Quadratic Word Problems WS 2 Solutions #110 inches #2 #3 #4 #5 #6 15 cm by 7 cm a. 156.25 ftb. 6.25...
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Transcript of Quadratic Word Problems WS 2 Solutions #110 inches #2 #3 #4 #5 #6 15 cm by 7 cm a. 156.25 ftb. 6.25...
Quadratic Word Problems WS 2 Solutions
#1 10 inches
#2
#3
#4
#5
#6
15 cm by 7 cm
a. 156.25 ft b. 6.25 sec c. 0.55 & 5.70 sec
72 units
96 cm
Height is 5.25 m or 5 m and 25 cmDistance is 3.79 meters or 3 m and 79 cm
About 10 inches
#1
2x + 15
2x + 20
(2x +15)(2x +20) −(15)(20) =60 x =
−35 ± 35 2 −4 • 2 • −30
2 • 2
4x2 + 70x + 300 −300 =60
4x2 + 70x −60 =0
2x2 + 35x −30 =0
x =
−35 ± 1465
4≈0.8188
0.8188 ft x 12 in/ft
#2
The dimensions of the rectangle are 15 cm by 7 cm
Perimeter = 44 cmArea = 105 sq. cm
Width = W
Length = L
2L + 2W = 44
L + W = 22
L = 22 - W
(L)(W) = 105
(22 - W)(W) = 105
22W – W2 = 105
W2 – 22W + 105 = 0
(W - 15)(W - 7) = 105
#3a h = -16t2 + 100t
t =−
b
2a=−100
−32=25
8
h=−16
25
8
⎛
⎝⎜
⎞
⎠⎟
2
+10025
8
⎛
⎝⎜
⎞
⎠⎟
First find Axis of Symmetry
Plug it in to get the vertexSolving for h will give the maximum point (or highest point)
h=−16
625
64
⎛
⎝⎜
⎞
⎠⎟ +25
25
2
⎛
⎝⎜
⎞
⎠⎟
h=−
625
4+625
2=625
4=156.25
#3b
Why does this work?
h = -16t2 + 100t
25
8• 2 =
25
4=6
1
4
Alternate solution
0 = -16t2 + 100t
0 = -4t(4t – 25)
t = 0 or 6.25
Two ways to solve …
#3c h = -16t2 + 100t
x =
50 ± 2500 −4g8g25
2(8)
50 = -16t2 + 100t
16t2 - 100t + 50 = 0
8t2 - 50t + 25 = 0
x =
50 ± 1700
16=50 ±10 17
16 ≈0.55 or 5.70
#4
x
x + 6
x
x + 9
(x)(x) =
1
2x + 6( ) x + 9( )
x2 =
1
2x2 +15x + 54( )
2x2 =x2 +15x + 54
x2 −15x −54 =0
(x −18)(x + 3) =0
x =18
Square's perimeter = 72 units
Volume = L x W x H
Width = w
Length = 3w
512 = (3w - 4)(w - 4)(2)
w - 4 3w - 4
256 = (3w - 4)(w – 4)
256 = 3w2 – 16w + 16
0 = 3w2 – 16w - 240
0 = (w – 12)(3w + 20)
Width = 12 cm
Perimeter = 96 cm
3 meters
Maximum height (at vertex)
height when I hit the water?
Distance from board when entering water?
Mr. Lomas (at least the way I remember me) ….
y =−x 2 + 3x + 3
#6
y =−x 2 + 3x + 3
x =
−b
2a=3
2
y =−
3
2
⎛
⎝⎜
⎞
⎠⎟
2
+ 33
2
⎛
⎝⎜
⎞
⎠⎟ + 3
y =−
9
4+9
2+ 3
y =−
9
4+18
4+12
4=21
4=5.25
Maximum height of 525 cm
0 =−x 2 + 3x + 3
x2 −3x −3 =0
x =3 ± −3( )
2
−4 1( ) −3( )
2 1( )
x =
3 ± 21
2≈3.79 meters
#7Equation: y = x2 - 10x + 15 Direct: UpWidth: StandardL of S: x = 5Vertex: (5, -10)y-int: (0, 15)
Roots: 5± 10
#8Equation: y = -2x2 + 8 Direct: DownWidth: NarrowL of S: x = 0Vertex: (0, 8)y-int: (0, 8)Roots: 2 or -2
#9Equation: y = (1/4)x2 - 2x - 5 Direct: UpWidth: WideL of S: x = 4Vertex: (4, -9)y-int: (0, -5)Roots: 10 or -2