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faltersainse [42]
4 years ago
5

Part A

Mathematics
1 answer:
olasank [31]4 years ago
8 0
<h3>Answer:</h3>
  • B. f(x) = 3,000(0.85)^x
  • $1566.02
<h3>Step-by-step explanation:</h3>

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.

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Answer:

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or x = (-9-√57)/2

Step-by-step explanation:

1. Move all terms to one side

x^2+9x+12-6=0

x^2+9x+6=0

2. Plug it in to the quadratic formula

x=(-9±√9^2-4*1*6)/(2)

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tamaranim1 [39]

Answer:

\huge{x =  \frac{236164}{9477}  = 24 \frac{8716}{9477}}

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\boxed{\green{x =  \frac{236164}{9477}  = 24 \frac{8716}{9477}}}

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