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alexandr402 [8]
3 years ago
5

3. Russell is filling a cylinder-shaped swimming pool that

Mathematics
2 answers:
djverab [1.8K]3 years ago
6 0

Answer:

942.4777961 or 942

Step-by-step explanation:

First find the area of the circle using \pi r^{2} . Radius is half of diameter so the radius is 10.

\pi 10^{2} = 314.1592654

Then multiply the depth to the area of the circle.

314.1592654 x 3 = 942.4777961

pickupchik [31]3 years ago
3 0

Answer:

942 cubic feet

Step-by-step explanation:

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22. An employee joined a company in 2017 with a starting salary of $50,000. Every year this employee receives a raise of $1000 p
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Answer:

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  c) s[n] = 70000·1.05^n -20000

Step-by-step explanation:

a) Let s[n] represents the employee's salary in dollars n years after 2017. The salary in 2017 is given as $50,000, so ...

  s[0] = 50000 . . . . the recursion relation initial condition

The next year, the salary is multiplied by 1.05 and 1000 is added:

  s[1] = 1.05·s[0] +1000

This pattern repeats, so the recursion relation is ...

  s[n] = 1.05·s[n-1] +1000

__

c) It is convenient to find a formula for the salary before trying to compute the salary 8 years on. So, we work part (c) before part (b).

The base salary gets multiplied by 1.05 each year, so can be described by the exponential function ...

  base salary after n years = 50,000·1.05^n

The add-on to the raise becomes a geometric sequence whose common ratio is 1.05. The sum can be described by the formula for the sum of such a sequence:

  sum = a1(r^n -1)/(r -1) . . . . . (see note below)

  sum of add-ons = 1000(1.05^n -1)/(1.05 -1) = 20000(1.05^n -1)

So, the total salary after n years is ...

  s[n] = 50000·1.05^n + 20000(1.05^n -1)

The exponential terms can be combined, so we have the explicit formula ...

  s[n] = 70000·1.05^n -20000

__

b) The year 2025 is 8 years after 2017, so we want to find s[8].

  s[8] = 70000·1.05^8 -20000 = 70000·1.4774554 -20000

     = 103,421.88 -20000 = 83,421.88

In 2025, the employee's salary will be $83,421.88.

_____

<em>Note on the sum of the add-ons</em>

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