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toxicsugar22
 5 years ago
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
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.015*12*...maybe something like this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol thinker, still here? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=4000+0.15x \rightarrow (*)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To 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
 5 years ago
Best ResponseYou've already chosen the best response.0you mean solve for y, right anwar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. I mean solve for x, since x represents the months and y represents the number of websites.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but she wants the number of websites and not months

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, nvm lol, I misread the question >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let's take the first one, reaching 8000.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0wait can you tell me the formula aginain with T being the input variable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just put T instead of x.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0so now will you help me with pluging the nubers in

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0OMG.. I got a large number that does not seem right to me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just let me check my steps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh I did a big mistake. My first step wasn't right. The problem is much more complicated.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. 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
 5 years ago
Best ResponseYou've already chosen the best response.0I meant: \[y={0.15y \over T}+4000\]

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0tha is the formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah!! the first one was wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can simplify it more, or just leave it as it is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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
 5 years ago
Best ResponseYou've already chosen the best response.0Similarly, You can do it for 16000.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good!! I hope it makes sense to you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it is increasing by 15%per month then how can it double in just 9 days

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0OK 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok i will try but it may take few minutes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0TW= I(1 + r/q)nq TWtotal web Iinitial web qnumber of months nnumber of years

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0At 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
 5 years ago
Best ResponseYou've already chosen the best response.0can u help me with this qetion please jas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i m sorry i m not sure about it
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