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3 years ago

PLEASE HELP ME I NEED THIS DONE TODAY PLEASE HELP I KNOW NO ONE WILL BUT IT'S 7TH GRADE MATH PLEASE HELP

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer:

Overall answer: 10 1/9 - 1 5/18x (Most simplified answer)

Step-by-step explanation:

First you have to do all the Distributive property which the answer would be:

2 1/2 - 1/2x + 3 1/9 + 8/9x + 1/6x - 5/18x + 2 14/18 + 1 13/18

Then you have to find the common demoninator which would be 18

After that you have to put them in the correct order since there is two different terms which would be:

2 9/18 + 3 2/18 + 2 14/18 + 1 13/18 - 9/18x + 16/18x + 3/18x - 5/18x

After that you solve the problem which in the end after solving both terms it ends up as..:

10 2/18 - 1 5/18x

and I think that is as simplified as you can go term wise.

I hope this helps and I hope I didn't make a mistake :D

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We have the matrix $A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]$

To find the eigenvalues of A we need find the zeros of the polynomial characteristic $p(\lambda)=det(A-\lambda I_3)$

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$p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda$

Now, we fin the zeros of $p(\lambda)$ .

$p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4$

Then, the eigenvalues of A are $\lambda_{1}=0$ of multiplicity 1 and $\lambda{2}=-4$ of multiplicity 2.

Let's find the eigenspaces of A. For $\lambda_{1}=0$ : $E_0 = Null(A- 0I_3)=Null(A)$ .Then, we use row operations to find the echelon form of the matrix