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?

- toxicsugar22

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- anonymous

15*12*...maybe something like this

- anonymous

lol thinker, still here? :)

- anonymous

going going lol

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## More answers

- anonymous

scoot =P

- anonymous

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.

- anonymous

\[y=4000+0.15x \rightarrow (*)\]

- anonymous

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.

- anonymous

you mean solve for y, right anwar?

- anonymous

No. I mean solve for x, since x represents the months and y represents the number of websites.

- anonymous

but she wants the number of websites and not months

- anonymous

wait, nvm lol, I misread the question >_<

- anonymous

my bad ^_^

- anonymous

Let's take the first one, reaching 8000.

- toxicsugar22

wait can you tell me the formula aginain with T being the input variable

- anonymous

Just put T instead of x.

- toxicsugar22

ok

- toxicsugar22

so now will you help me with pluging the nubers in

- anonymous

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

- anonymous

OMG.. I got a large number that does not seem right to me.

- anonymous

Just let me check my steps.

- anonymous

Oh I did a big mistake. My first step wasn't right. The problem is much more complicated.

- anonymous

Ok. Focus now. let y be the number of websites, and T the number of months, the relation is given by:
\[y={0.15y \over x}+4000 \rightarrow(1)\]

- anonymous

I meant:
\[y={0.15y \over T}+4000\]

- toxicsugar22

tha is the formula

- anonymous

Yeah!! the first one was wrong.

- anonymous

Sorry about that.

- anonymous

you can simplify it more, or just leave it as it is.

- anonymous

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.

- anonymous

Similarly, You can do it for 16000.

- toxicsugar22

OK I GOT IT

- anonymous

Good!! I hope it makes sense to you :)

- anonymous

if it is increasing by 15%per month then how can it double in just 9 days

- anonymous

^^ point :(

- toxicsugar22

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

- anonymous

ok i will try but it may take few minutes

- anonymous

TW= I(1 + r/q)nq
TW-total web
I-initial web
q-number of months
n-number of years

- anonymous

r-rate

- toxicsugar22

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.

- toxicsugar22

can u help me with this qetion please jas

- anonymous

i m sorry
i m not sure about it

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