Austin deposits $2,250 into a one-year CD at an interest rate of

goldenfox [79]

<h3>Answer: Choice A) circle</h3>

Explanation:

Imagine that white rectangle as a blade that cuts the cylinder as the diagram shows. If you pull the top cylinder off and examine the bottom of that upper piece, then you'll see a circle forms. It's congruent to the circular face of the original cylinder. This is because the cutting plane is parallel to both base faces of the cylinder. Any sort of tilt will make an ellipse form. Keep in mind that any circle is an ellipse, but not vice versa.

Another example of a cross section: cut an orange along its center and notice that a circle (more or less) forms showing the inner part of the orange.

Yet another example of a cross section: Imagine an egyptian pyramid cut from the top most point on downward such that you vertically slice it in half. If you pull away one half, you should see a triangular cross section forms.

3 0

3 years ago

Read 2 more answers

Determine the measure of the angle labeled (4x+7)°.

Murrr4er [49]

Answer:

32

Step-by-step explanation:

First, combine like terms

4x+7=6x-9

+9 +9

4x+16=-9

-16 -16

Then divide by four to isolate x

4x=-25

x=-6.25

Plug in X to get the angle measure .

3 0

2 years ago

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Explain how to find the surface area of a 3-D figure

german

Well, for any prism, it is the lateral area + the area of two ends.

8 0

2 years ago

What are both of the ranges

MArishka [77]

Answer:

0.8 < x < 16.8

Step-by-step explanation:

The third side of the triangle is always less than the other two sides combined. I think of it like if it was greater than the other two sides added, there would be no triangle- it'd be a line.

The other rule is that it has to be greater than the difference between the two lines.

I hope this helped!

6 0

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