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sineoko [7]
3 years ago
12

Consider the exponential function f ( x ) = 13 , 500 ⋅ 0.89^x, which models the value of Mikayla's scooter, where x represents t

he number of years since she purchased the scooter.
What is the value of Mikayla's scooter after 4 years? Explain.
Mathematics
1 answer:
Elden [556K]3 years ago
7 0

Answer: $8470

Step-by-step explanation:

To solve this, you can substitute 4 for x:

f(x) = 13500*.89^4\\f(x) = 13500*0.6274\\f(x) = 8470

Mikayla's scooter is worth around $8470 after 4 years.

It was originally worth 13,500 dollars and depreciated by 11% every year.

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Give the first ten terms of the following sequences. You can assume that the sequences start with an index of 1. Logs are to bas
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Answer:

a)

a1 = log(1) = 0 (2⁰ = 1)

a2 = log(2) = 1 (2¹ = 2)

a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849

a4 = log(4) = 2 (2² = 4)

a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322

a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)

a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807

a8 = log(8) = 3 (2³ = 8)

a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )

a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322

b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:

log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that

a1 = 1

a2 = 2

a3 = 3

a4 = 4

a5 = 5

a6 = 6

a7 = 7

a8 = 8

a9 = 9

a10 = 10

I hope this works for you!!

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3 years ago
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