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Warm up Solve 1. 12 – 6r = 2r + 36 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

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Page 1: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Warm up

Solve

1. 12 – 6r = 2r + 361. 12 – 6r = 2r + 36

2. 3k – 5 = 7k + 7 2. 3k – 5 = 7k + 7

3. 2(x + 4) = 6x3. 2(x + 4) = 6x

r = -3

k = -3

x = 2

Page 2: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Review the HWReview the HW

Page 3: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

11, yx 22 , yx

2 22 1 2 1( ) ( )D x x y y

The Distance The Distance FormulaFormula

Page 4: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Example Find the distance between (1, 4) and (-2, 3). Round to the nearest hundredths.

D = 3.16

2122

12 )()(D yyxx

22 )43()12(D

22 )1()3(D

19D

10D

Page 5: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Example

Find the distance between the points, (10, 5) and (40, 45). Round to the nearest hundredths.

2122

12 )()(D yyxx

22 )545()1040(D

22 )40()30(D

1600900D

2500D

D = 50

Page 6: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

3. Find the distance between the points. Round to the nearest tenths.

( ) ( )4 1 2 02 2

9 413

2122

12 yyxx

Page 7: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

4. Find the distance between the points. Round to the nearest tenths.

2122

12 yyxx

Page 8: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Pythagorean Pythagorean TheoremTheorem2 2 2leg leg hyp

Page 9: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Pythagorean Theorem Word Pythagorean Theorem Word ProblemsProblems

• A square has a diagonal with length of 20 cm. What is the measure of each side? Round to the nearest tenths.

x = 14.1 cm

Page 10: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Pythagorean Theorem Word Pythagorean Theorem Word ProblemsProblems

• A 25 foot ladder is leaning against a building. The foot of the ladder is 15 feet from the base of the building. How high is the top of the ladder along the building? Round to the nearest tenths.

x = 20 ft

Page 11: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Pythagorean Theorem Word Pythagorean Theorem Word ProblemsProblems

• Ashley travels 42 miles east, then 19 miles south. How far is Ashley from the starting point? Round to the nearest tenths.

x = 46.1 miles

Page 12: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Pythagorean Theorem Word Pythagorean Theorem Word ProblemsProblems

• What is the length of the altitude of an equilateral triangle if a side is 12 cm? Round to the nearest tenths.

x = 10.4 cm

Page 13: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

ClassworkClasswork Round to the nearest tenths.

Worksheet

Page 14: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

Classwork / Classwork / HomeworkHomework

Quadrilaterals Task

Page 15: Warm up Solve 1. 12 – 6r = 2r + 36 2. 3k – 5 = 7k + 7 3. 2(x + 4) = 6x r = -3 k = -3 x = 2

ConclusionsConclusionsParallel•Same slope means sides are parallel•Opposite reciprocal slopes mean perpendicular segments (90)

Distance•Same distance means segments are congruent