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Studentka2010 [4]
3 years ago
9

An office building rents for $40,000 the first year, then increases by 0.5% each year. Use summation notation to represent the t

otal payments for the first three years. Explain your reasoning. Marcus’s response is given below: If rent increases by 0.5% each year, the year’s rent is 1.005 times the previous year’s rent. So, an exponential expression can be used. The summation is 1.005 raised to the year times $40,000, where the years go from 0 to 3.
Mathematics
2 answers:
Sholpan [36]3 years ago
4 0
Given the initial rent is $40000, and the rent is increased by 0.5% each year, the rent in the third year will be given by:
FV=P(1+r)^n
FV=40000(1.05)^n
Thus the answer  for the sum of payments for 3 years will be:
The summation is 1.05 raised to n years times  $40000, where the years is from 0 to 3.
erica [24]3 years ago
3 0

Answer:

The indices Marcus used are not correct.

As is, the indices create a summation with four terms, not three.

The upper index should be 2 instead of 3.

Step-by-step explanation:

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Explanation

Problem #2

We must find the solution to the following system of inequalities:

\begin{gathered} 3x-2y\leq4, \\ x+3y\leq6. \end{gathered}

(1) We solve for y the first inequality:

-2y\leq4-3x.

Now, we multiply both sides of the inequality by (-1), this changes the signs on both sides and inverts the inequality symbol:

\begin{gathered} 2y\ge-4+3x, \\ y\ge\frac{3}{2}x-2. \end{gathered}

The solution to this inequality is the set of all the points (x, y) over the line:

y=\frac{3}{2}x-2.

This line has:

• slope m = 3/2,

,

• y-intercept b = -2.

(2) We solve for y the second inequality:

\begin{gathered} x+3y\leq6, \\ 3y\leq6-x, \\ y\leq-\frac{1}{3}x+2. \end{gathered}

The solution to this inequality is the set of all the points (x, y) below the line:

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,

• y-intercept b = 2.

(3) Plotting the lines of points (1) and (2), and painting the region:

• over the line from point (1),

,

• and below the line from point (2),

we get the following graph:

Answer

The points that satisfy both inequalities are given by the intersection of the blue and red regions:

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