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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?

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  1. anonymous
    • 5 years ago
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    15*12*...maybe something like this

  2. anonymous
    • 5 years ago
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    lol thinker, still here? :)

  3. anonymous
    • 5 years ago
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    going going lol

  4. anonymous
    • 5 years ago
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    scoot =P

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    \[y=4000+0.15x \rightarrow (*)\]

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    you mean solve for y, right anwar?

  9. anonymous
    • 5 years ago
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    No. I mean solve for x, since x represents the months and y represents the number of websites.

  10. anonymous
    • 5 years ago
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    but she wants the number of websites and not months

  11. anonymous
    • 5 years ago
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    wait, nvm lol, I misread the question >_<

  12. anonymous
    • 5 years ago
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    my bad ^_^

  13. anonymous
    • 5 years ago
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    Let's take the first one, reaching 8000.

  14. toxicsugar22
    • 5 years ago
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    wait can you tell me the formula aginain with T being the input variable

  15. anonymous
    • 5 years ago
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    Just put T instead of x.

  16. toxicsugar22
    • 5 years ago
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    ok

  17. toxicsugar22
    • 5 years ago
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    so now will you help me with pluging the nubers in

  18. anonymous
    • 5 years ago
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    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

  19. anonymous
    • 5 years ago
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    OMG.. I got a large number that does not seem right to me.

  20. anonymous
    • 5 years ago
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    Just let me check my steps.

  21. anonymous
    • 5 years ago
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    Oh I did a big mistake. My first step wasn't right. The problem is much more complicated.

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

  23. anonymous
    • 5 years ago
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    I meant: \[y={0.15y \over T}+4000\]

  24. toxicsugar22
    • 5 years ago
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    tha is the formula

  25. anonymous
    • 5 years ago
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    Yeah!! the first one was wrong.

  26. anonymous
    • 5 years ago
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    Sorry about that.

  27. anonymous
    • 5 years ago
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    you can simplify it more, or just leave it as it is.

  28. anonymous
    • 5 years ago
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    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.

  29. anonymous
    • 5 years ago
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    Similarly, You can do it for 16000.

  30. toxicsugar22
    • 5 years ago
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    OK I GOT IT

  31. anonymous
    • 5 years ago
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    Good!! I hope it makes sense to you :)

  32. anonymous
    • 5 years ago
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    if it is increasing by 15%per month then how can it double in just 9 days

  33. anonymous
    • 5 years ago
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    ^^ point :(

  34. toxicsugar22
    • 5 years ago
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    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

  35. anonymous
    • 5 years ago
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    ok i will try but it may take few minutes

  36. anonymous
    • 5 years ago
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    TW= I(1 + r/q)nq TW-total web I-initial web q-number of months n-number of years

  37. anonymous
    • 5 years ago
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    r-rate

  38. toxicsugar22
    • 5 years ago
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    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.

  39. toxicsugar22
    • 5 years ago
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    can u help me with this qetion please jas

  40. anonymous
    • 5 years ago
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    i m sorry i m not sure about it

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