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Bumek [7]
3 years ago
13

You want to save $500 for a school trip. You begin by saving a penny on the first day. You save an additional penny each day aft

er that. For example, you will save two pennies on the second day, three pennies on the third day, and so on. a. How much money will you have saved after 100 days?b. Use a series to determine how many days it takes you to save $500?
Mathematics
1 answer:
irga5000 [103]3 years ago
6 0

Answer:

Step-by-step explanation:

The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the arithmetic sequence.

d represents the common difference of the terms in the arithmetic sequence.

a represents the first term of the arithmetic sequence.

From the information given,

a = 1 penny = 1/100 = $0.01

d = 0.01

a) For 100 days, the sum of the first 100 terms, S100 would be

S100 = 100/2[2 × 0.01 + (100 - 1)0.01]

S100 = 50[0.02 + 0.99)

S100 = 50 × 1.01 = $50.5

b) when Sn = $500, then

500 = n/2[2 × 0.01 + (n - 1)0.01]

Multiplying through by 2, it becomes

500 × 2 = n[2 × 0.01 + (n - 1)0.01]

1000 = n[0.02 + 0.01n - 0.01]

1000 = n[0.01 + 0.01n]

1000 = 0.01n + 0.01n²

0.01n² + 0.01n - 1000 = 0

Applying the general formula for quadratic equations,

x = [-b±√(b² - 4ac)]/2a

n = - 0.01±√0.01²-4(0.01 × - 1000)]/2 × 0.01

n = (- 0.01 ± √40.001)/0.02

n = (- 0.01 + 6.32)/0.02 or

n = (- 0.01 - 6.32)/0.02

n = 315.5 or n = - 316.5

Since n cannot be negative, then n = 315.5

It will take approximately 316 days to save $500

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Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation det(A-\lambdaI)=0 where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6

So the characteristic polynomial is \lambda^2-5\lambda+6=0.

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0

So \lambda=3, \lambda=2

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If \lambda=3 we get the following matrix

\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right].

Since both rows are equal, we have the equation

x-2y=0. Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case \lambda=2, using the same process, we get the vector (1,1).

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P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

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