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Math 137 Assignment 3 Chap 3 and 4 Key y Sec. 3.1 1. 3 x + y = 5 -> #4. x y = −3 x + 5 –5 –4 –3 –2 –1 2. x − 2 y = 4 -> y = 1 2 3 4 5 1 x−2 2 Sol’n: (2,-1) y #14. 1. y − x = 5 -> y = x + 5 x 2. 2 x − 2 y = 10 -> −2 y = −2 x + 10 -> y = x − 5 –5 –4 –3 –2 –1 1 2 3 4 5 Sol’n: No Solution (lines are parallel) Sec. 3.2 #4. 1. 9 x − 2 y = 3 2. 3 x − 6 = y -> y = 3 x − 6 9 x − 2(3x − 6) = 3 9 x − 6 x + 12 = 3 Sub 1 into 2 3x = −9 y = 3(−3) − 6 Sub x into 2 = −9 − 6 Sol’n: (-3,-15) = −15 x = −3 #14 1. 5 x + 3 y = 4 2. x − 4 y = 3 -> x = 4 y + 3 Sub 2 into 1 −11 x = 4 +3 23 5(4 y + 3) + 3 y = 4 −44 20 y + 15 + 3 y = 4 = +3 25 −11 23 Sub y into 2 Sol’n: , 23 y = −11 −44 69 23 23 = + −11 y= 23 23 23 25 = 23 #20 Let and y represent the complementary angles 1. x + y = 90 2. y = 5 x + 6 x + (5 x + 6) = 90 Sub 2 into 1 6 x + 6 = 90 y = 5(14) + 6 sub into 2 = 70 + 6 6 x = 84 x = 14 Sol’n: The angles are 14° and 76° = 76 Sec. 3.3 #10 1. 3 x − 2 y = 1 -> 6x − 4 y = 2 2. −6 x + 4 y = −2 -> −6 x + 4 y = −2 Add 1 + 2 0=0 (true) Sol’n: Infinite number of solutions #28 1. 1.3 x − 0.2 y = 12− > 13 x − 2 y = 120 1. x 85 sub into 1. 13(10) -2y=120 1105x-170y=10 200 2. 0.4 x + 17 y = 89 − > 4 x + 170 y = 890 130 – 2y=120 4x +170y=890 -2y=-10 Add 1. + 2. 1109x=11090 y=5 x=10 Sol’n: (10,5) Sec. 3.4 #8 Let x= amt of Deep Thought Granola Item DT OT Mix 1. x+y=20 -> y=20-x Let y= amount of Oat Dream Granola Quantity 1 x y 20 2. 0.25x+0.10y=38 -> 25x + 10y = 380 Sub 1 into 2 15x=360 Let x= # quarters y= 8 Sol’n: 12 lbs of Deep Thought and 8 lbs of Oat Dream should be mixed. x=12 Let y = # loonies Item quarters loonies Mix 1. x+y=19 -> y=19-x Quantity 1 x y 19 2. 0.25x+ y=10.75 -> 25x+100y=1075 Sub 1 into 2 Quantity 2 0.25x 0.10y 20 i 0.19 = 3.8 Sub into 1. y=20-12 25x+ 10(20-x)=380 25x + 200-10x=380 #18 Rate 0.25 0.10 .19 25x + 100(19-x)=1075 Sub into 1. Quantity 2 0.25x y 20-9.25 y=19-11 y=8 25x + 1900-100x=1075 -75x= -825 Rate 0.25 0.50 Sol’n: There are 11 quarters and 8 loonies. x=11 Sec 3.7 14. #22. x ≤ −4 2x+3y<6 The line 2x+3y=6 has x intercept: y 2x+3(0)=6 5 4 2x=6 3 2 y=3 (3,0) and y-intercept: 2(0)+3y=6 y 3y=6 5 y=2 (0,2) 2 x 1 –5 –4 –3 –2 –1 -1-2-3-4 1 2 3 4 5 -5 [Note: please fill in the region on a test] 4.1 4) Degree of polynomial: 7 Leading coefficient: -1 -1-2-3-4 -5 4 3 x 1 –5 –4 –3 –2 –1 Leading Term: -a4b3 Constant term: -11 1 2 3 4 5 76) (10 xy − 4 x 2 y 2 − 3 y 3 ) − (− 9 x 2 y 2 + 4 y3 − 7 xy) = 10 xy − 4 x 2 y 2 − 3 y 3 + 9 x 2 y 2 − 4 y 3 + 7 xy = 5 x 2 y 2 − 7 y 3 +17 xy 4.2 30) (2 x 2 + y 2 − 2 xy )( x2 − 2 y2 − xy ) = 2 x 4 − 4 x 2 y 2 − 2 x3 y + x2 y2 − 2 y4 − xy3 − 2 x3 y + 4 xy3 + 2 x2 y2 = 2 x 4 − x 2 y 2 − 4 x3 y − 2 y4 + 3 xy3 80) (2 x − y )(2 x + y )(4 x 2 − y 2 ) = (4 x 2 − y 2 )(4 x2 − y2 ) = 16 x 4 − 4 x 2 y 2 − 4 x2 y2 + y4 = 16 x 4 − 8 x 2 y 2 + y 4 90) f ( x ) = 4x − 2x 2 f (a + h ) − f (a ) = (4(a + h ) − 2(a + h )2 ) − (4(a ) − 2(a )2 ) = (4 a + 4 h − 2(a 2 + 2ah + h 2 )) − (4 a − 2 a 2 ) = 4 a + 4 h − 2 a 2 − 4 ah − 2h 2 − 4 a + 2 a 2 = −4ah − 2h 2 + 4h 4.3 8) 48) 5 x 2 y 3 + 15 x 3 y 2 t 3 + 6t 2 − 2t − 12 = t 2 (t + 6) − 2(t + 6) = 5 x 2 y 2 ( y + 3 x) = (t + 6)(t 2 − 2) 4.4 22) 30) 32 + 4 y − y 2 p 2 − 5 pq − 24 q 2 = ( p − 8q )( p + 3q ) = − y 2 + 4 y + 32 = −1( y 2 − 4 y − 32) = −1( y − 8)( y + 4) 4.5 20) 70 x 4 − 68 x 3 + 16 x 2 = 2 x 2 (35 x 2 − 34 x + 8) 2 32) 15 y 2 − 10 − 15 y = 15 y 2 − 15 y − 10 2 = 2 x (35 x − 20 x −14 x + 8) 2 = 2 x (5 x(7 x − 4) − 2(7x − 4)) = 2 x 2 (7 x − 4)(5 x − 2) 44) 8m 2 − 6 mn − 9 n 2 = 8 m 2 − 12 mn + 6 mn − 9 n 2 = 4 m (2 m − 3 n ) + 3 n (2 m − 3 n ) = (2 m − 3 n )(4 m + 3 n ) = 5(3 y 2 − 3 y − 2) 4.6 26) 49 p 2 − 84 pq + 36 q 2 42) 25 ab 4 − 25 az 4 = 49 p 2 − 42 pq − 42 pq + 36 q 2 = 7 p (7 p − 6q ) − 6q (7 p − 6q ) = 25 a (b 4 − z 4 ) = (7 p − 6 q )2 = 25 a (b − z )(b + z )(b 2 + z 2) 62) = 25 a (b 2 − z 2 )(b 2 + z 2 ) 84) x 2 − 2 xy + y 2 − 25 27 y 3 + 64 = ( x − y )( x − y) − 25 = (3 y + 4)(9 y 2 − 12 y + 16) = ( x − y ) 2 − 25 = ( x − y + 5)( x − y − 5) 4.7 38) 250 a 3 + 54b 3 47) 5 x3 − 5 x2 y − 5 xy2 + 5 y3 = 2(125 a 3 + 27 b 3 ) = 5( x3 − x 2 y − xy 2 + y 3 ) = 2(5 a + 3b)(25 a 2 −15 ab + 9 b2 ) = 5( x 2 ( x − y) − y 2 ( x − y)) = 5( x − y)( x2 − y2 ) = 5( x − y)( x + y)( x − y) = 5( x − y) 2 ( x + y) 4.8 147 y = 3 y3 38) 0 = 3 y 3 −147 y 0 = 3 y ( y 2 − 49) 0 = 3 y ( y + 7)( y − 7) y=0 or y=7 or y=-7 66) A triangular sail is 9m taller than it is wide. The area is 56 m^2. Find the height and the base of the sail. 1 A = bh 2 1 56 = b (9 + b ) 2 2 2(56) = 9 b + b 112 = 9b + b 2 Solution: Base is 7 m and height is 16 m. (b=-16 is rejected because length can’t be negative) 0 = b 2 + 9 b −112 0 = (b + 16)(b − 7) 80) One leg of a right triangle has length 10 cm. The other sides have lengths that are consecutive EVEN integers. Find these lengths. x x+2 10 2 = x 2 + 102 x + 4 x + 4 = x 2 + 100 4 x = 96 x = 24 The hypotenuse is x + 2 = 24 + 2 = 26 ( x + 2) 2 Solution: The other leg is 24cm cm and the hypotenuse is 26 cm.