anonymous
  • anonymous
*(FAN AND MEDAL) HELP!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@pinkbubbles @ganeshie8
anonymous
  • anonymous
I NEED THE LAST 3 PLEASE I REALLY NEED HELP GUYS
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anonymous
  • anonymous
i appreciate please help me!!!

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anonymous
  • anonymous
@Vocaloid
anonymous
  • anonymous
@Haseeb96 ??? please :(
anonymous
  • anonymous
@campbell_st
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
@undeadknight26 @mathmath333
anonymous
  • anonymous
help :(
undeadknight26
  • undeadknight26
All of them need answered?
Michele_Laino
  • Michele_Laino
question 4) hwre we have to evaluate this quantities: \[\Large f\left( {10} \right) = ...?\;f\left( 3 \right) = ...?\]
Michele_Laino
  • Michele_Laino
oops..these*
anonymous
  • anonymous
(10)(3)
anonymous
  • anonymous
?
anonymous
  • anonymous
30?
Michele_Laino
  • Michele_Laino
you have to use your function: \[\Large f\left( x \right) = 2 \times {3^x}\]
anonymous
  • anonymous
ok, wait
anonymous
  • anonymous
idk ;/
Michele_Laino
  • Michele_Laino
you have to replace x with 3 and x with 10
anonymous
  • anonymous
oh
anonymous
  • anonymous
f (3) = 2 x 3^10
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
are you there @Michele_Laino ?
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
hint: \[\Large \begin{gathered} f\left( 3 \right) = 2 \times {3^3} = 2 \times 27 = ...? \hfill \\ f\left( {10} \right) = 2 \times {3^{10}} = ...? \hfill \\ \end{gathered} \]
anonymous
  • anonymous
wait
Michele_Laino
  • Michele_Laino
f(10) is a very big number, so please use a calculator
anonymous
  • anonymous
i use wolfram i copy and paste: 0?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
so whats nexts?
Michele_Laino
  • Michele_Laino
now we have to compute the same quantities, using this function: \[\Large g\left( x \right) = 3 \times {4^x}\]
anonymous
  • anonymous
where its the formula, or its the same formula?
Michele_Laino
  • Michele_Laino
namely we have to evaluate these quantities: \[\Large \begin{gathered} g\left( 2 \right) = 3 \times {4^2} = 3 \times 16 = ...? \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = ...? \hfill \\ \end{gathered} \]
anonymous
  • anonymous
g(2) = 3x4^2 = 3x16 = 48? g(10) = 3x4^10 = 0?
anonymous
  • anonymous
im right?
Michele_Laino
  • Michele_Laino
48 is right! the second is a very big number g(10)=3*1048576=...?
Michele_Laino
  • Michele_Laino
since 4^10=1,048,576
anonymous
  • anonymous
what i need to do now?
Michele_Laino
  • Michele_Laino
next ste: we have to write the function of Carter
anonymous
  • anonymous
how?
Michele_Laino
  • Michele_Laino
we have to read the text of your problem
Michele_Laino
  • Michele_Laino
I think that the function used by Carter is: \[\Large h\left( x \right) = 10 \times {2^x}\] am I right?
anonymous
  • anonymous
yes i think so
Michele_Laino
  • Michele_Laino
ok! now we have to compute these quantities: \[\Large \begin{gathered} h\left( 2 \right) = 10 \times {2^2} = 10 \times 4 = ... \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
carter: 10(2)^x?
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
sorry I have made an error, we have to compute g(3) and h(3) not g(2) and h(2)
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} g\left( 3 \right) = 3 \times {4^3} = 3 \times 64 = ...? \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = ...? \hfill \\ \end{gathered} \] \[\Large \begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = ... \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
h(2)=10x2^2=10x4= 40? h(10)=10x2^10x1024= 10485760?
anonymous
  • anonymous
im not sure about the last one, but im right?
Michele_Laino
  • Michele_Laino
we have: \[\Large \begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = 80 \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = 10240 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
so...
Michele_Laino
  • Michele_Laino
so, we have completed the question 4)
anonymous
  • anonymous
yea, but whats nexts? i did something wrong?
Michele_Laino
  • Michele_Laino
I think you have done right!
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
the new function is: \[\Large {g_2}\left( x \right) = 3 \times {4^x} + 45\] whereas the old function is: \[\Large {g_1}\left( x \right) = 3 \times {4^x}\]
anonymous
  • anonymous
i need to solve that? right?
Michele_Laino
  • Michele_Laino
I think that we have to draw both those graphs
anonymous
  • anonymous
im not good doing graph ;(
Michele_Laino
  • Michele_Laino
it is simple, the graph of the old function is like below: |dw:1437499913652:dw|
anonymous
  • anonymous
i need to graph now the new function?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
the new graph is shifted up by 45 units with respect to the old one, like this: |dw:1437500095683:dw|
anonymous
  • anonymous
you already did the graph
Michele_Laino
  • Michele_Laino
yes! as you can see it is simple
anonymous
  • anonymous
yea i see, so we done with number 4?
Michele_Laino
  • Michele_Laino
we have done with number 5)
anonymous
  • anonymous
wait wait i dont see the answer for # 4 :(
Michele_Laino
  • Michele_Laino
the answer of #4 are given by the subsequent quantites: f(3), f(10) g(3), g(10) h(3), h(10)
anonymous
  • anonymous
ok, so the asnwer for # 5 are the graphs? right
Michele_Laino
  • Michele_Laino
yes! right!
anonymous
  • anonymous
so we can move to # 6?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
can you continue? ^_^
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • 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)
anonymous
  • anonymous
how? like that: f(3)(10) g(3)(10) h(3)(10)
Michele_Laino
  • Michele_Laino
their values
anonymous
  • anonymous
idk how to compare the quantities
Michele_Laino
  • Michele_Laino
we have to compare these values: \[\begin{gathered} f\left( 3 \right) = 2 \times {3^3} = 2 \times 27 = 54 \hfill \\ f\left( {10} \right) = 2 \times {3^{10}} = 118098 \hfill \\ \end{gathered} \] \[\begin{gathered} g\left( 3 \right) = 3 \times {4^3} = 3 \times 64 = 192 \hfill \\ g\left( {10} \right) = 3 \times {4^{10}} = 3145728 \hfill \\ \end{gathered} \] \[\begin{gathered} h\left( 3 \right) = 10 \times {2^3} = 10 \times 8 = 80 \hfill \\ h\left( {10} \right) = 10 \times {2^{10}} = 10 \times 1024 = 10240 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
yes ik, but my question its how because if im right, i see you already did, right?
Michele_Laino
  • Michele_Laino
we have to establish which student increased fastly the number of shared posts
anonymous
  • anonymous
oh
Michele_Laino
  • Michele_Laino
from those numbers above, can you see the requested student
Michele_Laino
  • Michele_Laino
please, look at the posts shared by Amber, her function is g(x)
anonymous
  • anonymous
g(3)=3×43=3×64=192 g(10)=3×410=3145728
anonymous
  • anonymous
?
anonymous
  • anonymous
g(3), g(10)
Michele_Laino
  • Michele_Laino
within the same time, namely 1 week her shared posts went from 192, to more than 3 millions, right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
whereas the other two students have not been so fast, right?
anonymous
  • anonymous
f(3), f(10) h(3), h(10)?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
so, who is the fastest student in sharing the posts?
anonymous
  • anonymous
g(10)? right?
Michele_Laino
  • Michele_Laino
yes! better if we answer saying that the fastest student is Amber
anonymous
  • anonymous
so the answer for # 6 its: the fastest student is Amber
anonymous
  • anonymous
right?
Michele_Laino
  • Michele_Laino
yes! Please wait Look at these formulas: |dw:1437502126622:dw|
Michele_Laino
  • Michele_Laino
we have three powers. Now i ask you, which power has the highest base?
anonymous
  • anonymous
f(x)?
Michele_Laino
  • Michele_Laino
are you sure? we have: 2 < 3 < 4
anonymous
  • anonymous
no wait let me see again
anonymous
  • anonymous
so, its, g(x)? :/
Michele_Laino
  • Michele_Laino
yes! that's right! That is why Amber was the fastest student in sharing her posts
Michele_Laino
  • Michele_Laino
we have answered to question #6
anonymous
  • anonymous
ok, for make sure the answer for # 6 its:
anonymous
  • anonymous
6) g(x) = 3.4^x the fastest student is Amber?
anonymous
  • anonymous
i write good?
Michele_Laino
  • 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
anonymous
  • anonymous
thank you so much!!!! can we now move to the last one?
Michele_Laino
  • Michele_Laino
yes! sure! :)
Michele_Laino
  • 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
anonymous
  • anonymous
so the answer for #7 its the same answer of #6?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
really? omg
Michele_Laino
  • Michele_Laino
yes! really!
anonymous
  • anonymous
but i want to write in others words
Michele_Laino
  • 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"
anonymous
  • anonymous
thank you so much, you help me a lot, you're awesome!!!!! ^_^
Michele_Laino
  • Michele_Laino
:)

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