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Step2247 [10]
2 years ago
12

A computer is worth $4000 when it is new. After each year it is worth half what it was the previous year. What will its worth be

after 4 years? Round your answer to the nearest dollar.
Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
8 0

Answer: $250

Step-by-step explanation:

The computer reduces by half of its value every year. It is originally valued at $4,000 so the value in 4 years can be calculated by the formula:

= 4,000 * 0.5 ^ number of years

= 4,000 * 0.5⁴

= $250

Another method is to reduce by half every year:

First year = 4,000 / 2 = $2,000

Second year = 2,000 / 2 = $1,000

Third year = 1,000 / 2 = $500

Fourth year = 500 / 2 = $250

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<h2>Answer:</h2>

<em><u>Recursive equation for the pattern followed is given by,</u></em>

a_{n}=a_{n-1}+(n-1)^{2}

<h2>Step-by-step explanation:</h2>

In the question,

The number of interaction for 1 child = 0

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We need to find out the pattern for the recursive equation for the given conditions.

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We see that,

a_{1}=0\\a_{2}=1\\a_{3}=5\\a_{4}=14\\

Therefore, on checking, we observe that,

a_{n}=a_{n-1}+(n-1)^{2}

On checking the equation at the given values of 'n' of, 1, 2, 3 and 4.

<u>At, </u>

<u>n = 1</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{1}=a_{1-1}+(1-1)^{2}\\a_{1}=0+0=0\\a_{1}=0

which is true.

<u>At, </u>

<u>n = 2</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{2}=a_{2-1}+(2-1)^{2}\\a_{2}=a_{1}+1\\a_{2}=1

Which is also true.

<u>At, </u>

<u>n = 3</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{3}=a_{3-1}+(3-1)^{2}\\a_{3}=a_{2}+4\\a_{3}=5

Which is true.

<u>At, </u>

<u>n = 4</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{4}=a_{4-1}+(4-1)^{2}\\a_{4}=a_{3}+9\\a_{4}=14

This is also true at the given value of 'n'.

<em><u>Therefore, the recursive equation for the pattern followed is given by,</u></em>

a_{n}=a_{n-1}+(n-1)^{2}

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mars1129 [50]

Answer:

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Step-by-step explanation:

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You can solve this by subtracting 2x (from both sides) ...

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Step-by-step explanation:

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