*(FAN AND MEDAL) HELP!!!

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- anonymous

*(FAN AND MEDAL) HELP!!!

- chestercat

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- anonymous

- anonymous

I NEED THE LAST 3 PLEASE I REALLY NEED HELP GUYS

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- anonymous

i appreciate please help me!!!

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## More answers

- anonymous

- anonymous

@Haseeb96 ??? please :(

- anonymous

- anonymous

- anonymous

- anonymous

- anonymous

help :(

- undeadknight26

All of them need answered?

- Michele_Laino

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

- Michele_Laino

oops..these*

- anonymous

(10)(3)

- anonymous

?

- anonymous

30?

- Michele_Laino

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

- anonymous

ok, wait

- anonymous

idk ;/

- Michele_Laino

you have to replace x with 3 and x with 10

- anonymous

oh

- anonymous

f (3) = 2 x 3^10

- anonymous

- anonymous

are you there @Michele_Laino ?

- anonymous

- 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

wait

- Michele_Laino

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

- anonymous

i use wolfram i copy and paste: 0?

- Michele_Laino

yes!

- anonymous

so whats nexts?

- Michele_Laino

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

- anonymous

where its the formula, or its the same formula?

- 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

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

- anonymous

im right?

- Michele_Laino

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

- Michele_Laino

since 4^10=1,048,576

- anonymous

what i need to do now?

- Michele_Laino

next ste:
we have to write the function of Carter

- anonymous

how?

- Michele_Laino

we have to read the text of your problem

- Michele_Laino

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

- anonymous

yes i think so

- 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

carter: 10(2)^x?

- anonymous

ok

- 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

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

- anonymous

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

- anonymous

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

- 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

so...

- Michele_Laino

so, we have completed the question 4)

- anonymous

yea, but whats nexts? i did something wrong?

- Michele_Laino

I think you have done right!

- 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

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

i need to solve that? right?

- Michele_Laino

I think that we have to draw both those graphs

- anonymous

im not good doing graph ;(

- Michele_Laino

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

- anonymous

i need to graph now the new function?

- Michele_Laino

yes!

- Michele_Laino

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

- anonymous

you already did the graph

- Michele_Laino

yes! as you can see it is simple

- anonymous

yea i see, so we done with number 4?

- Michele_Laino

we have done with number 5)

- anonymous

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

- 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

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

- Michele_Laino

yes! right!

- anonymous

so we can move to # 6?

- Michele_Laino

yes!

- anonymous

can you continue? ^_^

- Michele_Laino

ok!

- 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

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

- Michele_Laino

their values

- anonymous

idk how to compare the quantities

- 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

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

- Michele_Laino

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

- anonymous

oh

- Michele_Laino

from those numbers above, can you see the requested student

- Michele_Laino

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

- anonymous

g(3)=3Ã—43=3Ã—64=192
g(10)=3Ã—410=3145728

- anonymous

?

- anonymous

g(3), g(10)

- Michele_Laino

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

- anonymous

yes

- Michele_Laino

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

- anonymous

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

- anonymous

yes

- Michele_Laino

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

- anonymous

g(10)? right?

- Michele_Laino

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

- anonymous

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

- anonymous

right?

- Michele_Laino

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

- Michele_Laino

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

- anonymous

f(x)?

- Michele_Laino

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

- anonymous

no wait let me see again

- anonymous

so, its, g(x)? :/

- Michele_Laino

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

- Michele_Laino

we have answered to question #6

- anonymous

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

- anonymous

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

- anonymous

i write good?

- 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

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

- Michele_Laino

yes! sure! :)

- 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

so the answer for #7 its the same answer of #6?

- Michele_Laino

yes!

- anonymous

really? omg

- Michele_Laino

yes! really!

- anonymous

but i want to write in others words

- 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

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

- Michele_Laino

:)

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