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____ [38]
3 years ago
14

Which correctly gives the location of the point (–19, 0)?

Mathematics
2 answers:
mafiozo [28]3 years ago
8 0
(-19,0) is located on the x-axis
xxTIMURxx [149]3 years ago
6 0
<span>A. x-axis</span>
.....................
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River C is 100 miles longer than River D. If the sum of their lengths is 5,570 ​miles, what is the length of each​ river?
lyudmila [28]

Let the Length of River D be : P

Given : River C is 100 Miles Longer than River D

⇒ Length of River C = P + 100

Given : The Sum of Both Rivers is 5570 Miles

⇒ P + (P + 100) = 5570

⇒ 2P + 100 = 5570

⇒ 2P = 5570 - 100

⇒ 2P = 5470

⇒ P = 2735

⇒ Length of River D is : 2735 Miles

⇒ Length of River C is : (P + 100) = (2735 + 100) = 2835 Miles

7 0
4 years ago
Suppose that v is an eigenvector of matrix A with eigenvalue λA, and it is also an eigenvector of matrix B with eigenvalue λB. (
galben [10]

Answer:

(a) Yes, λ_{A}+λ_{B}

(b) Yes, λ_{A}λ_{B}

Step-by-step explanation:

First, lets understand what are eigenvectors and eigenvalues?

Note: I am using the notation λ_{A} to denote Lambda(A) sign.

v is an eigenvector of matrix A with eigenvalue λ_{A}

v is also eigenvector of matrix B with eigenvalue λ_{B}

So we can write this in equation form as

Av=λ_{A}v

So what does this equation say?

When you multiply any vector by A they do change their direction. any vector  that is in the same direction as of Av, then this v  is called the eigenvector of A. Av is λ_{A} times the original v. The number λ_{A} is the eigenvalue of A.

λ_{A} this number is very important and tells us what is happening when we multiply Av. Is it shrinking or expanding or reversed or something else?

It tells us everything we need to know!

Bonus:

By the way you can find out the eigenvalue of Av by using the following equation:

det(A-λI)=0

where I is identity matrix of the size of same as A.

Now lets come to the solution!

(a) Show that v is an eigenvector of A + B and find its associated eigenvalue.

The eigenvalues of A and B are λ_{A} and λ_{B}, then

(A+B)(v)=Av+Bv=(λ_{A})v + (λ_{B})v=(λ_{A}+λ_{B})(v)

so,  (A+B)(v)=(λ_{A}+λ_{B})(v)

which means that v is also an eigenvector of A+B and the associated eigenvalues are λ_{A}+λ_{B}

(b) Show that v is an eigenvector of AB and find its associated eigenvalue.

The eigenvalues of A and B are λ_{A} and λ_{B}, then

(AB)(v)=A(Bv)=A(λ_{B})=λ_{B}(Av)=λ_{B}λ_{A}(v)=λ_{A}λ_{B}(v)

so,  

(AB)(v)=λ_{A}λ_{B}(v)

which means that v is also an eigenvector of AB and the associated eigenvalues are λ_{A}λ_{B}

5 0
3 years ago
Consider the procedure used below to solve the given equation.
slavikrds [6]

Answer is D you can plug it in to check it as well


5 0
3 years ago
Gage's math teacher entered the seventh-grade student in a math competition. There was an enrollment fee of $30 and also $11 cha
Amanda [17]

Answer:

110 test

Step-by-step explanation:

Let

x ----> the number  of packet of 10 test

y ----> the total cost

we know that

The number of packet of 10 test purchase multiplied by $11 plus the enrollment fee of $30 must be equal to $151

so

The linear equation that represent this scenario is

y=11x+30

we have

y=\$151

substitute

151=11x+30

Solve for x

subtract 30 both sides

11x=151-30

11x=121

Divide by 11 both sides

x=11

The number of packets purchase was 11

To find out the number of test, multiply the number of packets by 10

11(10)=110\ test

4 0
3 years ago
Adiciones racinales 1/7+3/7+2/7
Digiron [165]
6/7 is the awnser for this one
7 0
3 years ago
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