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Artist 52 [7]
3 years ago
8

I don’t know where to start someone helpp

Mathematics
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

8 0
3 years ago
HCF and LCM of 24 and 30
IgorC [24]
Green beans, cheese and hamburger
5 0
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
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