Part a a business purchases a computer system for $3,000. the value of the system decreases at a rate of 15% per year. which exponential function models this situation? a. f(x) = 3,000(0.15)x b. f(x) = 3,000(0.85)x c. f(x) = 3,000 – 0.15x d. f(x) = 3,000 – 0.85x part b the function f(x) = 3,000(0.85)x models the value of a computer system. find the value of the computer after 4 years. value after 4 years = $

The exponential function that models this situation will be:

From the information given, the function f(x) = 3,000(0.85)x models the value of a computer system, the value of the computer after 4 years will be:

= f(x) = 3,000(0.85)x

= f(x) = 3,000(0.85)⁴

= 1566.02.

Therefore, the value of the computer after 4 years will be $1566.02.

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Part A

At the end of the year, the value of the computer system is ...

... (beginning value) - 15% · (beginning value) = (beginning value) · (1 - 0.15)

... = 0.85 · (beginning value)

Since the same is true for the next year and the next, the multiplier after x years will be 0.85^x. Then the value after x years is ...

... f(x) = (beginning value) · 0.85^x

The beginning value is given as $3000, so this is ...

... f(x) = 3000·0.85^x

Part B

For x=4, this is ...

... f(4) = 3000·0.85^4 = 3000·0.52200625 ≈ 1566.02

The value after 4 years is $1566.02.

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,