All categories

3 years ago

Why yall keep deleting my stuff lol. ∵

Mathematics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

If you ask any non study question then anyone can report,mate .

You might be interested in

I need fast help!!!!!!!!! Please

german

Answer:

x= −5/ 2 + 1 /2 √21 or x= −5 /2 + −1/ 2 √21

Step-by-step explanation:

7 0

3 years ago

A student paid at least 51$ for a computer diskette. The total charge includes the price and 2% tax. Find minimum price of the d

BartSMP [9]

52.02 at is the minimum

8 0

3 years ago

HELP ME PLEASEEEEEEEEEE

Daniel [21]

Answer:

1 and 4 i think

Step-by-step explanation:

3 0

3 years ago

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

Prove that a rectangle has congruent diagonals. Be sure to create and name the appropriate geometric figures.

Anni [7]

Answer:

The first way to prove that the diagonals of a rectangle are congruent is to show that triangle ABC is congruent to triangle DCB

Here is what is given: Rectangle ABCD

Here is what you need to prove: segment AC ≅ segment BD

Since ABCD is a rectangle, it is also a parallelogram.

Since ABCD is a parallelogram, segment AB ≅ segment DC because opposite sides of a parallelogram are congruent.

5 0

4 years ago