## Kathatesmath94 Group Title can someone check my work?? one year ago one year ago

1. Kathatesmath94 Group Title

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

2. Kathatesmath94 Group Title

|dw:1354245450082:dw|

3. Lime Group Title

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 Group Title

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

5. Kathatesmath94 Group Title

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

6. Kathatesmath94 Group Title

whoah where did that come from

7. scarydoor Group Title

|dw:1354246002329:dw|

8. Kathatesmath94 Group Title

ohh i see i took the sqaure root of 8 wrong

9. Kathatesmath94 Group Title

can i ask ya'll another question??

10. scarydoor Group Title

yep.

11. Kathatesmath94 Group Title

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 Group Title

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

13. Kathatesmath94 Group Title

sqrt of 128= 8sqrt 2?

14. scarydoor Group Title

That's right.

15. Kathatesmath94 Group Title

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

16. scarydoor Group Title

yep.

17. Kathatesmath94 Group Title

what do I do from there?

18. Kathatesmath94 Group Title

19. scarydoor Group Title

perimeter of a rectangle.

20. Kathatesmath94 Group Title

how do i do that?

21. scarydoor Group Title

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

22. Kathatesmath94 Group Title

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

23. Kathatesmath94 Group Title

are you still there?

24. Kathatesmath94 Group Title

am I still doing this right? hello?

25. scarydoor Group Title

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

26. Kathatesmath94 Group Title

after you multiply?

27. scarydoor Group Title

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

28. Kathatesmath94 Group Title

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

29. Kathatesmath94 Group Title

am i right so far?

30. Kathatesmath94 Group Title

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

31. jim_thompson5910 Group Title

|dw:1354248044017:dw|

32. jim_thompson5910 Group Title

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33. jim_thompson5910 Group Title

|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 Group Title

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

35. Kathatesmath94 Group Title

oh so I was suppose to add them not multiply

36. jim_thompson5910 Group Title

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 Group Title

oh ok that what i was thinking of

38. jim_thompson5910 Group Title

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 Group Title

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

40. jim_thompson5910 Group Title

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 Group Title

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

42. Kathatesmath94 Group Title

the funny equal sign is a approximate sign right?

43. jim_thompson5910 Group Title

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

44. jim_thompson5910 Group Title

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

45. Kathatesmath94 Group Title

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

46. jim_thompson5910 Group Title

yeah or a large table/counter of some sort

47. Kathatesmath94 Group Title

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 Group Title

sure go for it

49. Kathatesmath94 Group Title

awesome that way I know I really understand it

50. Kathatesmath94 Group Title

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 Group Title

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

52. jim_thompson5910 Group Title

and not the perimeter

53. Kathatesmath94 Group Title

isn't area multiplying the sides together?

54. jim_thompson5910 Group Title

exactly

55. jim_thompson5910 Group Title

Area = Length times Width

56. Kathatesmath94 Group Title

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

57. Kathatesmath94 Group Title

9sqrt 2?

58. jim_thompson5910 Group Title

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

59. Kathatesmath94 Group Title

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 Group Title

you can't do that

61. jim_thompson5910 Group Title

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

62. Kathatesmath94 Group Title

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

63. jim_thompson5910 Group Title

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

64. jim_thompson5910 Group Title

so sqrt(6) can't be simplified

65. Kathatesmath94 Group Title

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

66. jim_thompson5910 Group Title

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

67. Ammarah Group Title

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

68. Ammarah Group Title

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

69. jim_thompson5910 Group Title

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

70. Ammarah Group Title

yes dont forget!! Simplify!

71. Kathatesmath94 Group Title

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

72. Ammarah Group Title

yes.

73. Kathatesmath94 Group Title

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

74. Ammarah Group Title

?

75. jim_thompson5910 Group Title

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

76. Kathatesmath94 Group Title

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

77. jim_thompson5910 Group Title

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

78. Kathatesmath94 Group Title

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 Group Title

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

80. jim_thompson5910 Group Title

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

81. Kathatesmath94 Group Title

and 4sqqrt of 2 is approximately 5.65685424949238

82. jim_thompson5910 Group Title

correct

83. Kathatesmath94 Group Title

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

84. jim_thompson5910 Group Title

yeah that's a reasonably sized whiteboard

85. Kathatesmath94 Group Title

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 Group Title

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

87. Kathatesmath94 Group Title

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 Group Title

youre WELCOME

89. Ammarah Group Title

no thanks