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first one

ok... well our function doesn't have denominators so there are no restrictions.

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the domain is at the x-axis
the range is at the y-axis

we need to graph this function t-t

Okay..

note: you can also re-write your function as:
|dw:1436850224394:dw|

we need x = 1,2,3... and so on..
for
\[\large f(x)=49(\frac{1}{7})^x \]

we can use desmos. easier XD

Ohh okay you have to graph to see it..

I wasn't graphing it, so that is why it was really difficult to see

sure

Right, so A and B are correct..

\[\large f(2)=49(\frac{1}{7})^2\]
\[\large f(2)=49(\frac{1}{49})\]
\[\large f(2)=1\]

when my x is increasing I have
f(0) = 49
f(1) = 7
f(2) = 1

So Y is decreasing..

x y
0 49
1 7
2 1

and it claims that when x is increasing the previous y is 1/7 th of the previous y value.

?!

Right...? That is not true though. so we can eliminate that one

let's try f(4) maybe that question is talking about that?
\[\large f(4)=49(\frac{1}{7})^4\]

\[\large f(4)=\frac{49}{2401}\]

\[f(4) = \frac{1}{49}\]

first two are fine
last one is when x = 0 , so it's not 7

Okay, and that is all?

ugh trying to think to those 3 ...

second to the last one is confusing... x.x

We'll just say no to that one...SO now we are left with C and D.

if it's all reals then that would have the domain and range at >0

I see it well on the range though. when y > 0

on the graph

but the domain is smoosh... like it's just on the line

I think after x = 3 the previous is 1/7 due to
1/7
(1/7)(1/7)
(1/7)(1/7)(1/7)

Okay thank you

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