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Obj. 10 Deductive Reasoning and Biconditional Statements September 26, 2013

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Homework: Khan Academy Logical Arguments and Deductive Reasoning

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Obj. 9 Inductive Reasoning and Conditional Statements September 24, 2013

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Homework: (after notes) Khan Academy Conditional Statements and Conditional Statments and Truth Value

I was never able to get the audio uploaded. Sorry.

Obj. 8 Angles September 21, 2013

Posted by Ms. Miller in Geometry.
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Homework: Khan Academy Angle Types and Exploring Angle Pairs

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)

PAP Unit 1 Test Review Solutions September 19, 2013

Posted by Ms. Miller in Geometry.
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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}

Obj. 7 Midpoint and Distance Formulas September 17, 2013

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Homework: Khan Academy Midpoint Formula and Distance Formula

Obj. 6 Pythagorean Theorem (Pt. 1) September 12, 2013

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Homework: Obj. 6 Pythagorean Theorem (1) WS (If you don’t understand reducing the radicals to simplest form, just round your answers to the nearest tenth.)

We have sound! You can pause the presentation as you take notes, but if you advance or go back a slide, that doesn’t affect the narration, unfortunately.

Obj. 5 Notes – Measuring Segments September 10, 2013

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Homework: Khan Academy Measuring Segments

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Obj. 4 Points, Lines, and Planes September 6, 2013

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Let’s try this up here:

Homework: Khan Academy: Points, Lines, and Planes

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Obj. 3 Equations of Lines September 4, 2013

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Homework: Khan Academy Point-Slope Form and Converting Between Point-Slope and Slope-Intercept