## ashontae19 one year ago anybody in geomtry in ga virtual school?

1. k_lynn

2. ashontae19

i need help with this

3. k_lynn

@freckles , @mathmate , or @Michele_Laino would probably be able to help you with that. I haven't gotten to that yet.

4. ashontae19

@Michele_Laino @mathmate @freckles could any one of yall help me

5. ashontae19

oh are you not in 10th grade?

6. Michele_Laino

question #1a the requested inequality is: $\Large - 12{x^2} + 960x \geqslant 18000$

7. ashontae19

thank you but i need help pn 3a 3b

8. ashontae19

@Michele_Laino could you help me

9. Michele_Laino

question #3a the requested inequality is: $\Large - 12{x^2} + 960x \leqslant 18432$

10. Michele_Laino

dividing both sides by 12, we can rewrite that inequality as follows: $\Large {x^2} - 80x + 1536 \geqslant 0$

11. ashontae19

so thats the inequality?

12. ashontae19

would the answer be </ 32 and </ 48?

13. Michele_Laino

the correspnding equation is: $\Large 12{x^2} - 960x + 18432 = 0$ whereas the critical values are: x=32, and x=48

14. Michele_Laino

finally the solution to that inequality is: $\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x$

15. ashontae19

ok thank you so for 3a where as it says write an inequailty when she will not be taxes will it be >/ 50 because on the graph

16. Michele_Laino

ok!

17. ashontae19

18. Michele_Laino

the requested inequality is $\Large x \geqslant 48$

19. ashontae19

ok thank you. could you help me on more or are you stressed out?

20. Michele_Laino

21. ashontae19

ok thankyou here are the other ones

22. Michele_Laino

question #4a the requested inequality is: $\Large - 12{x^2} + 960x \leqslant 14400$

23. ashontae19

how would you do b?

24. Michele_Laino

here is the graph:

25. ashontae19

THANK YOU ! so i would shade below the graph correct?

26. Michele_Laino

yes! correct!

27. Michele_Laino

28. ashontae19

ok for E it says are any of the values not in the orginal domain explain your answer and write your final solution as inequality.

29. Michele_Laino

yes all the values x, such that: $\Large 0 \leqslant x < 30$ don't belong to the original domain of our function: $\Large y=- 12{x^2} + 960x$

30. Michele_Laino

so the solution is: $\Large x \geqslant 60$

31. ashontae19

wait so how would x> 60 belong in the equation ?

32. Michele_Laino

since the intersection between the graph: $y = - 12{x^2} + 960x$ and the line: $y = 14400$ occurs at x=20 and x=60. Now all x-values, such that 0</ x</20 don't belong to the domain of our function, so we have towrite x>/ 60

33. ashontae19

ok thank you what about these

34. Michele_Laino

question #5 as we can see from the drawing below, our solution is: $x \leqslant 1 \cup x \geqslant 5$

35. ashontae19

what does u stand for?

36. Michele_Laino

it is the union of sets

37. ashontae19

do i put and or or ? or u

38. Michele_Laino

that writing is equivalent to this one: $( - \infty ,1] \cup [5, + \infty )$ you can put "or"

39. Michele_Laino

Question #6 as we can see from the drawing below, our solution is: $- 1 \leqslant x \leqslant 5$

40. Michele_Laino

Question #7 here you have to solve the quadratic equation associated to your inequality, first. For example, if we consider case a), we get: ${x^2} - 4x + 3 < 0$ so you have to solve this equation: ${x^2} - 4x + 3 = 0$ what do you get?

41. ashontae19

x=3,1?

42. Michele_Laino

ok!

43. Michele_Laino

so we can write the solution as an inequality below: $1 < x < 3$

44. ashontae19

45. Michele_Laino

it is: |dw:1434396507609:dw| and x=1 , x=3 don't belong to the solution interval

46. Michele_Laino

case b) here after a simplification, we get: $3{x^2} - 5x - 12 > 0$ so we have to solve this equation: $3{x^2} - 5x - 12 = 0$ what do you get?

47. ashontae19

1.93?

48. Michele_Laino

are you sure?

49. Michele_Laino

50. ashontae19

ya...

51. Michele_Laino

I got: x=3, and x=-4/3 so our solution is: $x < - 4/3,\;x > 3$ |dw:1434399244525:dw|

52. Michele_Laino

x=3, and x=-4/3 don't belong to the solution interval

53. Michele_Laino

case c) we can write: ${x^2} + 6x + 9 \geqslant 0$ now we have to nothe that: ${x^2} + 6x + 9 = {\left( {x + 3} \right)^2}$

54. Michele_Laino

so our inequality is checked for all real values of x: |dw:1434399797411:dw| in x=-3 we have: ${x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0$

55. ashontae19

i got x>/ -3?

56. ashontae19

ok thank you for 1d on paula peaches i got 30 </ x</ 50 is that correct?

57. Michele_Laino

the inequality: ${x^2} + 6x + 9 \geqslant 0$ is always true, and the equal sign holds, when x=-3

58. Michele_Laino

30</x</50 is not correct, since the production of peaches can not be higher than 14400, so the solution is x>/60

59. Michele_Laino

more precisely: 60</x</80

60. ashontae19

for 2b i got x would equal 50, 30 and c im a little lost on thats the last two

61. Michele_Laino

there we have to solve this inequality: $- 12{x^2} + 960x \geqslant 18000$

62. Michele_Laino

which, after a simplification, becomes: ${x^2} - 80x + 1500 \leqslant 0$

63. Michele_Laino

now the roots of the associated quadratic equation: ${x^2} - 80x + 1500 = 0$ are: x=50 and x=30

64. Michele_Laino

so the requested solution is: $30 \leqslant x \leqslant 50$

65. Michele_Laino

so we have the subsequent graph: |dw:1434401181442:dw|

66. ashontae19