Algebra: Graphing, Distance between Points and Equations of Line
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Page 187
1. a) 12, b) 14, c) 16, d) 18 (Check for all four of our symmetries SY, SX, SO, SI; consult in WEEK7 NOTES, in COURSE CONTENT. Practice graphing these using the downloaded graphing utility GraphCalc.)
2. a) 20, b) 24
3. 28
4. 32
P. 203
5. a) 22, b) 24
6. a) 26, b) 30
7. a) 36, b) 42
P. 219
8. a) 8, b) 18
9. a) 36, b) 38
10. a) 46, b) 54, c) 58, d) 64, e) 78
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Solution Summary
Graphing problems involving linear equations are answered.
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P187.
12.)
y = 1/2 x + 1 has no symmetry about either x-axis or y-axis or origin.
(see attached file). --Answer
14.)
y = 2x is unsymmetric about either x-axis or y-axis or origin. --Answer
16.)
|y| = -x
for y> 0 => y = -x
for y < 0 => -y = -x => y = x
=> symmetric about y-axis --Answer
18.)
y = -x => symmetric about origin.
20.)
Distance between points (-6,4) and (2,-1)
s = sqrt ((-6 - 2)^2 + (4 - (-1))^2)
s = sqrt((-8)^2 + (5)^2) = sqrt(64 + 25) = sqrt(89) --Answer
24.)
C =(x1,y1) =(0,0); r = 6
eqn of circle,
(x-x1)^2 + (y-y1)^2 = r^2
=> (x-0)^2 + (y-0)^2 = 6^2
=> x^2 + y^2 = 36 --Answer
28.)
C(-1,-3) = (x1, y1) ; r = sqrt(5)
Eqn of circle,
(x - x1)^2 + (y - y1)^2 = r^2
=> (x - (-1))^2 + (y - (-3))^2 = (sqrt(5))^2
=> (x + 1)^2 + (y + 3)^2 = 5
=> x^2 + 2x + 1 + y^2 + 6y + 9 = 5
=> x^2 + Y^2 + 2x + 6y + 5 = 0 --Answer
32.)
Figure shows the symmetry about y-axis only --Answer(B)
P203.)
22.)
slope (m)= -1, y-intercept (c)= 7
Eqn of straight line:
y = mx + c
=> y = -1*x + 7
=> x + y = 7 ...
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