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2 years ago

5

For the matrix a, find (if possible) a nonsingular matrix p such that p−1ap is diagonal.

1 answer:

Flura [38]2 years ago

4 0

Sorry but can you please show like a diagram or something for me to answer it for you. Thank You

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(−t 4 −5t 3 −10t 2 )+(9t 3 +3t 2 −1)

klemol [59]

Answer:

$\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1$

Step-by-step explanation:

Given the expression

$\left(-t^4\:-5t^3\:-10t^2\:\right)+\left(9t^3\:+3t^2\:-1\right)$

Remove parentheses: (a)=a

$=-t^4-5t^3-10t^2+9t^3+3t^2-1$

Group like terms

$=-t^4-5t^3+9t^3-10t^2+3t^2-1$

Add similar elements

$=-t^4-5t^3+9t^3-7t^2-1$ ∵ $-10t^2+3t^2=-7t^2$

Add similar elements

$=-t^4+4t^3-7t^2-1$ ∵ $-5t^3+9t^3=4t^3$

Thus, the equivalent expression in simplified form:

$\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1$

4 0

3 years ago

How many solutions does the problem 4x-6=2(x-1)+x

luda_lava [24]

......................

5 0

2 years ago

HELP MEEE!!! PLEASE!!!

lakkis [162]

Easy multiply the first number and do keep change flip

3 0

3 years ago

Read 2 more answers

Mrs.Lynch has noted that 7 out of every 10 pieces of mail she gets are junk mail. What percent of her mail is junk?

AlladinOne [14]

Answer:

70%

Step-by-step explanation:

create fraction pieces/whole.

7 0

2 years ago

Graph<img src="https://tex.z-dn.net/?f=fx%3D%20%7Cx%2B1%7C%20%2B2" id="TexFormula1" title="fx= |x+1| +2" alt="fx= |x+1| +2" alig

AURORKA [14]

Answer:

The graph is shown in the following image

Step-by-step explanation:

The parent function of $f(x) = | x + 1 | +2$ is $y = | x |$

The graph of $y = | x |$ is the union of two lines:

$y = x$ for $x\geq0$

$y = -x$ for $x$

The vertex of this graph is in point (0,0)

<em>The graphic is shown in the attached image</em>.

Then, if we do $y = f(x + 1)$ to the parent function we get:

$y = | x + 1 |$

This transformation shifts the graph of $y = |x|$ 1 unit to the left

Then, if we make the transformation $y = f(x) + 2$ to the function $y = | x + 1 |$ we obtain:

$y = | x + 1 | +2$

This transformation displaces the graph of the function $y = | x + 1 |$ 2 units up on the y axis.

So, if the vertex of the main function $y = | x |$ was at point (0,0), the vertex of the function $f(x) = |x + 1| +2$ has its vertex in the point (-1, 2).

<em>The graph is shown in the following image</em>

5 0

3 years ago