Mathematics, 21.04.2020 04:24 spetgrave069

An employee joined a company in 2009 with a starting salary of $50,000. Every year this employee receives a raise of $1000 plus 5% of the salary of the previous year.
a) Set up a recurrence relation for the salary of this employee n years after 2009.
b) What will the salary of this employee be in 2017?
c) Find an explicit formula for the salary of this employee n years after 2009.

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Answer from: anthony4034

(a) The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)The salary of the employee will be $83421.88 in 2017.

Step-by-step explanation:

Summation of a G.P series

$\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}$

(a)

Every year the salary is increasing 5% of the salary of the previous year plus $1000.

Let a_n represents the salary of the employee of n years after 2009.

Then $a_{n-1}$ represents the salary of the employee of (n-1) years after 2009.

Then $a_n= a_{n-1}+5\%.a_{n-1}+1000$

$=a_{n-1}+0.05a_{n-1}+1000$

$=(1+0.05)a_{n-1}+1000$

$=1.05a_{n-1}+1000$

The required recurrence relation for the salary of the employee of n years after 2009 is $a_n=1.05a_{n-1}+1000$ .

(b)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

Since 2017 is 8 years after 2009.

So, n=8.

∴ a_8

=1.05 a_7+1000

=1.05(1.05a_6+1000)+1000

$=1.05^2a_6+1.05\times 1000+1000$

$=1.05^2(1.05a_5+1000)+1.05\times 1000+1000$

$=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^5(1.05a_2+1000)+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6a_2+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^6(1.05a_1+1000)+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7a_1+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^7(1.05a_0+1000)+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

$=1.05^8a_0+1.05^7\times1000+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000$

=1.05^8a_0+(1.05^7+1.05^6+1.05^5+1.05^4+1.05^3+1.05^2+1.05+1)1000

$=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000$

$=1.05^8\times 50,000+20,000(1.58^8-1)$

$=70,000\times 1.05^8-20,000$

≈$83421.88

The salary of the employee will be $83421.88 in 2017.

(c)

Given, $a_0=\$50,000$

$a_n=1.05a_{n-1}+1000$

We successively apply the recurrence relation

$a_n=1.05a_{n-1}+1000$

$=1.05^1a_{n-1}+1.05^0.1000$

$=1.05^1(1.05a_{n-2}+1000)+1.05^0.1000$

$=1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000$

$=1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)$

$=1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)$

...............................

.................................

$=1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000$

$=1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i$

$=1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}$

=1.05^n.50,000+20,000.(1.05^n-1)

=(50,000+20,000)1.05^n-20,000

$=70,000 . \ 1.05^n-20,000$

$\therefore a_n=70,000 . \ 1.05^n-20,000$

Answer from: Quest

step-by-step explanation:

most likely the answer is a.

alice wins if a total of 5 is rolled. finn wins if a total of 9 is rolled.

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

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