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anonymous

  • one year ago

Belinda wants to invest $1000. The table below shows the value of her investment under two different options for three different years: Number of years 1 2 3 Option 1 (amount in dollars) 1300 1690 2197 Option 2 (amount in dollars) 1300 1600 1900 Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer. (2 points) Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years. (4 points) Part C: Beli

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  1. anonymous
    • one year ago
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    Part C: Belinda wants to invest in an option that would help to increase her investment value by the greatest amount in 20 years. Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1? Explain your answer, and show the investment value after 20 years for each option. (4 points)

  2. anonymous
    • one year ago
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    I ONLY NEED PART C IGIVE MEDALS :)

  3. anonymous
    • one year ago
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    @Luigi0210

  4. anonymous
    • one year ago
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    @mathmate

  5. anonymous
    • one year ago
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    @Nnesha

  6. anonymous
    • one year ago
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    @sammixboo

  7. anonymous
    • one year ago
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    @DanJS

  8. anonymous
    • one year ago
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    @Kainui

  9. anonymous
    • one year ago
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    help please?

  10. DanJS
    • one year ago
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    20 year value, n = 20, put that in both of your functions from part B

  11. anonymous
    • one year ago
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    can you show me how to solve it

  12. DanJS
    • one year ago
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    what are your functions from part b ?

  13. anonymous
    • one year ago
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    1300+300n 1300+300n* 1.3n

  14. DanJS
    • one year ago
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    hmm The first option is exponential form (1000)*(r)^n The second option is linear form 1000 + n*d have to figure what r and d are

  15. DanJS
    • one year ago
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    Option 1) THe function looks like (initial investment)*(r)^n \[f(n) = 1000*(r)^n\] r is the common ratio of consecutive terms 1300/1000 = 1.3 or 1690/1300 = 1.3 or 2197/1690 = 1.3 r is 1.3

  16. DanJS
    • one year ago
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    f(n) = 1000(1.3)^n

  17. DanJS
    • one year ago
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    Each next term is the previous multiplied by 1.3

  18. DanJS
    • one year ago
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    Option 2) Each term increases by the same amount over the previous term, it is always 300 more

  19. DanJS
    • one year ago
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    f(n) = (starting value) + n*(common difference) f(n) = 1000 + 300*n

  20. anonymous
    • one year ago
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    so i use this to solve for the difference right?

  21. DanJS
    • one year ago
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    1) f(n) = 1000(1.3)^n 2) f(n) = 1000 + 300*n yep, put n=20 in both options and calculate the values

  22. anonymous
    • one year ago
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    do i subtract both of the answers i get for the answer

  23. DanJS
    • one year ago
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    you dont have to, just compare them

  24. DanJS
    • one year ago
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    option 1 should be way larger

  25. anonymous
    • one year ago
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    yes it was that you so much :)

  26. anonymous
    • one year ago
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    *thank

  27. DanJS
    • one year ago
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    welcome

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