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kykrilka [37]
2 years ago
5

Which of these expressions are equivalent to p/3?

Mathematics
2 answers:
aleksandrvk [35]2 years ago
5 0
<h3>Answer: Choice B.  1/3p</h3>

This is assuming that p is not in the denominator. To make it more clear that p is not in the denominator, I recommend you write (1/3)p.

Taking 1/3 of a number is the same as dividing by 3.

Example:

27/3 = 9

27*(1/3) = 9

evablogger [386]2 years ago
4 0
1/3p

This is expressed as equívalent because 1/1 = 1 this 1/3=.33, thus 1/3 is p/3
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(8,3)

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3 years ago
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
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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.

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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

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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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