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anonymous

  • one year ago

HELP?!

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  1. mathstudent55
    • one year ago
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    How do you show growth?

  2. mathstudent55
    • one year ago
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    For example, if the growth is a certain percentage each year, what equation shows growth?

  3. mathstudent55
    • one year ago
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    Let's say the growth is 10%. Since 100% + 10% = 110% = 1.1, then each year, multiply the previous year's amount by 1.1 Year 0 a Year 1 1.1a Year 2 1.1(1.1a) = 1.1^2 * a Year 3 1.1(1.1^2)a) = 1.1^3 * a Year x 1.1^x * a

  4. mathstudent55
    • one year ago
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    In your case, the growth is 20%, so we use 1.2 Neighborhood A \(y = 30 \times 1.2^x\)

  5. anonymous
    • one year ago
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    and I would just replace x with 5?

  6. mathstudent55
    • one year ago
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    In Neighborhood B, you start with 45 and the growth is linear, so its: Year 0 45 houses Year 1 45 + 3 * 1 Year 2 45 + 3 * 2 Year 4 45 + 3 * 3 etc. Year x 45 + 3 * x \(y = 45 + 3x\)

  7. mathstudent55
    • one year ago
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    Part A. Neighborhood A. \(y = 30 \times 1.2^x\) Neighborhood B. \(y = 45 + 3x\)

  8. mathstudent55
    • one year ago
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    Part B. Let x equal 5 in each of the two functions above to find the number of homes in the two neighborhoods after 5 years.

  9. mathstudent55
    • one year ago
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    Part C. Set the two functions equal and solve for x.

  10. anonymous
    • one year ago
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    so y=30*1.2^5 and y=45+3(5) for B?

  11. mathstudent55
    • one year ago
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    Yes, and evaluate each expression.

  12. anonymous
    • one year ago
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    y=74.6496 y=60

  13. mathstudent55
    • one year ago
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    Good. 75 & 60

  14. mathstudent55
    • one year ago
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    Notice that Neighborhood A started with 30 homes and B with 45. By year 5, Neighborhood A already has more homes than B. This gives you a hint as to the solution of part C.

  15. anonymous
    • one year ago
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    what would I set my function equal to?

  16. anonymous
    • one year ago
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    would I make a function table and use the formula until I get the same number of houses?

  17. mathstudent55
    • one year ago
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    For Part C?

  18. anonymous
    • one year ago
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    yes

  19. mathstudent55
    • one year ago
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    \(30 \times 1.2^x = 45 + 3x\)

  20. anonymous
    • one year ago
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    ohhh ok... and so that's all I need for part c, is to solve that?

  21. mathstudent55
    • one year ago
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    You can use a table. The hint I mentioned above is that since by year 5, Neighborhood A already has more homes than Neighborhood B, that means the year they have the same number of homes is between years 0 and 5.

  22. mathstudent55
    • one year ago
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    Yes, up need to solve the equation, but your idea of letting x = 1, 2, 3, 4, 5 is good.

  23. mathstudent55
    • one year ago
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    Make a table and see when Neighborhood A overtakes B.

  24. anonymous
    • one year ago
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    awee yes I get it now

  25. mathstudent55
    • one year ago
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    Are you given choices?

  26. anonymous
    • one year ago
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    no it's open response

  27. anonymous
    • one year ago
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    I'm making a table rn

  28. mathstudent55
    • one year ago
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    OK. Then do what you mentioned above. Make a table.

  29. mathstudent55
    • one year ago
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    We already know this: |dw:1438567229776:dw|

  30. mathstudent55
    • one year ago
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    It happens between year 3 and year 4. |dw:1438567359681:dw|

  31. anonymous
    • one year ago
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    ok thanks for that because I was plugging in the different numbers and didn't see when they were equal, I didn't know that it could be in between years. So it would be after approximately 3 and a half yrs?

  32. mathstudent55
    • one year ago
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    The smallest difference between A and B is at 3.3 years, so that is a close approximation. |dw:1438567705900:dw|

  33. mathstudent55
    • one year ago
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    Your guess of approximately 3.5 years is also good.

  34. anonymous
    • one year ago
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    Oh my gosh! You were more than helpful and I honestly understand and appreciate everything you've told me. Thank you sooooo much!!! If I were there I'd give you a hug! Absolute lifesaver!!!

  35. mathstudent55
    • one year ago
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    You are very welcome. I just want to mention one point. I don't know if you've learned logarithms yet. Usually, when you have an equation with a variable in the exponent, you use logarithms to find the solution. In this case, the equation we had does not lend itself to taking logarithms of both sides and simplifying easily. That is why I recommended you use a table with values (which was your idea) to see when the functions are of equal value to then see at which time the number of homes are the same.

  36. anonymous
    • one year ago
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    You are correct, I have not learned logarithms yet, but I will remember your advice.

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