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jeka94
3 years ago
11

Please help!!show work!!please!!

Mathematics
1 answer:
Advocard [28]3 years ago
4 0
\frac{1}{1} *  \frac{1}{3} would work. 
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I need help please!?!!!
11111nata11111 [884]
8(p-2) = 32
8p - 16 = 32
+16 on each side
8p= 48
Divide by 8 on each side
P= 6
7 0
3 years ago
Read 2 more answers
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
Whats the height of the prism
zheka24 [161]

Answer:

oh, you may have forgotten to add the file, you can edit and add it, which i will edit my response and respond.

Step-by-step explanation:

8 0
3 years ago
The Quadratic Equation c² + 6c+ 8 = 0<br><br> Show work please
Bad White [126]
The formula for the quadratic formula is x (c in this case) = (-b(+/-)√(b²-4ac))/2a
This is used for an equation in standard quadratic form: ax² + bx + c = 0
1.) Put it in the correct form, if not already in it.
Ex. c² + 6c + 8 = 0
2.) Identify each part of the equation:
a = 1 (the leading coefficient), b = 6 (the coefficient in front of the second variable), c = 8
3.) Plug in each variable answer
c = (-6(+/-)√(6²-4(1)(8))/2(1)
4.) Simplify
c = (-6(+/-)√(36-(4*8))/2
c = (-6(+/-)√(36-32))/2
c = (-6(+/-)√(4))/2
c = (-6(+/-)2)/2
*Here, the equation splits in two. It becomes:
c = (-6+2)/2 AND c = (-6-2)/2
*Simplify again:
c = -4/2 AND c = -8/2
c = -2 AND c = -4
The answers c = -2 and c = -4 would solve the given equation.
Hope this helps! :)
6 0
4 years ago
Read 2 more answers
Evaluate the follow expression
STatiana [176]
I see the answers,but where is the problem (equation that needs to be solved)
4 0
3 years ago
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