anonymous
  • anonymous
Need Help, medal + fan: Which are true of the function f(x) = 49 (1/7)x? Check all that apply. A:The domain is the set of all real numbers. B:The range is the set of all real numbers. C:The domain is x > 0. D:The range is y > 0. E:As x increases by 1, each y-value is one-seventh of the previous y-value. F:The initial value for the function is
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
\[\large f(x)=49(\frac{1}{7})^x \] is this the function for the question? or \[\large f(x)=49(\frac{1}{7})x \]
anonymous
  • anonymous
first one
UsukiDoll
  • UsukiDoll
ok... well our function doesn't have denominators so there are no restrictions.

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More answers

UsukiDoll
  • UsukiDoll
the domain is at the x-axis the range is at the y-axis
UsukiDoll
  • UsukiDoll
we need to graph this function t-t
anonymous
  • anonymous
Okay..
butterflydreamer
  • butterflydreamer
note: you can also re-write your function as: |dw:1436850224394:dw|
UsukiDoll
  • UsukiDoll
we need x = 1,2,3... and so on.. for \[\large f(x)=49(\frac{1}{7})^x \]
UsukiDoll
  • UsukiDoll
we can use desmos. easier XD
anonymous
  • anonymous
Ohh okay you have to graph to see it..
UsukiDoll
  • UsukiDoll
anonymous
  • anonymous
I wasn't graphing it, so that is why it was really difficult to see
UsukiDoll
  • UsukiDoll
yeah... sometimes questions are easier when you graph but since this function is too screwed up (I don't feel like doing it manually lol) we use desmos which graphs it for us
anonymous
  • anonymous
So from the graph, I already concluded what some of the answers could be, but still not sure if I am right, could you check to see? I will screenshot it
UsukiDoll
  • UsukiDoll
sure
anonymous
  • anonymous
1 Attachment
UsukiDoll
  • UsukiDoll
The first 2 is indeed all real numbers as I mentioned earlier we don't have a function that's like a fraction ex. \[f(x) =\frac{1}{x-2} \] which means we can't have x be 2
anonymous
  • anonymous
Right, so A and B are correct..
UsukiDoll
  • UsukiDoll
yes... hmm interesting on the last one.. I mean that would work if x = 1, but I could've sworn that initial value would mean that x = 0 .
UsukiDoll
  • UsukiDoll
like.. if x = 1 \[\large f(1)=49(\frac{1}{7})^1\] \[\large f(1)=49(\frac{1}{7})\] \[\large f(1)=(\frac{49}{7})=7\] If x = 0 \[\large f(0)=49(\frac{1}{7})^0\] (due to the zero exponent rule.. anything to the 0 results in 1 ) \[\large f(0)=49(1) = 49\]
anonymous
  • anonymous
Okay, we are getting somewhere haha. So we already know A and B are correct, anything else I am missing, or something I marked incorrectly?
UsukiDoll
  • UsukiDoll
\[\large f(2)=49(\frac{1}{7})^2\] \[\large f(2)=49(\frac{1}{49})\] \[\large f(2)=1\]
UsukiDoll
  • UsukiDoll
when my x is increasing I have f(0) = 49 f(1) = 7 f(2) = 1
anonymous
  • anonymous
So Y is decreasing..
UsukiDoll
  • UsukiDoll
I'm confused at that wording though... for the second to the last .. but yeah my y is going down.. fast.
UsukiDoll
  • UsukiDoll
x y 0 49 1 7 2 1
UsukiDoll
  • UsukiDoll
and it claims that when x is increasing the previous y is 1/7 th of the previous y value.
UsukiDoll
  • UsukiDoll
?!
anonymous
  • anonymous
Right...? That is not true though. so we can eliminate that one
UsukiDoll
  • UsukiDoll
I don't think so it's like 49 ./ 7 =7 7/7 = 1 Then again maybe if I calculate f(3) \[\large f(3)=49(\frac{1}{7})^3 \] \[\large f(3)=(\frac{49}{343}) \] \[\large f(3)=(\frac{1}{7}) \]
UsukiDoll
  • UsukiDoll
let's try f(4) maybe that question is talking about that? \[\large f(4)=49(\frac{1}{7})^4\]
anonymous
  • anonymous
Hey I'm sorry I am running really short on time, So lets just sum it up. What do you think are the answers?
UsukiDoll
  • UsukiDoll
\[\large f(4)=\frac{49}{2401}\]
UsukiDoll
  • UsukiDoll
\[f(4) = \frac{1}{49}\]
UsukiDoll
  • UsukiDoll
first two are fine last one is when x = 0 , so it's not 7
anonymous
  • anonymous
Okay, and that is all?
UsukiDoll
  • UsukiDoll
ugh trying to think to those 3 ...
UsukiDoll
  • UsukiDoll
second to the last one is confusing... x.x
anonymous
  • anonymous
We'll just say no to that one...SO now we are left with C and D.
UsukiDoll
  • UsukiDoll
if it's all reals then that would have the domain and range at >0
UsukiDoll
  • UsukiDoll
I see it well on the range though. when y > 0
UsukiDoll
  • UsukiDoll
on the graph
UsukiDoll
  • UsukiDoll
but the domain is smoosh... like it's just on the line
UsukiDoll
  • UsukiDoll
so first two are fine domain is smoosh it's just 0 flat range is greater than 0 no idea - split on that one it's not 7 because x = 0 for initial conditions.
UsukiDoll
  • UsukiDoll
I think after x = 3 the previous is 1/7 due to 1/7 (1/7)(1/7) (1/7)(1/7)(1/7)
anonymous
  • anonymous
Okay thank you

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