A has coordinates a b while B is a a + b, c so the, c b, b, line containing A and B is the horizontal line yc/, and the distance from A to B is a + b - b a. [ 0, 0] by [ 0, 0] From the graphs, it appears that a is half the -intercept and b is half the y-intercept when c. Proof: When c, we can divide both sides by and we have By part (d) the -intercept a + y b. is a and the y-intercept is b. (f) By a similar argument, when c, a is the opposite of the -intercept and b is the opposite of the y-intercept. 68. (a) Section P. Solving Equations Graphically, Numerically, and Algebraically Eploration. [, ] by [, 0]. Using the numerical zoom, we find the zeros to be 0.79 and... [ 8, 8] by [, ] These graphs all pass through the origin. They have different slopes. (b) If m>0, then the graphs of ym and y m have the same steepness, but one increases from left to right, and the other decreases from left to right. (c) [, ] by [, 0] [, ] by [, 0] By this method we have zeros at 0.79 and... [ 8, 8] by [, ] These graphs have the same slope, but different y-intercepts. 69. As in the diagram, we can choose one point to be the origin, and another to be on the -ais. The midpoints of the sides, starting from the origin and working around counterclockwise in the diagram, are then Aa a a + b, 0b, Ba, c b + d b, Ca, c + e b, and Da d The opposite sides are therefore parallel, since, e b. the slopes of the four lines connecting those points are: m m CD c b ; m e AB c DA b ; m e BC d - a ; d - a. [.0,.6] by [ 0., 0.] Copyright 0 Pearson Education, Inc. [0.6, 0.9] by [ 0.9, 0.] Zooming in and tracing reveals the same zeros, correct to two decimal places.. The answers in parts,, and are the same. 6. On a calculator, evaluating -+7 when 0.79 gives y0.06 and when. gives y0.06, so the numbers 0.79 and. are approimate zeros. 7. [.7,.] by [ 0., 0.] [0.7, 0.8] by [ 0., 0.] Zooming in and tracing reveals zeros of and.0707 accurate to si decimal places. If rounded to two decimal places, these would be the same as the answers found in part.

15 Section P. Solving Equations Graphically, Numerically, and Algebraically Quick Review P.. (-) (+) (+)(-) (y-)(y+)y +y-y- y +7y-. -0+(-)(-)(-) ( -+8) (-)(-) (+)-(+) (+)( -) 8. y -y +6(y -)(y -9) (y-)(y+)(y-)(y+) ( + ) if Z ( - )( + ) Section P. Eercises ( + ) ( + )( + ) - ( + ) ( + )( + ) ( + )( + ) ( - )( + ) ( + )( + ) ( + + ) - ( + - ) ( - )( - )( + ) + ( - )( - ) - + ( - )( + ) ( + )( + ) ( - )( - )( + ) - ( + )( - ) ( - )( - )( + ) ( - )( - )( + ) -( + - ) ( - )( - )( + ) -( + )( - ) ( - )( - )( + ) - - ( + )( + ). [, ] by [ 0, 0] or 0. The left side factors to (+)(-)0: +0 or [, ] by [, ] 0. or. The left side factors to (-)(-)0: -0 or [ 6, 6] by [, ] or Rewrite as -8+0; the left side factors to (-)(-)0: -0 or -0. [ 6, 6] by [ 0, 0] - or Rewrite as -7-60; the left side factors to (+)(-)0: +0 or -0 - [ 0, 0] by [ 0, 0] or The left side factors to (+)(-)0: +0 or -0 Copyright 0 Pearson Education, Inc.

16 6 Chapter P Prerequisites 6. [ 0, 0] by [ 0, 0] or Rewrite as +-00; the left side factors to (-)(+)0: -0 or Rewrite as () ; then, or ;. 8. Divide both sides by to get (-) 8.. Then - ;8. and ; Divide both sides by to get ( + ) 8 Then. + ; 8 A and 0. Divide both sides by to get (u+).. Then u + ;. and u - ;... Adding y +8 to both sides gives y. Divide both 7 sides by to get y 7 so y ;, A.. +_ so (- ; ), which gives 8 or (+) 6 + ; 6 _ 7 or a b 9 + a b (+.) ; L -6. or L a - 7 b - + a - 7 b a - 7 b - ; A ; 7 ; 7 - L 0.8 or 7 + L a 6 b + a 6 b (+) +9 - ; - - L -6.6 or - + L ( ) +( ) (-) - _ or a - b or - 6 L a, b8, and c : -8 ; 8 - ()(-) () -8 ; 6 - ; L -8. or L a, b, and c: ; (-) - ()() () or. --0, so a, b, and c : ; (-) - ()(-) () or , so a, b -, and c : ; (- ) - ()(-) () ; + ; a - b L -. or L a - b ; 6 A 9 ; 6 ; + 6 L.9-8 ; 7 ; ; ; ; Copyright 0 Pearson Education, Inc.

17 Section P. Solving Equations Graphically, Numerically, and Algebraically , so a, b, c. - ; () - ()(-) () - ; 7 - ; 7 L or L ,so a, b, c : ; ; 6 or 8. -intercept: ; y-intercept: 6. -intercepts:, ; y-intercept: 7. -intercepts:, 0, ; y-intercept: 0 8. no -intercepts; no y-intercepts (-) ; (-) - ()(-) () [, ] by [, ] [, ] by [, ] [, ] by [, ] [, ] by [, ] [, ] by [, ] [, ] by [, ] [, ] by [, ] [, ] by [, ]. [, ] by [, ] [, ] by [, ]. +-0; L ; L Using TblStart.6 and Tbl0.00 gives a zero at.6. Using TblStart 0.6 and Tbl0.00 gives a zero at Using TblStart. and Tbl0.00 gives a zero at.. 9. Graph y ƒ - 8ƒ and y: t6 or t0 0. Graph y ƒ + ƒ and y: or. Graph y ƒ + ƒ and y7: or 6. Graph y ƒ - ƒ and y: - or 7. Graph y ƒ - ƒ and y : or. Graph y ƒ + ƒ and y-:. (a) The two functions are y + (the one that begins on the -ais) and y -. (b) This is the graph of y (c) The -coordinates of the intersections in the first picture are the same as the -coordinates where the second graph crosses the -ais. 6. Any number between. and. must have the digit in its thousandths position. Such a number would round to.. 7. The left side factors to (+)(-)0: +0 or Graphing y -8 in (e.g.) [ 0, 0]*[ 0, 0] and looking for -intercepts gives L -. or L or (+) 8-8 or 8-8 is an etraneous solution, 8 L.8 Copyright 0 Pearson Education, Inc.

18 8 Chapter P Prerequisites. From the graph of y + -- on [ 0, 0]*[ 0, 0], the solutions of the equation (-intercepts of the graph) are L -.6, L -0.,.. From the graph of y -+ on [ 0, 0]*[ 0, 0], the solutions of the equation (-intercepts of the graph) are L -., L 0., and L or ; ; ; no real solutions to this equation.. Graph y ƒ + ƒ - ƒ - ƒ:. Graph y ƒ0. + ƒ and y -: L -. or L.9 6. Graph y + 7 and y +: L -.6 or L. 7. (a) There must be two distinct real zeros, because b -ac>0 implies that ; b - ac are two distinct real numbers. (b) There must be one real zero, because b -ac0 implies that ; b - ac 0, so the root must be - b a. (c) There must be no real zeros, because b -ac<0 implies that ; b - ac are not real numbers. 8. For (a) (c), answers may vary. (a) +- has discriminant () -()( )6, so it has two distinct real zeros. The graph (or factoring) shows the zeros are at and. (b) ++ has discriminant () -()()0, so it has one real zero. The graph (or factoring) shows the zero is at. (c) ++ has discriminant () -()(), so it has no real zeros. The graph lies entirely above the -ais. 9. Let be the width of the field (in yd); the length is +0. Then the field is 80 yd wide and yd long. 8800(+0) (+0)(-80) 0+0 or or Solving +(+) 8, or , gives L 9.98 or L The ladder is about + L.98 ft up the wall. 6. The area of the square is. The area of the semicircle is since the radius of the semicircle is pr pa b. Then 00 + Solving this (graphically is pa. b easiest) gives L.98 ft (since must be positive). 6. True. If is an -intercept of the graph of ya +b+c, then y0 when. That is, a +b+c0 when. 6. False. Notice that for, ( ) 8. So could also be. 6. (-)0 when 0 and when -0 or. The answer is D. 6. For -+? to be a perfect square,? must be replaced by the square of half of, which is a -. b The answer is B. 66. By the quadratic formula, the solutions are -(-) ; (-) - ()(-) ; 7 () The answer is B. 67. Since an absolute value cannot be negative, there are no solutions. The answer is E. 68. (a) a +b+c0 a +b c (b) (c) + b a + a b a b a + b a ba + b a b -ac a a + b a b b - ac a + b a ; b - ac A a + b a ; b - ac a Copyright 0 Pearson Education, Inc. + b a -c a + b a + a # b a b - c a + a # b a b - c a + b a - b a ; b - ac a -b ; b - ac a 69. Graph y ƒ - ƒ and yc for several values of c. (a) Let c. The graph suggests y intersects y ƒ - ƒ four times. ƒ - ƒ Q - or ;6 + b a ; ƒ - ƒ has four solutions: {;, ; 6}. (b) Let c. The graph suggests y intersects y ƒ - ƒ three times. ƒ - ƒ Q - or ;8 0 (c) Let c. The graph suggest y intersects y ƒ - ƒ twice. ƒ - ƒ Q - or - 9 no solution ƒ - ƒ has two solutions: { }.

19 Section P.6 Comple Numbers 9 (d) Let c. The graph suggests y does not intersect y ƒ - ƒ. Since absolute value is never negative, ƒ - ƒ has no solutions. (e) There is no other possible number of solutions of this equation. For any c, the solution involves solving two quadratic equations, each of which can have 0,, or solutions. 70. (a) Let Db -b ; D -ac. The two solutions are ; a adding them gives -b + D -b - D -b + D - D + a a a -b a -b a (b) Let Db -b ; D -ac. The two solutions are ; a multiplying them gives -b + D# -b - D a a 7. From #70(a), + - b Since a, this means a. b 0. From #70(b), # c ; since a, this a 0 ; 00-8 means c6. The solutions are ; this reduces to. ; or approimately and.0., Section P.6 Comple Numbers Quick Review P y. a+d. z Section P.6 Eercises (-b) - (D) a b - (b - ac) a c a In # 8, add or subtract the real and imaginary parts separately.. (-i)+(6+i)(+6)+( +)i8+i. (-i)+(-i)(+)+( -)i-7i. (7-i)+(6-i)(7+6)+( -)i-i. (+i)-(9i-)(+)+(-9)i-8i. (-i)+( - -) (+)+ (- - )i -( + ) i 6. ( - i) + (- + -9) ( - ) + ( +)i ( - ) + 0i 7. (i +)-(7+i )( +)-(7-i) (-7)+i +i 8. (7 + i )-(6 - -8) (7 - ) -(6-9i)( ) +9i (7-7) +9i In #9 6, multiply out and simplify, recalling that i. 9. (+i)(-i)-i+6i-i +i+7+i 0. (-i)(+i)+6i-i-i +i++i. (-i)(-i)-i-i+8i -i-8 -i. (i-)(i+)0i +i-6i- 0-i- -i. (7i-)(+6i)i+i -6-8i -6-i 8-i. (- + i) (6-i)(i)(6-i)8i-i +8i. ( -i)(+i) -6i-i-8i -0i+8-0i 6. (- + i) (6+i)( + ) i(6+i) 6( + ) i+ ( + ) i (0 + ) + ( + 6) i i 8. - i 9. - i 0. - i In #, equate the real and imaginary parts.., y., y 7., y. 7, y 7/ In # 8, multiply out and simplify, recalling that i.. (+i) 9+i+i +i 6. (-i) (-i+i )(-i)( i)(-i) i+i -i 7. a a (+i) + ib b (+i+i ) ( ) +0i (i) 8. a + ib a b ( + i) 8 ( + i + i ) ( + i) ( + i) ( + i) ( + i + i + i ) (i) 0 + i In #9, recall that (a+bi)(a-bi)a +b () Copyright 0 Pearson Education, Inc.

20 0 Chapter P Prerequisites In # 0, multiply both the numerator and denominator by the comple conjugate of the denominator, recalling that (a+bi)(a-bi)a +b ( + i) (-i) 7. + i ( + i)(- - i) ( - i)( + i) 8. + i ( + i)( - i) i ( - i)( - i) 9. # + i - i + i ( - i - i + i )( + i) ( - i)( + i) + i - i - 6i 7 - i ( - i)( + i) 0. # - i + i - i ( + i - i - i )( - i) [ + + ( - )i]( - i) + - ( + )i + ( - )i - ( - )i (- - )i In #, use the quadratic formula.. - ; i. # - i + i - i - i - i i # + i i + i - i + i + i # + i + i + i - i + i + i i # -i -6i - i -i ; 6 i # - i - i ( + i + i )(-i + i ) # - i - i ( + i - i - i )( - i) i - + i + i - i - - i - i - i 0-8i + i - 6i 9-7 i. 7 8 ; 8 i. ; i. False. When a0, za+bi becomes zbi, and then -z ( bi)biz. 6. True. Because i, i i(i ) i, and i (i ), we obtain i+i +i +i i+( )+( i) (+i)(-i) is a product of conjugates and equals + +0i. The answer is E. 8. # -i The answer is E. i i -i -i - + 0i. 9. Comple, nonreal solutions of polynomials with real coefficients always come in conjugate pairs. So another solution is +i, and the answer is A. 0. ( - i) (-i)( - i) -i + i - - i. The answer is C.. (a) ii i i # i i i i 6 i # i i ( )i i i 7 i # i i i ( ) i 8 i # i (b) i # i i i # i i i i i i i i i 6 # i i i i # i i 7 - # - i i i i i i i i # ( )( ) i 8 # # i i i i (c) i 0 (d) Answers will vary.. Answers will vary. One possibility: The graph has the shape of a parabola, but does not cross the -ais when plotted in the real plane, beacuse it does not have any real zeros. As a result, the function will always be positive or always be negative.. Let a and b be any two real numbers. Then (a+bi) -(a-bi)(a-a)+(b+b)i0+bibi.. (a + bi) (a + bi) (a + bi) (a - bi) a + b, imaginary part is zero.. (a + bi)(c + di) (ac - bd) + (ad + bc)i (ac - bd) - (ad + bc)i and (a + bi) (c + di) (a - bi)# (c - di) (ac - bd) - (ad + bc)i are equal. 6. (a + bi) + (c + di) (a + c) + (b + d)i (a + c) - (b + d)i and (a + bi) + (c + di) (a - bi) + (c - di) (a + c) - (b + d)i are equal. 7. (-i) - i(-i) + 0 but (i) - i(i) + Z 0. Because the coefficient of in -i+0 is not a real number, the comple conjugate, i, of i, need not be a solution. Section P.7 Solving Inequalities Algebraically and Graphically Quick Review P.7 #. 7<-<7 <<0 <<. - Ú 7+ Ú or + or. -9(-)(+). -( -)(-)(+) Copyright 0 Pearson Education, Inc.

21 Section P.7 Solving Inequalities Algebraically and Graphically y (-y)(+y) z - (z - )(z + ) 7. z z + - z z(z - ) z + - ( + 7)( - ) ( - )( - ) ( - ) ( - )( - ) - - ( - )( - ) - 0. ( - )( + ) + - ( - )( - ) ( - )( - ) + ( - )( + ) ( - )( + )( - ) ( - + ) + ( - - ) ( - )( + )( - ) - - ( + )( - ) ( - )( + )( - ) ( - )( + )( - ) ( + ) if Z ( + )( - ) Section P.7 Eercises. (- q, -9] h [, q) : + Ú or + Ú or (- q, -.) h (., q) : ->.6 or -<.6 >.6 or <.6 >. or <.. (, ): <-< < <. [ 8, ]: a - : - <6, 0 b 6<- <6 0< < 0 > > - 6. (- q, 0) h (, q) : - > -> or -< >0 or < 6 <0 or > + ( + )( - ) ( - )( - ) (- q, -] h [7, q) : + + or Ú + 9 or + Ú 9 or Ú [ 9, 9]: Copyright 0 Pearson Education, Inc (+)(+7)0 +0 or or 7 The graph of y +7+ lies below the -ais for 7 - Hence c -7, - is the solution since. d the endpoints are included (-)(-)0-0 or -0 or The graph of y6 -+6 lies above the -ais for 6 and for Hence a - q, is the d h c., q b solution since the endpoints are included (-)(+)0-0 or +0 or The graph of y +7- lies above the -ais for < and for 7 Hence (- q, -) h a is., q b the solution (-)(-)0-0 or -0 or The graph of y -9+ lies below the -ais for Hence a is the solution. 6 6., b (+)(-)0 +0 or -0 or The graph of y-- lies below the -ais for < and for 7 Hence (- q, -) h a is., q b the solution.

22 Chapter P Prerequisites (7-)(+)0 7-0 or +0 7 or The graph of y+- lies above the -ais for <<7. Hence (, 7) is the solution.. -0 ( -)0 (+)(-)0 0 or +0 or -0 0 or or The graph of y - lies above the -ais for > and for <<0. Hence [, 0] h [, q) is the solution ( --0)0 (-6)(+)0 0 or -60 or +0 0 or 6 or The graph of y - -0 lies below the -ais for < and for 0<<6. Hence ( q, ] h [0, 6] is the solution. 7. The graph of y -- is zero for L -0. and L., and lies below the -ais for 0.<<.. Hence ( 0.,.) is the approimate solution. 8. The graph of y -+ is zero for and and lies above the -ais for 6 and for 7. Hence a - q, is the solution. d h c, q b (-)(+)0-0 or +0 or - The graph of y6 -- lies above the -ais for < and for > Hence. a - q, - is the solution. b h a, q b (+)(-)0 +0 or -0 - or The graph of y - lies below the -ais for - <<. Hence c - is the solution., d. The graph of y9 +- appears to be zero for L. and L 0.08 and lies above the -ais for <. and >0.08. Hence ( q,.] h [0.08, q) is the approimate solution.. The graph of y -+7 appears to be zero for L 0.79 and L. and lies below the -ais for 0.79<<.. Hence (0.79,.) is the approimate solution. Copyright 0 Pearson Education, Inc (-)(-)0 (-) 0-0 The graph of y -+ lies entirely above the -ais, ecept at Hence a - q, is b h a., q b the solution set (-)(-)0 (-) 0-0 The graph of y -6+9 lies entirely above the -ais, ecept at. Hence is the only solution (-)(-)0 (-) 0-0 The graph of y -8+6 lies entirely above the -ais, ecept at. Hence there is no solution (+)(+)0 (+) The graph of y9 ++ lies entirely above the -ais, ecept at - Hence every real number. satisfies the inequality. The solution is (- q, q). 7. The graph of y -+ is zero for L.08, L 0.7, and L.9 and lies above the -ais for.08<<0.7 and >.9. Hence, [.08, 0.7] h [.9, q) is the approimate solution. 8. The graph of y8- - is zero for L -.06, L 0., and L.9 and lies below the -ais for.06<<0. and >.9. Hence, (.06, 0.) h (.9, q) is the approimate solution. 9. +> is equivalent to +->0.The graph of y +- is zero for L. and lies above the -ais for >..So,(., q) is the approimate solution is equivalent to Ú 0. The graph of y +8- is zero for L 0.7 and lies above the -ais for >0.7.So,[0.7, q) is the approimate solution.. Answers may vary. Here are some possibilities. (a) +>0 (b) +<0 (c) 0 (d) (+)(-) 0 (e) (+)(-)>0 (f) (-) Ú 0

23 Section P.7 Solving Inequalities Algebraically and Graphically. 6t +88t-0 t -8t+70 (t-6)(t-)0 t-60 or t-0 t6 or t The graph of 6t +88t- lies above the t-ais for 6<t<. Hence [6, ] is the solution. This agrees with the result obtained in Eample 0.. s 6t +6t (a) 6t +6t768 6t +6t-7680 t -6t+80 (t-)(t-)0 t-0 or t-0 t or t The projectile is 768 ft above ground twice: at t sec, on the way up, and t sec, on the way down. (b) The graph of s 6t +6t lies above the graph of s768 for <t<. Hence the projectile s height will be at least 768 ft when t is in the interval [, ]. (c) The graph of s 6t +6t lies below the graph of s768 for 0<t< and <t<6. Hence the projectile s height will be less than or equal to 768 ft when t is in the interval (0, ] or [, 6).. s 6t +7t (a) 6t +7t960 6t +7t-9600 t -7t+600 (t-)(t-)0 t-0 or t-0 t or t The projectile is 960 ft above ground twice: at t sec, on the way up, and t sec, on the way down. (b) The graph of s 6t +7t lies above the graph of s960 for <t<. Hence the projectile s height will be more than 960 ft when t is in the interval (, ). (c) The graph of s 6t +7t lies below the graph of s960 for 0<t< and <t<7. Hence the projectile s height will be less than or equal to 960 ft when t is in the interval (0, ] or [, 7).. Solving the corresponding equation in the process of solving an inequality reveals the boundaries of the solution set. For eample, to solve the inequality - 0, we first solve the corresponding equation -0 and find that. The solution, [, ],of inequality has as its boundaries. 6. Let be her average speed; then 0<. Solving this gives >., so her least average speed is. mph. 7. (a) Let >0 be the width of a rectangle; then the length is - and the perimeter is P[+(-)]. Solving P<00 and ->0 gives in.<< in. [+(-)]<00 and ->0 (-)<00 > 6-<00 > 6<0 < (b) The area is A(-). We already know > from (a). Solve A 00. (-) (-)(+)0-0 or +0 or The graph of y --00 lies below the -ais for <<, so A 00 when is in the interval (, ]. 8. Substitute 0 and 0 into the equation P 00 to find V the range for P: P 00 and P 00 The pressure can range from 0 to 0, or 0 P 0. Alternatively, solve graphically: graph y 00 on [0, 0] μ [0, 0] and observe that all y-values are between 0 and 0. 00, Let be the amount borrowed; then 0,000 + Ú. Solving for reveals that the company can borrow no more than $00, False. If b is negative, there are no solutions, because the absolute value of a number is always nonnegative and every nonnegative real number is greater than any negative real number.. True. The absolute value of any real number is always nonnegative, i.e., greater than or equal to zero.. ƒ - ƒ (-, ) The answer is E.. The graph of y -+ lies entirely above the -ais, so - + Ú 0 for all real numbers.the answer is D.. > is true for all negative, and for positive when >. So the solution is (- q, 0) h (, q). The answer is A.. implies -, so the solution is [, ]. The answer is D. 6. (a) The lengths of the sides of the bo are, -,and -, so the volume is (-)(-).To solve (-)(-), graph y(-)(-) and y and find where the graphs intersect: Either L 0.9 in or L.78 in. (b) The graph of y(-)(-) lies above the graph of y for 0.9<y<.78 (approimately). So choosing in the interval (0.9,.78) will yield a bo with volume greater than in. (c) The graph of y(-)(-) lies below the graph of y for 0<y<0.9 and for.78<<6 (approimately). So choosing in either interval (0, 0.9) or interval (.78, 6) will yield a bo with volume at most in. Copyright 0 Pearson Education, Inc.

24 Chapter P Prerequisites or The graph of The graph of y +7- y +7- appears to be zero for appears to be zero for L -.69 and L.9 L -. and L 0.6 Now look at the graphs of y +7- and y0. The graph of y +7- lies below the graph of y0 when.69<<. and when 0.6<<.9. Hence (.69,.) h (0.6,.9) is the approimate solution or The graph of The graph of y +-0 y +-0 appears to be zero for appears to be zero for L.69 and L.9 L. and L.6 Now look at the graphs of y +-0 and y0. The graph of y +7-0 lies above the graph of y0 when <.69,.<<.6, and >.9. Hence ( q,.69] h [.,.6] h [.9, q) is the (approimate) solution. Chapter P Review. Endpoints 0 and ; bounded. Endpoint ; unbounded. ( -) -. + # + # (+). (uv ) v u u v 6 u v v 6. ( y ) 7..68* *0 6 9.,000,000, (7 zeros between the decimal point and the first 9). (a).*0 (b).6*0 8 (c).*0 0 (d) 9.7*0 7 (e) 7.* (repeating). (a) Distance: ƒ - (-)ƒ ƒ9ƒ 9 (b) Midpoint:. (a) Distance: ( y ) ( ) (y ) 9 y [ - (-)] + (- - ) 9 + (-) L 9.8 (b) Midpoint: a - +, + (-) b a, b a, b. The three side lengths (distances between pairs of points) are [7 - (-)] + (9 - ) Since () + () 0 + () the sum of the squares of the two shorter side lengths equals the square of the long side length the points determine a right triangle. 6. The three side lengths (distances between pairs of points) are ( - ) + [ - ( - )] } + () + 6. Since all three sides have the same length, the figure is an equilateral triangle. 7. (-0) +(y-0), or +y 8. (-) +[y-( )], or (-) +(y+) 6 9. [-( )] +[y-( )], so the center is (, ) and the radius is. 0. (-0) +(y-0), so the center is (0, 0) and the radius is.. (a) Distance between (, ) and (, ): (- - ) + [- - (-)] (-) + () L.7 Distance between (, ) and (, 6): (6 - ) + [ - (-)] L 8.9 Distance between (, 6) and (, ): (b) (0) + (80) , so the Pythagorean Theorem guarantees the triangle is a right triangle.. ƒz - (-)ƒ, or ƒz + ƒ.. [ - (-)] + ( - ) a + 00 (7 - ) + (9 - ) + (-) ( - 0) + ( - ) ( - 0) + [( - ) - ] + (-) (- - 6) + (- - ) (-8) + (-6) and + b +a6 +b0 a7 b9 m y + - ( - ) + 6 Copyright 0 Pearson Education, Inc.

25 Chapter P Review 9 A 6. The slope is m so we can choose A9 7 B, and B7. Since, y solves 9+7y+C0, C must equal 7: 9+7y+70. Note that the coefficients can be multiplied by any nonzero number, e.g., another answer would be 8+y Beginning with point-slope form: y + ( - ), so y m + so in point-slope form, +, y + ( + ), and therefore y y 0. Solve for y: y The slope of the given line is the same as the line we want: m - so y + - ( - ), and therefore,. The slope of the given line is - so the slope of the line, we seek is m Then y + ( - ), and. therefore y (a) y - -. [990, 0] by [7, 0] (b) Slope of the line between the points (99, 06) and 0-06 (00, 0) is m Using the point-slope form equation for the line, we have y - 06.( - 99), so y.( - 99) [990, 0] by [7, 0] (c) Using y.( - 99) + 06 and 0, the model estimates the average SAT score in 007 was. (d) No; the superimposed graphs in (b) show that data for 00 0 do not follow the pattern for 99 00, so it is unlikely that the average SAT math score in 0 will be nearly as high as.. (a) -y, or y + (b) +y 6, or y - -. m Both graphs look the same, but the graph on the left has slope less than the slope of the one on the right, which is The different horizontal and vertical. scales for the two windows make it difficult to judge by looking at the graphs (-)+(+) (-y)-(-y)y+ 0-y-+yy+ 7-yy+ y 6 y 0. (-) (-) 7 - ; or L () +() (-) 7 - ; or L L Using the quadratic formula: ; - (6)(7) (6) - ; L 0.0 ; or (-)(+)0-0 or L. + L.0 or - Copyright 0 Pearson Education, Inc.

26 6 Chapter P Prerequisites. +80 (+)0 0 or +0 0 or. (+)(+7) (-7)(+)0-70 or +0 7 or 6. + or + or or (-)(-)0 (-) (-)(-)0 (-) (-)0 0 or -0 0 or 0. Solving -+0 by using the quadratic formula with a, b, and c gives ; (-) - ()() () ; i 8. Solving -6+0 by using the quadratic formula with a, b 6, and c gives 6 ; (-6) - ()() () 6 ; i. Solving -+0 by using the quadratic formula with a, b, and c gives ; (-) - ()() () ; i ; i ; i ; i ; ; -6 ; L The graph of y -9 - is zero for 0, - and 7., or or 6. The graph of y is zero for, and. 7. The graph of y - - is zero for L The graph of y ƒ - ƒ - + is zero for and for L.. 9. <+ 7 6< Hence ( 6, ] is the solution Ú - Hence c - is the solution., q b a - b + a - b - ; L -. a - b ;7 ; 7 - ; () - ()(-) () - ; Hence a - q, is the solution. d 6. 7<-<7 << <<6 Hence (, 6) is the solution. + 7 L L 0. Copyright 0 Pearson Education, Inc.

27 Chapter P Review or + or 6 Ú - or Hence a - q, - d h c - is the solution., q b (-)(+)0-0 or +0 or The graph of y +-0 lies above the -ais for < and for 7 Hence (- q, -) h a is., q b the solution. 6. The graph of y -- is zero for L -0.7 and L.7, and lies above the -ais for < 0.7 and for >.7. Hence (- q, -0.7) h (.7, q) is the approimate solution. 66. The graph of y9 -- is zero for L -0.08, and L., and lies below the -ais for 0.08<<.. Hence [ 0.08,.] is the approimate solution is equivalent to The graph of y -9- is zero for L -.8, L -0., and L., and lies below the -ais for <.8 and for 0.<<.. Hence the approimate solution is (- q, -.8] h [-0.,.]. 68. The graph of y -9+ is zero for L -.60, L 0., and L.7, and lies above the -ais for.60<<0. and for >.7. Hence the approimate solution is (-.60, 0.) h (.7, q) > or < +7>0 or +7< 0 > or < 7 Hence (- q, -7) h (, q) is the solution (-7)(+)0-70 or +0 7 or The graph of y +- lies below the -ais for Hence a -, 7 is the solution.. b (+)(+)0 (+) The graph of y ++9 lies entirely above the -ais ecept for - Hence all real numbers satisfy. the inequality. So (- q, q) is the solution (-)(-)0 (-) 0-0 The graph of y -6+9 lies entirely above the -ais ecept for. Hence no real number satisfies the inequality. There is no solution ( + i) ( - i) ( + i + i )( - i + i ) 78. i 9 i 8 i (i ) i (-) i 0 + i ( - i) + (- + i) ( - ) + (- + )i + i ( - 7i) - ( - i) ( - ) + (-7 + )i - i ( + i)( - i) - i + 6i - i + i i ( + i) (( + i)( + i))( + i) ( + i + i )( + i) i( + i) i + i -6 (6)(-) i + i - i + i # + i - i + i - + i 6 8. s 6t +0t (a) 6t +0t8 6t +0t-80 The graph of s 6t +0t-8 is zero at t -0 ; 0 - (-6)(-8) (-6) So t - + i -0 ; (- + i)(- - i) 9 - i + i - 6i + 0i - + i + 0i + i + i + i - i - i 0 ; 6. L 8.0 sec or t L.97 sec. The projectile is 8 ft above ground twice: at t L 8 sec, on the way up, and at t L sec, on the way down. (b) The graph of s 6t +0t lies below the graph of s8 for 0<t<8 and for <t<0 (approimately). Hence the projectile s height will be at most 8 ft when t is in the interval (0, 8] or [, 0) (approimately). (c) The graph of s 6t +0t lies above the graph of s8 for 8<t< (approimately). Hence the projectile s height will be greater than or equal to 8 when t is in the interval [8, ] (approimately). Copyright 0 Pearson Education, Inc.