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andrey2020 [161]
3 years ago
9

NEED HELP ASAP ON GEOMETRY!! ONLY 3 QUESTIONS ATTACHED WITH PHOTOS. THANK YOU

Mathematics
1 answer:
ankoles [38]3 years ago
4 0
3. Look at the picture.

Use Pythagorean theorem:

x^2=3.6^2+5.5^2\\\\x^2=12.96+30.25\\\\x^2=43.21\to x=\sqrt{43.21}\\\\\boxed{x\approx6.6}


4.
Use Pythagorean theorem:

3^2+\left(\dfrac{x}{2}\right)^2=6.5^2\\\\9+\dfrac{x^2}{4}=42.25\ \ \ |-9\\\\\dfrac{x^2}{4}=33.25\ \ \ \ |\cdot4\\\\x^2=133\to x=\sqrt{133}\\\\\boxed{x\approx11.5}


5.
\widehat{AB}\ \cong\ \widehat{DE}

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Answer:

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Read 2 more answers
Suppose that v is an eigenvector of matrix A with eigenvalue λA, and it is also an eigenvector of matrix B with eigenvalue λB. (
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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}

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