jump to navigation

PAP Unit 1 Test Review Solutions September 19, 2013

Posted by Ms. Miller in Geometry.
trackback

Note: These are the answers to the PAP Test Review.

Proof #1:

1. Given equation
2. Dist. prop.
3. 3x-13=3 3. Subtr. prop. =
4. 3x=16 4. Add. prop. =
5. x=\frac{16}{3} 5. Div. prop. =
Proof #2:

1. 7x+11=3x-5 1. Given equation
2. 4x+11=-5 2. Subtr. prop. =
3. 4x=-16 3. Add. prop. =
4. x=-4 4. Div. prop. =
1. C 2. m=\frac{-6}{4}=\frac{-3}{2}
3. m=\frac{0-b}{3b-0}=\frac{-b}{3b}=-\frac{1}{3} 4. \frac{a-17}{2-4}=6
a-17=-12
a=5
5. -4y=-5x+10
y=\frac{-5}{-4}x+\frac{10}{-4}
m=\frac{5}{4}
6. y+3=-2(x-7)
7. y-0=3(x+2)
y=3x+6
8. y-1=\frac{2}{3}(x-3)
9. perpendicular slope: m=\frac{1}{3},
therefore C
10.
11. \overrightarrow{}TW 12. collinear; coplanar
13. 2\frac{1}{4}"; 6.1 \mathrm{cm} 14. 6
15. omit 16. omit
17. 15\sqrt{2} 18. 5\sqrt{3}=\sqrt{(5)(5)(3)}=\sqrt{75}
19. c^2=8^2+14^2
c^2=260
c=\sqrt{260}=2\sqrt{65}
20. a^2+5^2=18^2
a^2=299
a=\sqrt{299}
21. 7 (Pythagorean triple) 22. LS^2+5^2=10^2
LS^2=75
LS=\sqrt{75}=5\sqrt{3}
23. (\frac{-1+3}{2},\frac{7-4}{2})=(1,\frac{3}{2}) 24. \frac{2+y}{2}=0
y=-2
25. (\frac{a+a-6}{2},\frac{b+3b}{2})=(\frac{2a-6}{2},\frac{4b}{2})
=(a-3,2b)
26. (7,11)
27. d=\sqrt{(7-3)^2+(1+1)^2}=\sqrt{20}=2\sqrt{5}
28. d=\sqrt{(2-1)^2+(-7-7)^2}=\sqrt{197}
29. C(-3,4) 30. d=\sqrt{(-3-4)^2+(9-0)^2}
\sqrt{130}\approx 11.4
31. d=\sqrt{(2a-3a)^2+(3b-4b)^2}=\sqrt{(-a)^2+(-b)^2}
=\sqrt{a^2+b^2}
Advertisements
%d bloggers like this: