anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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)
anonymous
  • anonymous
I ONLY NEED PART C IGIVE MEDALS :)
anonymous
  • anonymous
@Luigi0210

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More answers

anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@Nnesha
anonymous
  • anonymous
@sammixboo
anonymous
  • anonymous
@DanJS
anonymous
  • anonymous
@Kainui
anonymous
  • anonymous
help please?
DanJS
  • DanJS
20 year value, n = 20, put that in both of your functions from part B
anonymous
  • anonymous
can you show me how to solve it
DanJS
  • DanJS
what are your functions from part b ?
anonymous
  • anonymous
1300+300n 1300+300n* 1.3n
DanJS
  • DanJS
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
DanJS
  • DanJS
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
DanJS
  • DanJS
f(n) = 1000(1.3)^n
DanJS
  • DanJS
Each next term is the previous multiplied by 1.3
DanJS
  • DanJS
Option 2) Each term increases by the same amount over the previous term, it is always 300 more
DanJS
  • DanJS
f(n) = (starting value) + n*(common difference) f(n) = 1000 + 300*n
anonymous
  • anonymous
so i use this to solve for the difference right?
DanJS
  • DanJS
1) f(n) = 1000(1.3)^n 2) f(n) = 1000 + 300*n yep, put n=20 in both options and calculate the values
anonymous
  • anonymous
do i subtract both of the answers i get for the answer
DanJS
  • DanJS
you dont have to, just compare them
DanJS
  • DanJS
option 1 should be way larger
anonymous
  • anonymous
yes it was that you so much :)
anonymous
  • anonymous
*thank
DanJS
  • DanJS
welcome

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