## bobobox one year ago I have seven math questions I need help answering please, It's my very very last assignement of the year and I need to pass.

1. bobobox

2. anonymous

Hit me with em (;

3. anonymous

I'm quite sure the answer is C on that one

4. bobobox

okay

5. bobobox

6. mathstudent55

|dw:1435638657119:dw|

7. mathstudent55

1. \((3 + 2x)(5 + 2x) = 19.25\) \((2x + 3)(2x + 5) = 19.25\) \(4x^2 + 16x + 15 = 19.25\) \(4x^2 + 16x - 4.25 = 0\)

8. bobobox

Yay thanks so much for the explanation! It helps!

9. bobobox

Could you help me with the soccer one?

10. bobobox

@mathstudent55 @willb1220

11. mathstudent55

The acceleration due to gravity is g in the down direction, so let;'s call it -g. \(a = -g\) The velocity is \(v(t) = -gt + v_0\) where \(v_0\) is the initial velocity. The position (height) is \(h = -\dfrac{1}{2}gt^2 + v_0t + h_0\) where \(h_0\) is the initial height.

12. bobobox

okay

13. mathstudent55

We are told that the initial position is 0, so \(h_0 = 0\) We are also told the initial velocity is 42 ft/sec, so \(v_0 = 42\)

14. bobobox

yeah

15. mathstudent55

Also, g = 32 ft/sec^2

16. bobobox

yup

17. mathstudent55

\(h(t) = -16t^2 + 42t \) This is an inverted parabola. We can take the first derivative to find where the height is maximum: \(h'(t) = -32t + 42\) Now we set the derivative equal to zero and solve for t: \(-32t + 42 = 0\) \(-32t = -42\) \(t = 1.3125~sec\) At 1.3125 sec the ball reaches maximum height.

18. bobobox

So a since you round it

19. bobobox

Thank you for being awesome and walking me through it, do you think you could help me with the next?

20. mathstudent55

Now we plug in that time in the position (height) equation to find at what height the ball is at 1.3125 sec: \(h(t) = -16t^2 + 42t\) \(h(1.3125) = -16(1.3125)^2 + 42(1.3125) = 27.56\)

21. bobobox

oh wow

22. mathstudent55

Notice we first found the time was 1.3 sec. The question is what is the height, not the time, so we need to take the time in the position equation and find the position at that time. The answer is 27.6 ft, not 1.3 sec.

23. bobobox

Thank you

24. bobobox

COuld you help with another? It's simelar!

25. mathstudent55

It's so similar it's exactly the same. The answer is the same.

26. bobobox

it's 27.6? It's not the same question though

27. mathstudent55

28. mathstudent55

No. Wait.

29. mathstudent55

I finally read the whole question correctly.

30. bobobox

okay thanks! Can I pester you to help me with the last four questions? I really appreciate How hard you work to help others.

31. mathstudent55

Look at this graph: |dw:1435640162113:dw|

32. mathstudent55

|dw:1435640215914:dw|

33. mathstudent55

The ball is kicked at (0, 0), where time is zero, and the height of the ball is zero.

34. bobobox

okay

35. mathstudent55

As time goes by, the ball goes higher. It reaches a maximum height at 1.3 sec. |dw:1435640295295:dw|

36. bobobox

so is it 1.3?

37. mathstudent55

Notice that the shape of the curve is an inverted parabola. That means it is symmetric left to right. The amount of time it took for the ball to reach maximum height is the same amount of time it will take for the ball to hit the ground on the way down.

38. bobobox

okay so in total the ball was in the air for 2.6 seconds

39. mathstudent55

|dw:1435640433043:dw|

40. mathstudent55

Correct.

41. bobobox

thanks can I pester you to help with the last four?

42. mathstudent55

We can get this answer this way bec we already solved the part about maximum height. It's simply double the time it took to reach maximum height.

43. bobobox

Cool ill post the next

44. bobobox

A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0 feet. How long will it take the golf ball to hit the ground? (1 point) 13 secs 21 secs 42 secs 6 secs

45. mathstudent55

If we had not solved that part, then we could solve this problem on its own. The equation of the position is \(h(t) = -16t^2 + 42t\) The height is 0 at the beginning (when the ball is kicked) and at the end (when the ball falls on the ground) \(0 = -16t^2 + 42t\) \(t(-16t + 42) = 0\) \(t = 0\) or \(-16t + 42 = 0\) \(t = 0\) or \(-16t = -42\) \(t = 0\) or \(t = 2.6\) The ball is at zero height at 0 sec and at 2.6 sec. That means it was in the air for 2.6 sec.

46. bobobox

yup :)

47. mathstudent55

Use the general equation of height we have: \(h(t) = -16t^2 + v_0t + h_0\)

48. mathstudent55

\(h_0 = 0\) \(v_0 = 208\) Plug these values in and solve the equation for t.

49. bobobox

okay one sec

50. bobobox

I tried I got a fraction so I am not even sure I did it right

51. mathstudent55

\(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 208t = 0\) \(-16t^2 + 208t = 0\) \(-2t^2 + 26t = 0\) \(t(-2t + 26) = 0\) \(t = 0\) or \(-2t + 26 = 0\) \(~~~~~~~~~~~~~~~~~~-2t = -26\) \(~~~~~~~~~~~~~~~~~~~~~~~~~t = 13\)

52. bobobox

oohhhhh

53. bobobox

54. mathstudent55

The ball is at height zero at 0 seconds, when it is hit, and at 13 seconds when if falls on the ground.

55. bobobox

cool thanks! only three more left!

56. bobobox

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle? (1 point) 128 feet 224 feet 272 feet 484 feet

57. bobobox

58. bobobox

@mathstudent55 I will fan write a testimony and I will give you a medal please

59. bobobox

60. mathstudent55

We use the same height equation again, and we plug in the info of this problem. Here we have an initial position.

61. bobobox

is the answer A? because I tried figuring it out

62. bobobox

crap no I meant d

63. mathstudent55

\(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 144t + 160\) \(h'(t) = -32t + 144\) \(-32t + 144 = 0\) \(-32t = -144\) \(t = 4.5~sec\) Maximum height is achieved at 4.5 sec. Now we plug in 4.5 sec in the height equation to find the actual height.

64. mathstudent55

\(h(t) = -16(4.5)^2 + 144(4.5) + 160\)

65. mathstudent55

What DDDDDDDdo you get?

66. bobobox

let me see

67. bobobox

484

68. mathstudent55

You DDDDDDid get the correct answer.

69. bobobox

yay! Okay just two more questions okay?

70. mathstudent55

ok

71. bobobox

I think it's the one that I choose

72. bobobox

@mathstudent55

73. mathstudent55

correct

74. bobobox

Yay the last one I am not to sure about

75. mathstudent55

|dw:1435642461182:dw|

76. bobobox

Help

77. bobobox

@mathstudent55

78. mathstudent55

Can you solve it for c^2? Since c^2 is being multiplied by m, divide both sides by m. |dw:1435642705027:dw|

79. mathstudent55

|dw:1435642763475:dw|

80. mathstudent55

ok so far?

81. bobobox

omg yes

82. mathstudent55

Now you want c but you have c squared. The opposite of squaring is taking the square root, so you take the square root of both sides: |dw:1435642840101:dw|

83. bobobox

I just want to say thank you, I don't think I could have said it enough, really

84. mathstudent55

You're welcome. I hope you pass & graduate.

85. bobobox

I got a 100%!!!!!!!!!!!!!!!!!!!!!!!!! @mathstudent55

86. mathstudent55

Great!!!