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devlian [24]
3 years ago
9

Free points!!!I'm bored so enjoy.​

Mathematics
2 answers:
melamori03 [73]3 years ago
5 0

Answer:

Thank you so much man!! If you need help in the future just comment on this answer and I will gladly answer!  

IgorLugansk [536]3 years ago
3 0
Jane from breaking bad

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Help
Len [333]

I just took the test!

Answer:

16.2 m²

8 0
3 years ago
In which of the expressions below will x = 12.5? Select all
erma4kov [3.2K]

The answer is D

2x / 5 = 5

5 x 5 = 25

2x = 25

25/2

x= 12.5

4 0
3 years ago
Read 2 more answers
What is rhe sum of 4 5/6 3 5/6
irina [24]
Sum means add.

So, we'd need to add 

4 5/6 + 3 5/6

5/6 + 5/6 = 10/6, or 1 4/6, or 1 2/3.

4 + 3 = 7

7 + 1 =8

8 2/3 is your answer.

Glad I could help, and good luck!
4 0
3 years ago
Read 2 more answers
Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7
Ilia_Sergeevich [38]
I'll go out on a limb and guess the system is

\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x

with initial condition \mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top. The coefficient matrix has eigenvalues \lambda such that

\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12

The corresponding eigenvectors \eta are such that

\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}

\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}

So the characteristic solution to the ODE system is

\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

When t=0, we have

\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}

from which it follows that C_1=2 and C_2=5, making the particular solution to the IVP

\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}
5 0
4 years ago
Subtracting monomials
Arisa [49]

Answer:

55a^2

Step-by-step explanation:

4 0
3 years ago
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