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## Obj. 10 Deductive Reasoning and Biconditional StatementsSeptember 26, 2013

Posted by Ms. Miller in Geometry.
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Homework: Khan Academy Logical Arguments and Deductive Reasoning

## Obj. 9 Inductive Reasoning and Conditional StatementsSeptember 24, 2013

Posted by Ms. Miller in Geometry.
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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 AnglesSeptember 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 SolutionsSeptember 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 SolutionsSeptember 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 FormulasSeptember 17, 2013

Posted by Ms. Miller in Geometry.
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Homework: Khan Academy Midpoint Formula and Distance Formula

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

Posted by Ms. Miller in Geometry.
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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 SegmentsSeptember 10, 2013

Posted by Ms. Miller in Geometry.
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Homework: Khan Academy Measuring Segments

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## Obj. 4 Points, Lines, and PlanesSeptember 6, 2013

Posted by Ms. Miller in Geometry.
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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 LinesSeptember 4, 2013

Posted by Ms. Miller in Geometry.
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Homework: Khan Academy Point-Slope Form and Converting Between Point-Slope and Slope-Intercept