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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.
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