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Reg. Geometry Unit 1 Test Review Solutions September 19, 2013

Posted by Ms. Miller in Uncategorized.
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Note: These are the answers to the Regular Geometry 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. A 2. m=\frac{-6}{4}=\frac{-3}{2}
3. \frac{a-17}{2-4}=6
a-17=-12
a=5
4. -4y=-5x+10
y=\frac{-5}{-4}x+\frac{10}{-4}
m=\frac{5}{4}
5. y+3=-2(x-7)
6. y-0=3(x+2)
y=3x+6
7. y-1=\frac{2}{3}(x-3)
8. perpendicular slope: m=\frac{1}{3},
therefore C
9.
10. \overrightarrow{}TW 11. collinear; coplanar
12. 2\frac{1}{4}"; 6.1 \mathrm{cm} 13. 6
14. omit 15. omit
16. 15\sqrt{2} 17. 5\sqrt{3}=\sqrt{(5)(5)(3)}=\sqrt{75}
18. c^2=8^2+14^2
c^2=260
c=\sqrt{260}=2\sqrt{65}
19. a^2+5^2=18^2
a^2=299
a=\sqrt{299}
20. 7 (Pythagorean triple) 21. LS^2+5^2=10^2
LS^2=75
LS=\sqrt{75}=5\sqrt{3}
22. (\frac{-1+3}{2},\frac{7-4}{2})=(1,\frac{3}{2}) 23. \frac{2+y}{2}=0
y=-2
24. d=\sqrt{(7-3)^2+(1+1)^2}=\sqrt{20}=2\sqrt{5}
25. d=\sqrt{(2-1)^2+(-7-7)^2}=\sqrt{197}
26. d=\sqrt{(-3-4)^2+(9-0)^2}
\sqrt{130}\approx 11.4
27. (7,11)
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