In the year 2006, a person bought a new car for $18000. For each consecutive year after that, the value of the car depreciated by 15%. How much would the car be worth in the year 2009, to the nearest hundred dollars? Please i need answer now

Step-by-step explanation:

Use the exponential function

where a is the initial value of the car, b is the rate of decay, and t is the time in years. For us, a = 18000, b = (1 - .15), so the model for this car is:

 We can use this model now to find the value, v(t), of the car after 3 years has gone by (from 2006 to 2009 is 3 years):

and simplifying a bit:

v(t) = 18000(.614125) so

v(t) = 11054.25

Round however you need to.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: