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3 years ago

8

I don’t know where to start someone helpp

2 answers:

tia_tia [17]3 years ago

8 0

20x+10=total dollars

11111nata11111 [884]3 years ago

5 0

Donald will have 20x+10 dollars

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Solve the quadratic equations below.<br> 3x2 + 22x + 35 = 0

mestny [16]

Step-by-step explanation:

3x² + 22x + 35 = 0

(3x + 7)(x + 5) = 0

3x + 7 = 0

x = -7/3

or x+5 = 0

x = -5

3 0

2 years ago

How do you write an equation in standard form given point (2,2) (0,-4)

notka56 [123]

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (2, 2) and (0, -4). Substitute:

$m=\dfrac{-4-2}{0-2}=\dfrac{-6}{-2}=3\\\\y-2=3(x-2)$

The standard form:

$Ax+By=C$

$y-2=3(x-2)$ <em>use distributive property</em>

$y-2=3x-6$ <em>add 6 to both sides</em>

$y+4=3x$ <em>subtract y from both sides</em>

$4=3x-y$

<h3>Answer: 3x - y = 4</h3>

6 0

3 years ago

PLS HELP I BEG YOU I HAVE TO PASS

MrRa [10]

the range of the scores is 95-58 = 37

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3 years ago

HCF and LCM of 24 and 30

IgorC [24]

Green beans, cheese and hamburger

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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

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