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max2010maxim [7]
4 years ago
10

I'm not sure about 38 - 53 please help! :)

Mathematics
2 answers:
Maru [420]4 years ago
5 0
38. 6900 grams
39. 19600 grams
40.
27910 grams
41. 32840 grams
42. 
610 grams
43. 970 grams
44.
3712 grams
45. 8937 grams
46. 
37 grams
47. 69 grams
48.
1510 grams
49. 4700 grams
50. 150 grams
51. 15 grams
52. 15200 grams
53. 460

I'm sorry if these are wrong but I hope this could help.
SashulF [63]4 years ago
4 0
38. 6800, 39. 19600, 40. 27910, 41. 32840, 42. 610, 43. 970, 44. 3712, 45. 8937, 46. 37 47. 69, 48. 1510, 49. 4700, 50. 150, 51. 15, 52. 15200, 53. 460
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What are the zeros of the quadratic function f(x) = 6x ^ 2 - 24x + 1 ?
ZanzabumX [31]

Answer:

x = \frac{12+\sqrt{138} }{6} and x = \frac{12-\sqrt{138} }{6}

Step-by-step explanation:

Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: x=\frac{-b+\sqrt{b^2-4ac} }{2a} or x=\frac{-b-\sqrt{b^2-4ac} }{2a}.

Here, a = 6, b = -24, and c = 1. Plug these in:

x=\frac{-b+\sqrt{b^2-4ac} }{2a}

x=\frac{-(-24)+\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24+\sqrt{552} }{12}=\frac{24+2\sqrt{138} }{12} =\frac{12+\sqrt{138} }{6}

AND

x=\frac{-b-\sqrt{b^2-4ac} }{2a}

x=\frac{-(-24)-\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24-\sqrt{552} }{12}=\frac{24-2\sqrt{138} }{12} =\frac{12-\sqrt{138} }{6}

Thus, the zeroes are: x = \frac{12+\sqrt{138} }{6} and x = \frac{12-\sqrt{138} }{6}.

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3 years ago
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Answer:

x=4

Step-by-step explanation:

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Step-by-step explanation:

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