## anonymous one year ago Need help and a memory refresher on 8 problems.. Please do not give me answers. I want to remember how to solve them. This is my midterm review.

1. anonymous

11. Perform the indicated operations and write in standard form. (-10 + square root -9) ^2

2. anonymous

16. Solve the problem. A rain gutter is made from sheets of aluminum that are 24 inches wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 58 square inches. There are two solutions to this problem. Round to the nearest tenth of an inch.

3. anonymous

21. Solve the linear inequality. Other than not solution sign, use interval notation to express the solution set and graph the solution set on a number line. x-1 over 16 >/ x-4 over 20 + 1 over 80

4. anonymous

@jim_thompson5910 @welshfella @Nnesha

5. anonymous

Please help me finish. I still have a chapter test and midterm to do for another class. Im majorly behind since the system keeps glitching.

6. anonymous

@IrishBoy123

7. jim_thompson5910

Hopefully in your lesson somewhere you learned that $\Large i = \sqrt{-1}$ ? or no?

8. anonymous

Yes I did last term.

9. jim_thompson5910

ok great

10. jim_thompson5910

what we can do is rewrite -9 as -1*9 and simplify like this $\Large \sqrt{-9} = \sqrt{-1*9}$ $\Large \sqrt{-9} = \sqrt{-1}*\sqrt{9}$ $\Large \sqrt{-9} = i*3$ $\Large \sqrt{-9} = 3i$ make sense?

11. anonymous

Yes!

12. jim_thompson5910

$\Large \left(-10+\sqrt{-9}\right)^2$ turns into $\Large \left(-10+3i\right)^2$

13. anonymous

Yep I have that.

14. jim_thompson5910

Next you'll use the box method or FOIL or the distributive property to expand out $\Large \left(-10+3i\right)^2$

15. jim_thompson5910

personally I like the box method but you can choose which one you want to do

16. anonymous

So.. (-10+3i)(-10+3i)

17. jim_thompson5910

yes

18. jim_thompson5910

keep going

19. anonymous

I got.. 100-51i

20. jim_thompson5910

incorrect

21. anonymous

Yeah...

22. anonymous

Not sure where im going wrong.. (-10+3i)(-10+3i) 100-30i-30i+9i 100-60i+9i 100-51i

23. jim_thompson5910

3i times 3i = 9i^2 not just 9i

24. jim_thompson5910

since i = sqrt(-1), squaring both sides gives i^2 = -1

25. anonymous

Omg -.-

26. anonymous

91-60i

27. jim_thompson5910

perfect

28. anonymous

Okay. For number 16. Solve the problem. A rain gutter is made from sheets of aluminum that are 24 inches wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 58 square inches. There are two solutions to this problem. Round to the nearest tenth of an inch. I got to -20+- square root 296 all over -4

29. anonymous

the square root 296 can turn into 2 square root 74

30. anonymous

-20 +- 2 square root 74 all over -4 Then I can -20 and -4 --> 5

31. anonymous

I'm kinda stuck there.

32. jim_thompson5910

I'm not sure how you got that. What quadratic equation did you get to?

33. anonymous

Um.. -2x^2 +20x -13=0 -20+- square root (20)^2 -4(-2)(-13) all over 2(-2) Some girl earlier said to- 2x+y=20 y=20-2x x*y=13 x*(20-2x)=13

34. anonymous

Im really not sure. I followed what she said. And I responded saying I was stuck and she didnt reply.

35. jim_thompson5910

ok let's start with a 24 inch line |dw:1444423527504:dw|

36. jim_thompson5910

we will bend this line into a partial rectangle like this |dw:1444423553539:dw| I can't do animations on here but hopefully you can picture what is going on

37. anonymous

I can.

38. jim_thompson5910

after the bending is done, we have this left over |dw:1444423623561:dw|

39. jim_thompson5910

if we make x = depth, then we have this |dw:1444423774023:dw|

40. jim_thompson5910

what is the length of the bottom side?

41. anonymous

What do you mean? As in y?

42. jim_thompson5910

hint: |dw:1444423922673:dw|

43. jim_thompson5910

replace the question marks with some expression in terms of x

44. anonymous

I understand what your saying but I keep thinking Y. Unless its 58?

45. jim_thompson5910

|dw:1444424117414:dw|

46. anonymous

Or is it still 24 even though the edges have been bent.

47. anonymous

Oh..

48. jim_thompson5910

there are 2 xs being subtracted off, so that's how I got 24-2x

49. jim_thompson5910

|dw:1444424187334:dw|

50. anonymous

Yes I understand now.

51. jim_thompson5910

Area = Length * Width 58 = (24 - 2x) * x 58 = 24x - 2x^2 -2x^2 + 24x - 58 = 0 solve for x using the quadratic formula

52. anonymous

Okay. Im at.. -24 +- 4 square root 65 over -4

53. jim_thompson5910

you made an error somewhere

54. anonymous

-24 +- square root 576 +464 over -4

55. jim_thompson5910

it should be -464 not +464

56. jim_thompson5910

b^2 - 4ac (24)^2 - 4*(-2)*(-58) 576 - 464

57. anonymous

Oops I put a + in front of the 4...

58. anonymous

Now I have.. 6+- -1 square root 7

59. jim_thompson5910

good, $\Large x = 6 \pm \sqrt{7}$

60. anonymous

Answer-> 3.4 and 8.6

61. jim_thompson5910

both are correct

62. anonymous

Okay thank you.. 6 left.. 21. Solve the linear inequality. Other than not solution sign, use interval notation to express the solution set and graph the solution set on a number line. x-1 over 16 >/ x-4 over 20 + 1 over 80

63. jim_thompson5910

This? $\Large \frac{x-1}{16} \ge \frac{x-4}{20} + \frac{1}{80}$

64. anonymous

I know the answer is -10.. Its the only option.. My first attempt I canceled out the fractions by using 4... Yes thats it.

65. jim_thompson5910

I would use 80 instead

66. anonymous

True..

67. jim_thompson5910

Multiply every term by the LCD 80 to clear out the fractions $\Large \frac{x-1}{16} \ge \frac{x-4}{20} + \frac{1}{80}$ $\Large 80*\frac{x-1}{16} \ge 80*\frac{x-4}{20} + 80*\frac{1}{80}$ $\Large 5(x-1) \ge 4(x-4) + 1$

68. anonymous

Yep.. Should of figured that out :/ ... x</ -10

69. jim_thompson5910

correct

70. anonymous

Its [-10, infinity) correct?

71. jim_thompson5910

yes, sorry I misread your first inequality it should be $$\Large x \ge -10$$ which turns into [-10, infinity)

72. anonymous

Okay. Sorry typed the wrong one.

73. anonymous

Next problem.. 24. Determine whether the equation defines y as a function of x. xy + 5y = 1

74. anonymous

I don't remember doing any problems like this. And I couldn't find any examples from my notes.

75. jim_thompson5910

We solve for y to get xy + 5y = 1 y(x + 5) = 1 y = 1/(x + 5) which is definitely a function. You can graph to verify that it passes the vertical line test. Any given input (x) produces EXACTLY ONE output (y)

76. anonymous

Okay!

77. anonymous

26. Evaluate the function at the given value of the independent variable and simplify. f(x) = 3x2 - 4x + 4; f(x - 1)

78. anonymous

I'm not sure on this one either.. I was thinking plug (x-1) in for the x's but im not sure. If you want me to post in a new question I can.

79. jim_thompson5910

You're right. Replace each x with x-1 and simplify $\Large f(x) = 3x^2 - 4x + 4$ $\Large f(\color{red}{x}) = 3(\color{red}{x})^2 - 4(\color{red}{x}) + 4$ $\Large f(\color{red}{x-1}) = 3(\color{red}{x-1})^2 - 4(\color{red}{x-1}) + 4$ now simplify the right hand side

80. jim_thompson5910

 If you want me to post in a new question I can.  yes please. It's getting laggy with all of these questions. It's a good idea to have one question per post to avoid clutter and lag

81. anonymous

Okay Ill do it now..