1. Find the measure of 4FEB

an aircraft has 77 females on board. about 48% of all of the passengers are on board how many passengers are on board on the air

BabaBlast [244]

Answer:

160

Step-by-step explanation:

77 / 48 = 1.604

1.604 x 100 = 160.41 which is about 160 passengers in total.

4 0

3 years ago

Read 2 more answers

Annika is selling art prints at a comic convention.The convention charged her $50 for her table space, and she is selling her pr

asambeis [7]

If she wants to make 130.00, she should sell 90 prints. She is charged at first so she starts at -$50.00. And I don’t understand the first question.

5 0

3 years ago

Please help! Need help asap!

Sedbober [7]

Answer:

A

Step-by-step explanation:

The two angles in the triangle in the question are 58 degrees and 54 degrees.

Three angles in a triangle add up to 180 degrees

58 + 54 + x = 180

112 + x = 180

subtract 112 on both sides

x = 180 - 112 = 68 degrees

Only option A has a triangle that has the same angles, thus triangles are similar.

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6 0

2 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

2 years ago

Find the distance between the lines Y=3x+10 Y=3x-20 AND Y=3/2x+3/2 Y=3/2x-5

rosijanka [135]

At first we need to know how to find the distance between two parallel lines
The parallel lines which have the same slope
If we have two parallel lines ⇒ y₁ = mx + c₁ and y₂ = mx + c₂
So, the distance between y₁ and y₂ = d = $\frac{|c_{1} - c_{2}|}{\sqrt{1 + m^{2}}}$
========================================================

Part (1):
=======

The distance between the lines

 Y₁ = 3x + 10 and Y₂ = 3x - 20
∴ m = 3 , c₁ = 10 and c₂ = -20
∴ d = $\frac{| 10 - (-20)|}{\sqrt{1 + 3^{2}}}= \framebox{9.5}$

========================================================

Part (2):
=======

The distance between the lines

Y₁ = 3/2 x + 3/2 and Y₂ = 3/2 x - 5
∴ m = 3/2 = 1.5 , c₁ = 3/2 = 1.5 and c₂ = -5
∴ d = $\frac{| 1.5- (-5)|}{\sqrt{1 + 1.5^{2}}}= \framebox{3.6}$

6 0

2 years ago