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3 years ago

Simplify 25 × t⁻⁴ /5⁻³×10⁻²×t⁻⁸ plsssssssssssssss help

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

Answer:

Step-by-step explanation:

$\frac{25*t^-^4}{5^-^3*10^-^2*t^-^8}$

Could you confirm to me that this is in fact the problem? Do not report. I will edit this answer once you answer my question.

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3 years ago

What is 33/7 divided by 1 13/14

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4 years ago

The matrix A= (−3 0 1, 2 −4 2, −3 −2 1) has one real eigenvalue. Find this eigenvalue, its multiplicity, and the dimension of th

Aliun [14]

Answer:

a) -4

b) 1

c) 1

Step-by-step explanation:

a) The matrix A is given by:

$A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right]$

to find the eigenvalues of the matrix you use the following:

$det(A-\lambda I)=0$

where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:

$A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right]$

and by taking the determinant:

$[(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0$

and the roots of this polynomial is:

$\lambda_1=-4\\\\\lambda_2=-1+i\sqrt{3}\\\\\lambda_3=-1-i\sqrt{3}$

hence, the real eigenvalue of the matrix A is -4.

b) The multiplicity of the eigenvalue is 1.

c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)

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3 years ago