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Semenov [28]
3 years ago
5

IHELP if 1 out of 5 animals are cats then how many cats are there in an ecosystem of 200 animals???​

Mathematics
1 answer:
mote1985 [20]3 years ago
3 0
There are 40 cats in the ecosystem.

You set 1/5 equal to x/200. Then you cross multiply the fractions to get 5x=200. You divide both terms by 5 to get x, which equals 40.

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|3n – 6| = 6<br> absolute value equation
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Answer:

n=\{0,\: 4\}

Step-by-step explanation:

|3n-6| = 6\\\therefore 3n - 6 = \pm 6\\\therefore 3n = 6 \pm 6\\\therefore 3n =6+6\:\: or\:\: 3n = 6-6\\\therefore 3n =12\:\: or \:\:3n = 0\\\\\therefore n =\frac{12}{3}\:\: or\:\: n =\frac{0}{3}\\\\\therefore n = 4 \:\: or\:\: n = 0\\\\\therefore n=\{0,\: 4\}\\

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What is the simplest form of √324x^6y^8?
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Step-by-step explanation:

\sqrt{324 {x}^{6}  {y}^{8} }

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Jason considered two similar televisions at a local electronics store. The generic version was based off the brand name and was
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Any two rays will form angle. True False
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True

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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}

6 0
3 years ago
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