## NatalieandKayla 2 years ago solve the inequality. graph your solution. |x+3| > or equal to 15

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1. jim_thompson5910

Hint: |x| > k breaks into x < -k or x > k where k is some positive number

2. NatalieandKayla

is it still correct if the arrows go in the same direction when I graph it?

3. jim_thompson5910

no the arrows should be going in opposite directions

4. NatalieandKayla

oh so you flip the sign when you make k a negative?

5. jim_thompson5910

exactly

6. jim_thompson5910

what did you get

7. NatalieandKayla

x> or equal to 12 and x < or equal to -18

8. jim_thompson5910

good, you got it

9. NatalieandKayla

thanks :D

10. jim_thompson5910

you're welcome

11. NatalieandKayla

find the slope of the line passing through the given points. then tell whether the line rises, falls, is horizontal, or is vertical. (-7,4) (5,-3)

12. jim_thompson5910

use the slope formula m = (y2 - y1)/(x2 - x1)

13. jim_thompson5910

in this case (x1,y1) = (-7,4) (x2,y2) = (5,-3)

14. NatalieandKayla

how do I find out whether the line rises, falls..etc ?

15. jim_thompson5910

what did you get for the slope

16. NatalieandKayla

-7/12

17. NatalieandKayla

oh so the line decreases bc it has a negative slope?

18. jim_thompson5910

you nailed it

19. NatalieandKayla

Do you mind if i ask you questions from time to time ? I'm trying to study for a trig final on monday.

20. jim_thompson5910

sure I can help

21. NatalieandKayla

thanks! alright so this question has the same guidelines as the one we just did but with these coordinates (-5/4, 3); (2/3, 3) my friend had an error in her calculator trying to find the slope , but I got m=0 ...im confused

22. jim_thompson5910

you're correct, it's 0 how do I know so quickly? notice that the y coordinates are both equal to 3 since the y coordinates are equal, the slope will always be 0 because you have a horizontal line through y = 3

23. jim_thompson5910

all horizontal lines have a slope of 0

24. jim_thompson5910

using m = (y2 - y1)/(x2 - x1) we get m = (y2 - y1)/(x2 - x1) m = (3 - 3)/(2/3 - (-5/4)) m = 0/(2/3 - (-5/4)) m = 0

25. NatalieandKayla

oh ok i understand

26. NatalieandKayla

what if the x coordinates are equal? would that make the line vertical?

27. jim_thompson5910

exactly

28. jim_thompson5910

and the slope would be undefined

29. NatalieandKayla

oh okay

30. NatalieandKayla

how would i graph y=|x-3|?

31. jim_thompson5910

do you know how to graph y = | x |

32. NatalieandKayla

i think so? would there be 2 lines?

33. jim_thompson5910

yes here is your blank xy axis |dw:1359244103163:dw|

34. jim_thompson5910

the graph of y = | x | looks like this |dw:1359244135745:dw|

35. jim_thompson5910

to graph y=|x-3|, you just shift over y = |x| three units to the right

36. NatalieandKayla

im not sure i get you there...

37. jim_thompson5910

imagine you can pick up that v shaped graph and move it around well just move it over 3 to the right units like this |dw:1359244315265:dw|

38. NatalieandKayla

but how do i know the coordinates to create the v shape

39. NatalieandKayla

i understand the moving it to the right part

40. jim_thompson5910

the sharp point on the V of y = |x| is at (0,0) move that point over 3 units to the right to get (3,0)

41. jim_thompson5910

|dw:1359244469161:dw|

42. NatalieandKayla

yes i understand that but what are the other coordinates for that shape?

43. NatalieandKayla

for (3,0)..

44. jim_thompson5910

(1,1) is on the original graph y = |x| move that 3 units to the right to get (4,1)

45. jim_thompson5910

|dw:1359244638800:dw|

46. jim_thompson5910

(-1,1) is on the original graph y = |x| move that 3 units to the right to get (2,1)

47. jim_thompson5910

|dw:1359244675721:dw|

48. NatalieandKayla

ohhhh so the slope for it is rise 1 over 1?

49. jim_thompson5910

for the right half of it, yes

50. jim_thompson5910

|dw:1359244723365:dw|

51. jim_thompson5910

left half, slope is -1 |dw:1359244745927:dw|

52. NatalieandKayla

ohhh ok so on the left side for any problem i will always decrease 1 and rise 1?

53. NatalieandKayla

im trying to understand the concept so i can graph any absolute value problem

54. jim_thompson5910

on the left side, you go down 1 and over to the right 1 on the right side, you go up 1 and over to the right 1

55. NatalieandKayla

the problem they gave me was y=|x-3| would you have to move to graph from (0,0) to (0,-3) ? you moved it up 3

56. jim_thompson5910

if it was |x| - 3, then you would move (0,0) to (0,-3)

57. jim_thompson5910

but it's actually |x-3|

58. NatalieandKayla

oh

59. NatalieandKayla

would i be able to just plug this in to my calculator ?

60. jim_thompson5910

yes you should be able to

61. jim_thompson5910

do you know how to graph absolute value equations?

62. NatalieandKayla

I guess I forgot :/

63. jim_thompson5910

what kind of calculator do you have

64. NatalieandKayla

i just found the absolute value button on my TI-83

65. NatalieandKayla

and i put it into the y= and i got what you showed me earlier

66. jim_thompson5910

ok great, glad you did

67. NatalieandKayla

Solve the equation. \[4x ^{2}-6x+9=0\]

68. jim_thompson5910

you can either factor or use the quadratic formula

69. jim_thompson5910

oh wait, one sec

70. jim_thompson5910

no you can't factor, so you'll have to use the quadratic formula

71. NatalieandKayla

how do you know you cant factor this problem?

72. jim_thompson5910

D = b^2 - 4ac D = (-6)^2 - 4(4)(9) D = 36 - 4(4)(9) D = 36 - 144 D = -108 because the discriminant is negative, there are no real solutions so that's why you can't factor

73. NatalieandKayla

oh okay, and i can use the discriminant as a test on any problem that looks like this one?

74. jim_thompson5910

yes if the discriminant is a) positive or zero and b) a perfect square then the quadratic can be factored

75. NatalieandKayla

ohhhhh okay

76. NatalieandKayla

solve the system of linear equations using any method. 2x+3y=47 7x-8y=-2 2x-y+3z=-19

77. jim_thompson5910

solve the first two equations for x and y once you have x and y, use them to find z

78. NatalieandKayla

im not sure how i ended up with 111 for y... the answers say its wrong

79. jim_thompson5910

no y should be 9, try it again

80. NatalieandKayla

i did elimination with the first two equations

81. NatalieandKayla

to get them to equal to 14 and -14 so they can cancel out but i keep getting 111

82. jim_thompson5910

2x+3y=47 7x-8y=-2 8(2x+3y)=8*47 3(7x-8y)=3(-2) 16x+24y=376 21x-24y=-6 ------------- 72x+0y = 370 37x = 370 x = 370/37 x = 10 2x+3y=47 2(10)+3y=47 20+3y = 47 3y = 47-20 3y = 27 y = 27/3 y = 9 So x = 10 and y = 9 Use them to find z

83. NatalieandKayla

Oh! I understand now, thank you so much :)

84. jim_thompson5910

you're welcome

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