The number of websites is growing by 15% percent each month. In June 2005, there were 4000 websites. Write a formula for the number of websites as a fuction of months since june 2005. And hOW LONG WILL IT TAKE FOR THE NUMBER OF SITES TO reach 8000 and 16000?

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The number of websites is growing by 15% percent each month. In June 2005, there were 4000 websites. Write a formula for the number of websites as a fuction of months since june 2005. And hOW LONG WILL IT TAKE FOR THE NUMBER OF SITES TO reach 8000 and 16000?

Mathematics
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15*12*...maybe something like this
lol thinker, still here? :)
going going lol

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Other answers:

scoot =P
Let y be the number of websites, and x the number of months. A relationship that represents them is given by: y=4000+0.15x.
\[y=4000+0.15x \rightarrow (*)\]
To find how long it will take to reach a certain number of websites, just plug that number in y, and then solve for x.
you mean solve for y, right anwar?
No. I mean solve for x, since x represents the months and y represents the number of websites.
but she wants the number of websites and not months
wait, nvm lol, I misread the question >_<
my bad ^_^
Let's take the first one, reaching 8000.
wait can you tell me the formula aginain with T being the input variable
Just put T instead of x.
ok
so now will you help me with pluging the nubers in
So, let's do the part when it reaches 8000 websites. Substitute in (1) by "y=8000": \[y=4000+0.15x \implies 8000=4000+0.15x \rightarrow(1)\] Solving equation (1), by first subtracting 4000 from both sides and then dividing by 0.15
OMG.. I got a large number that does not seem right to me.
Just let me check my steps.
Oh I did a big mistake. My first step wasn't right. The problem is much more complicated.
Ok. Focus now. let y be the number of websites, and T the number of months, the relation is given by: \[y={0.15y \over x}+4000 \rightarrow(1)\]
I meant: \[y={0.15y \over T}+4000\]
tha is the formula
Yeah!! the first one was wrong.
Sorry about that.
you can simplify it more, or just leave it as it is.
Now plug y=8000, and solve for x: \[8000={0.15(8000) \over x}+4000 \implies 8000x=1200+4000x \implies 4000x=1200\] \[\implies x={1200 \over 4000} \implies x=0.3\] That means it would take 0.3 months (9 days) to reach to 8000 websites.
Similarly, You can do it for 16000.
OK I GOT IT
Good!! I hope it makes sense to you :)
if it is increasing by 15%per month then how can it double in just 9 days
^^ point :(
OK CAN YOU HELP ME WITH ANOTHER pROMBLEM . ACTUALLY it was solved earlier by someone else but They got it wrong I want to see if you will guide me the right way
ok i will try but it may take few minutes
TW= I(1 + r/q)nq TW-total web I-initial web q-number of months n-number of years
r-rate
At a concert By the Who in 1976, the sound level 50 meters from the stag registered 120 decibles. The treshold of pain for the human ear is 90 decibles. Calculate the sound intensity of the Who concert and the sound intenisity of the treshold of pain.
can u help me with this qetion please jas
i m sorry i m not sure about it

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