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Arlecino [84]
2 years ago
6

Is 365458 is divisible by 4.​

Mathematics
1 answer:
bija089 [108]2 years ago
5 0

Answer:

No. 365458 is not divisible by 4. hope this helps

Step-by-step explanation:

- Zombie

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Step-by-step explanation:  You put a line in the middle and it makes a triangle

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Mr. Rodriguez had his students line up in order from shortest to tallest. Tia was exactly in the middle of all the students. Whi
evablogger [386]

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Median

Step-by-step explanation:

The median value of a distribution is obtained by :

FIRSTLY : Arranging the data in order usually ascending order.

Then the 50th percentile of the data is determined , the value which falls in the 50th percentile is the median of the distribution. The 50th percentile is the exact mid value of the distribution.

This describes the exact same way Tia was chosen in the scenario described above. Hence the measure of the center described is the Median.

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A square has side length 4 cm. What is the volume
gayaneshka [121]

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Step-by-step explanation:

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Write an integer for the situation below: Greg has a debt of $55
Aleks [24]

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-55

5 0
2 years ago
Read 2 more answers
Use the method of undetermined coefficients to solve the given nonhomogeneous system. x' = −1 5 −1 1 x + sin(t) −2 cos(t)
AlekseyPX

It looks like the system is

x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}

Compute the eigenvalues of the coefficient matrix.

\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i

For \lambda = 2i, the corresponding eigenvector is \eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top such that

\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

Notice that the first row is 1 + 2i times the second row, so

(1+2i) \eta_1 - 5\eta_2 = 0

Let \eta_1 = 1-2i; then \eta_2=1, so that

\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}

The eigenvector corresponding to \lambda=-2i is the complex conjugate of \eta.

So, the characteristic solution to the homogeneous system is

x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}

The characteristic solution contains \cos(2t) and \sin(2t), both of which are linearly independent to \cos(t) and \sin(t). So for the nonhomogeneous part, we consider the ansatz particular solution

x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}

Differentiating this and substituting into the ODE system gives

-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}

\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}

Then the general solution to the system is

x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}

7 0
1 year ago
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