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Who Wants To Be A College Algebra Millionaire? Question 9 (Worth $1,000)

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Presentation on theme: "Who Wants To Be A College Algebra Millionaire? Question 9 (Worth $1,000)"— Presentation transcript:

1

2 Who Wants To Be A College Algebra Millionaire?

3 Question 9 (Worth $1,000)

4 Find B + E. A 6 B 3 C 9 D 20

5 A ( B ( C ( D ( The Solution:

6 $1,000

7 Question 11 (Worth $2,000)

8 11. Find -2A. A 6 B 3 C 9 D 20

9 Solution -2A: A 6 B 3 C 9 D 20

10 $2,000

11 Question 12 (Worth $4,000)

12 12. Find D - A. A 3 B 9 C 1 D 2

13 A 3 B 9 C 1 D 2 Solution D - A.

14 $4,000

15 Question 14 (Worth $8,000)

16 A -7 B -13 C -15 D -17 14. Find |D|.

17 A -7 B -13 C -15 D -17 Solution |D|.

18 $8,000

19 Question 15 (Worth $16,000)

20 A 26 B 24 C 50 D 74 15. Find |E|.

21 A 26 B 24 C 50 D 74 15. Solution |E|.

22 $16,000

23 Question 17 (Worth $32,000)

24 A (-17 B ( C ( D ( 17. Find D -1.

25 17. Solution D -1. A (-17 B ( C ( D ( Negate Switch

26 $32,000

27 Question 10 (Worth $64,000)

28 10. Find AC: A. B. C. D.

29 Solution AC: A D B D C D D

30 $64,000

31 Question 13 (Worth $125,000)

32 A y B y C y D y 13. Find D 2 + 2D:

33 A y B y C y D y Solution D 2 + 2D:

34 $125,000

35 Question 1 (Worth $250,000)

36 A No Intersection B (1,-3) C (-3,5) and (1, -3) D (3, 5) and (-1, -3) 1.Solve the system: y + 4 = x 2 2x + y = -1

37 A No Intersection B (1,-3) C (-3,5) and (1, -3) D (3, 5) and (-1, -3) 1.Solve the system: y + 4 = x 2 2x + y = -1

38 $250,000

39 Question 2 (Worth $500,000)

40 A No Intersection B (4,3) and (3,4) C (-4,3) and (-3,4) D (4,-3) and (3,-4) 2.Solve the system: x 2 + y 2 = 25 x - y = 7

41 A No Intersection B (4,3) and (3,4) C (-4,3) and (-3,4) D (4,-3) and (3,-4)

42 $500,000

43 Question 19 (Worth $1,000,000)

44 A 1 B 1 C 2 D 2 19. Use Cramer’s Rule to solve the system: 5x - 6y = 4 3x + 7y = 8

45 19. Use Cramer’s Rule to solve the system: 5x - 6y = 4 3x + 7y = 8 A 1 B 1 C 2 D 2

46 $1,000,000

47 Congratulations… You win!


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