anonymous one year ago *(FAN AND MEDAL) HELP!!!

1. anonymous

@pinkbubbles @ganeshie8

2. anonymous

I NEED THE LAST 3 PLEASE I REALLY NEED HELP GUYS

3. anonymous

4. anonymous

@Vocaloid

5. anonymous

6. anonymous

@campbell_st

7. anonymous

@dan815

8. anonymous

@Michele_Laino

9. anonymous

10. anonymous

help :(

11. anonymous

12. Michele_Laino

question 4) hwre we have to evaluate this quantities: $\Large f\left( {10} \right) = ...?\;f\left( 3 \right) = ...?$

13. Michele_Laino

oops..these*

14. anonymous

(10)(3)

15. anonymous

?

16. anonymous

30?

17. Michele_Laino

you have to use your function: $\Large f\left( x \right) = 2 \times {3^x}$

18. anonymous

ok, wait

19. anonymous

idk ;/

20. Michele_Laino

you have to replace x with 3 and x with 10

21. anonymous

oh

22. anonymous

f (3) = 2 x 3^10

23. anonymous

@Michele_Laino

24. anonymous

are you there @Michele_Laino ?

25. anonymous

@Michele_Laino

26. Michele_Laino

hint: $\Large \begin{gathered} f\left( 3 \right) = 2 \times {3^3} = 2 \times 27 = ...? \hfill \\ f\left( {10} \right) = 2 \times {3^{10}} = ...? \hfill \\ \end{gathered}$

27. anonymous

wait

28. Michele_Laino

f(10) is a very big number, so please use a calculator

29. anonymous

i use wolfram i copy and paste: 0?

30. Michele_Laino

yes!

31. anonymous

so whats nexts?

32. Michele_Laino

now we have to compute the same quantities, using this function: $\Large g\left( x \right) = 3 \times {4^x}$

33. anonymous

where its the formula, or its the same formula?

34. Michele_Laino

namely we have to evaluate these quantities: $\Large \begin{gathered} g\left( 2 \right) = 3 \times {4^2} = 3 \times 16 = ...? \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = ...? \hfill \\ \end{gathered}$

35. anonymous

g(2) = 3x4^2 = 3x16 = 48? g(10) = 3x4^10 = 0?

36. anonymous

im right?

37. Michele_Laino

48 is right! the second is a very big number g(10)=3*1048576=...?

38. Michele_Laino

since 4^10=1,048,576

39. anonymous

what i need to do now?

40. Michele_Laino

next ste: we have to write the function of Carter

41. anonymous

how?

42. Michele_Laino

43. Michele_Laino

I think that the function used by Carter is: $\Large h\left( x \right) = 10 \times {2^x}$ am I right?

44. anonymous

yes i think so

45. Michele_Laino

ok! now we have to compute these quantities: $\Large \begin{gathered} h\left( 2 \right) = 10 \times {2^2} = 10 \times 4 = ... \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = ... \hfill \\ \end{gathered}$

46. anonymous

carter: 10(2)^x?

47. anonymous

ok

48. Michele_Laino

sorry I have made an error, we have to compute g(3) and h(3) not g(2) and h(2)

49. Michele_Laino

$\Large \begin{gathered} g\left( 3 \right) = 3 \times {4^3} = 3 \times 64 = ...? \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = ...? \hfill \\ \end{gathered}$ $\Large \begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = ... \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = ... \hfill \\ \end{gathered}$

50. anonymous

h(2)=10x2^2=10x4= 40? h(10)=10x2^10x1024= 10485760?

51. anonymous

im not sure about the last one, but im right?

52. Michele_Laino

we have: $\Large \begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = 80 \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = 10240 \hfill \\ \end{gathered}$

53. anonymous

so...

54. Michele_Laino

so, we have completed the question 4)

55. anonymous

yea, but whats nexts? i did something wrong?

56. Michele_Laino

I think you have done right!

57. Michele_Laino

now we have to compare the graph of the new function of Amber, with the graph of the old function of the same Amber

58. Michele_Laino

the new function is: $\Large {g_2}\left( x \right) = 3 \times {4^x} + 45$ whereas the old function is: $\Large {g_1}\left( x \right) = 3 \times {4^x}$

59. anonymous

i need to solve that? right?

60. Michele_Laino

I think that we have to draw both those graphs

61. anonymous

im not good doing graph ;(

62. Michele_Laino

it is simple, the graph of the old function is like below: |dw:1437499913652:dw|

63. anonymous

i need to graph now the new function?

64. Michele_Laino

yes!

65. Michele_Laino

the new graph is shifted up by 45 units with respect to the old one, like this: |dw:1437500095683:dw|

66. anonymous

67. Michele_Laino

yes! as you can see it is simple

68. anonymous

yea i see, so we done with number 4?

69. Michele_Laino

we have done with number 5)

70. anonymous

wait wait i dont see the answer for # 4 :(

71. Michele_Laino

the answer of #4 are given by the subsequent quantites: f(3), f(10) g(3), g(10) h(3), h(10)

72. anonymous

ok, so the asnwer for # 5 are the graphs? right

73. Michele_Laino

yes! right!

74. anonymous

so we can move to # 6?

75. Michele_Laino

yes!

76. anonymous

can you continue? ^_^

77. Michele_Laino

ok!

78. Michele_Laino

for question #6 you have to compare the quantities above, namely f(3), f(10) g(3), g(10) h(3), h(10)

79. anonymous

how? like that: f(3)(10) g(3)(10) h(3)(10)

80. Michele_Laino

their values

81. anonymous

idk how to compare the quantities

82. Michele_Laino

we have to compare these values: $\begin{gathered} f\left( 3 \right) = 2 \times {3^3} = 2 \times 27 = 54 \hfill \\ f\left( {10} \right) = 2 \times {3^{10}} = 118098 \hfill \\ \end{gathered}$ $\begin{gathered} g\left( 3 \right) = 3 \times {4^3} = 3 \times 64 = 192 \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = 3145728 \hfill \\ \end{gathered}$ $\begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = 80 \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = 10240 \hfill \\ \end{gathered}$

83. anonymous

yes ik, but my question its how because if im right, i see you already did, right?

84. Michele_Laino

we have to establish which student increased fastly the number of shared posts

85. anonymous

oh

86. Michele_Laino

from those numbers above, can you see the requested student

87. Michele_Laino

please, look at the posts shared by Amber, her function is g(x)

88. anonymous

g(3)=3×43=3×64=192 g(10)=3×410=3145728

89. anonymous

?

90. anonymous

g(3), g(10)

91. Michele_Laino

within the same time, namely 1 week her shared posts went from 192, to more than 3 millions, right?

92. anonymous

yes

93. Michele_Laino

whereas the other two students have not been so fast, right?

94. anonymous

f(3), f(10) h(3), h(10)?

95. anonymous

yes

96. Michele_Laino

so, who is the fastest student in sharing the posts?

97. anonymous

g(10)? right?

98. Michele_Laino

yes! better if we answer saying that the fastest student is Amber

99. anonymous

so the answer for # 6 its: the fastest student is Amber

100. anonymous

right?

101. Michele_Laino

yes! Please wait Look at these formulas: |dw:1437502126622:dw|

102. Michele_Laino

we have three powers. Now i ask you, which power has the highest base?

103. anonymous

f(x)?

104. Michele_Laino

are you sure? we have: 2 < 3 < 4

105. anonymous

no wait let me see again

106. anonymous

so, its, g(x)? :/

107. Michele_Laino

yes! that's right! That is why Amber was the fastest student in sharing her posts

108. Michele_Laino

we have answered to question #6

109. anonymous

ok, for make sure the answer for # 6 its:

110. anonymous

6) g(x) = 3.4^x the fastest student is Amber?

111. anonymous

i write good?

112. Michele_Laino

yes! you are correct here is my answer equivalent to yours: the fastest student is Amber we can recognize that fact, since in the Amber's function, appears a power with the highest base

113. anonymous

thank you so much!!!! can we now move to the last one?

114. Michele_Laino

yes! sure! :)

115. Michele_Laino

the question #7 is equivalent to question #6, since the answer is the same, namely I prefer to choose Amber, since she is very fast in sharing her posts

116. anonymous

117. Michele_Laino

yes!

118. anonymous

really? omg

119. Michele_Laino

yes! really!

120. anonymous

but i want to write in others words

121. Michele_Laino

my answer is: "I prefer share my posts in the same manner of Amber, since she has more shares, so she is very fast in sharing her posts"

122. anonymous

thank you so much, you help me a lot, you're awesome!!!!! ^_^

123. Michele_Laino

:)