## Kathatesmath94 2 years ago can someone check my work??

1. Kathatesmath94

sqrt3(sqrt6+sqrt3)+(sqrt8-5)

2. Kathatesmath94

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3. Lime

I believe it is much easier to use a calculator for Square root, but let's break it down. $3 + 3 \sqrt2 +(\sqrt 8 - 5)$

4. Kathatesmath94

well from there i got 3+3sqrt2= 6sqrt2 + sqrt8-5

5. Kathatesmath94

and the sqrt of 8 is 4 and sqrt of 4 is 2 so it will be 6sqrt2+sqrt2-5

6. Kathatesmath94

whoah where did that come from

7. scarydoor

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8. Kathatesmath94

ohh i see i took the sqaure root of 8 wrong

9. Kathatesmath94

can i ask ya'll another question??

10. scarydoor

yep.

11. Kathatesmath94

Part 1: Find the perimeter of a rectangular object which has a length of sqrt 128 feet and a width of sqrt 200 feet. Part 2: Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form. Part 3: What type of object in your home or school might this be? i am not going to lie I don't even know where to start....

12. scarydoor

|dw:1354246494891:dw| Just add up the sides, for part one. Simplify the square roots a little.

13. Kathatesmath94

sqrt of 128= 8sqrt 2?

14. scarydoor

That's right.

15. Kathatesmath94

and the sqrt of 200 = 10sqrt 2???

16. scarydoor

yep.

17. Kathatesmath94

what do I do from there?

18. Kathatesmath94

19. scarydoor

perimeter of a rectangle.

20. Kathatesmath94

how do i do that?

21. scarydoor

You want to know how long the edge is... |dw:1354247036064:dw|

22. Kathatesmath94

so multiply 8sqrt2*8sqrt2 and 10sqrt2*10sqrt2? and then add them together?

23. Kathatesmath94

are you still there?

24. Kathatesmath94

am I still doing this right? hello?

25. scarydoor

Yeah to find the perimeter of a rectangle, you add the lengths of all sides together.

26. Kathatesmath94

after you multiply?

27. scarydoor

Just take the lengths of the sides and add them together.

28. Kathatesmath94

so 8sqrt2*8sqrt2=64sqrt2 10sqrt2*10sqrt2= 100sqrt 2 64sqrt2+100sqrt2= 164sqrt2?

29. Kathatesmath94

am i right so far?

30. Kathatesmath94

was I just suppose to add them all together or multiply? like i did above?

31. jim_thompson5910

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32. jim_thompson5910

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33. jim_thompson5910

|dw:1354248126981:dw| Perimeter = Sum of all sides P = s1 + s2 + s3 + s4 P = 8*sqrt(2) + 10*sqrt(2) + 8*sqrt(2) + 10*sqrt(2) P = 36*sqrt(2)

34. jim_thompson5910

So the perimeter is $\Large 36\sqrt{2}$

35. Kathatesmath94

oh so I was suppose to add them not multiply

36. jim_thompson5910

yes, or you can add the length and width and multiply by 2 this is because P = L + W + L + W P = 2L + 2W P = 2*(L + W)

37. Kathatesmath94

oh ok that what i was thinking of

38. jim_thompson5910

Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form you essentially simplify both sqrt(128) and sqrt(200) to get 8*sqrt(2) and 10*sqrt(2) you can either add up the four sides or you can add up the length and width and multiply by 2 either way, you get the answer of $\Large 36\sqrt{2}$

39. Kathatesmath94

ok thats what i wrote so far what is the next step?

40. jim_thompson5910

What type of object in your home or school might this be? $\Large 36\sqrt{2} \approx 36(1.41421356)$ $\Large 36\sqrt{2} \approx 50.91168816$ so the perimeter is roughly 50.91168816 feet

41. jim_thompson5910

$\Large 8\sqrt{2} \approx 8(1.41421356)$ $\Large 8\sqrt{2} \approx 11.31370848$ the width is roughly 11.31370848 feet

42. Kathatesmath94

the funny equal sign is a approximate sign right?

43. jim_thompson5910

$\Large 10\sqrt{2} \approx 10(1.41421356)$ $\Large 10\sqrt{2} \approx 14.1421356$ the length is roughly 14.1421356 feet

44. jim_thompson5910

so you just have to name an object that's roughly 14.1421356 ft by 11.31370848 ft

45. Kathatesmath94

so it could be a pool right? for the object?

46. jim_thompson5910

yeah or a large table/counter of some sort

47. Kathatesmath94

ohhh wow that wasnt too bad there is another one very similiar and if I post all my work will you tell me if I am right or not?

48. jim_thompson5910

sure go for it

49. Kathatesmath94

awesome that way I know I really understand it

50. Kathatesmath94

Part 1: Find the area of a rectangular object which has a length of 4sqrt2 inches and a width of 6sqrt6 inches. Part 2: Explain, in complete sentences, how you arrived at the simplified answer and give the final solution in simplified radical form. Part 3: What type of object in your home or school might this be? |dw:1354248883689:dw|

51. jim_thompson5910

ok it's a bit different because they want the area this time

52. jim_thompson5910

and not the perimeter

53. Kathatesmath94

isn't area multiplying the sides together?

54. jim_thompson5910

exactly

55. jim_thompson5910

Area = Length times Width

56. Kathatesmath94

ok so first i have to simplify the sqrts 6sqrt 6 = 6+3sqrt2?

57. Kathatesmath94

9sqrt 2?

58. jim_thompson5910

how are you getting 6sqrt 6 = 6+3sqrt2

59. Kathatesmath94

uhhh i might of factored wrong because i took the factors 2 and 3 and pulled the 3 out to add to the 6 and left the 2 inside the radical

60. jim_thompson5910

you can't do that

61. jim_thompson5910

6*sqrt(6) is as simple as it gets

62. Kathatesmath94

i was trying to make it so both of my radicals have the same square root thing of sqrt2

63. jim_thompson5910

notice that 6 has no factors that are perfect squares (ignore 1)

64. jim_thompson5910

so sqrt(6) can't be simplified

65. Kathatesmath94

oh..... i was wondering how that worked cause nothing times itself equals 6

66. jim_thompson5910

Multiply the coefficients: 6*4 = 24 Multiply the radicands: 6*2 = 12

67. Ammarah

what is ur question, i mean what r u trying to find exactly?

68. Ammarah

so u multiply and then plugg it back in to the equation.

69. jim_thompson5910

so 6*sqrt(6) times 4*sqrt(2) = 24*sqrt(12) simplify 24*sqrt(12)

70. Ammarah

yes dont forget!! Simplify!

71. Kathatesmath94

so I can go straight to part 2 right? and find the area?

72. Ammarah

yes.

73. Kathatesmath94

sorry idk why that post just showed up i have a major delay

74. Ammarah

?

75. jim_thompson5910

yes since you can't simplify the length/width just multiply them out

76. Kathatesmath94

now is 24sqrt12 the radical area that needs to be converted into the approximate length?

77. jim_thompson5910

simplify that 24*sqrt(12) 24*sqrt(4*3) 24*sqrt(4)*sqrt(3) 24*2*sqrt(3) 48*sqrt(3)

78. Kathatesmath94

and do you get the approximate length by putting the number in the calculator? i meant to ask you how you did that last time

79. jim_thompson5910

so the area is 48*sqrt(3) square inches

80. jim_thompson5910

you just type 6*sqrt(6) and you get 14.6969 roughly

81. Kathatesmath94

and 4sqqrt of 2 is approximately 5.65685424949238

82. jim_thompson5910

correct

83. Kathatesmath94

so the object could be like a table? o ra whiite board maybe?

84. jim_thompson5910

yeah that's a reasonably sized whiteboard

85. Kathatesmath94

you have helped me so much and thank you for helping me with this someone was helping me but they left and I didnt know where to go from there you are definitley a life SAVER!

86. jim_thompson5910

i'm glad i did and hopefully it's all making sense now

87. Kathatesmath94

yeah they were a little different but there steps were very similar and I took notes while we talked so hopefully that will help me if I have a problem like this later on in the course

88. danielcvalencia

youre WELCOME

89. Ammarah

no thanks