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3 years ago

What is this? Someone please help me

Mathematics

1 answer:

Reil [10]3 years ago

8 0

Answer and Step-by-step explanation:

Haven't done this in a while, so here we go:

The rule (I'm not really sure) would have to be negative number and negative number.

This means that:

(4, 0) would become (-4, 0) A'

(6, -2) becomes (-6, 2) B'

(8, -7) becomes (-8, 7) C'

and (0, -5) becomes (0, 5) D'

#teamtrees #WAP (Water And Plant)

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Miss Robin distributed a bag of skittles to her students she gave each student five skittles then ate the 8 that remained in the

Damm [24]

Answer:

Step-by-step explanation:

First, subtract 8 from 143 which would give you 135. You then divide 135 by 5 and you get 27

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2 years ago

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valina [46]

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3 years ago

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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

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3 years ago

Lainey bought a set of 20 markers for $6 dollar sign, 6. What is the cost of 1 marker?

marin [14]

Hello!

To find the price of one marker you divide the price paid by the amount of markers bought

6 / 20 = 0.30

A marker cost $0.30

The answer is $0.30

Hope this helps!

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Hope this helps! I apologize for my long response

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