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mr Goodwill [35]

3 years ago

11

Assessment items Jackie deposited $315 into a bank account that earned 1.5% simple interest each year. If no money was deposited

into or withdrawn from the account, how much money was in the account after 3 years?

1 answer:

Arturiano [62]3 years ago

4 0

The interest would be $14.17 so the total in the account is $329.17.

you find this by using the formula:
I = P*R*T
I =300*.015*3
I = 14.17

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Jim is hiking 4 1/2 mile trail.He has already hiked 2 5/8 miles.How much further does he have to hike?

musickatia [10]

Answer:

1 7/8

Step-by-step explanation:

2 5/8

+1 7/8= 4 1/2

7=3+4
5+3=8
2 5/8 + 3/8= 3
3 + 1= 4
4 + 4/8= 4 4/8
4 4/8 = 4 1/2

3 0

3 years ago

garri49 [273]

The area of the remaining board is [(L × B) - (l × b)].

According to the statement

We have to find that the area of the remaining board.

So, For this purpose, we know that the

Area of rectangle is the region occupied by a rectangle within its four sides or boundaries. The area of a rectangle depends on its sides.

From the given information:

Suppose the bigger rectangle is labelled as ABCD and the smaller rectangle is labelled as PQRS.

And

Consider that the length and breadth of the bigger rectangle are L and B respectively. And the length and breadth of the bigger rectangle are l and b respectively.

The area of any rectangle is:

Area = Length × Breadth

The area of the bigger rectangle is:

Area of ABCD = L × B

The area of the smaller rectangle is:

Area of PQRS = l × b

Then the area of the remaining board will be:

Area of remaining board = Area of ABCD - Area of PQRS

Area of remaining board= (L × B) - (l × b)

Thus, The area of the remaining board is [(L × B) - (l × b)].

Learn more about Area here

brainly.com/question/8409681

Disclaimer: This question was incomplete. Please find the full content below.

Question:

A rectangle is removed from the middle of a larger rectangular shaped board. What is the area of the remaining board?

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2 years ago

True or False: You can tell if two ratios form a proportion if their cross products are equal. True O False

daser333 [38]

Answer:

Step-by-step explanation:

False

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3 years ago

Find the value of $1000 deposited for 10 years in an account paying7% annual interest compounded yearly

Over [174]

Answer:

should be after 10 years , $1967.15

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3 years ago

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

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).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

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

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

4 0

3 years ago