Attempt 1: 6 attempts remaining.x+4y+2z4x=9-3y + z ==-3832-3x + y - 4zWrite the augmented matrix for the above system of equations and use the rref() function inOctave/MatLab to find the unique solution to this system of equations. Give your answer as a point(x,y,z): -I--I--x12x2 3x3-2x12x2 - 2x3=3= -3=-3(a) Find the reduced row echelon form of the augmented matrix for this system usingOctave/MatLab.Consider the system-2x1 + 3x2+2x3I-

Step 1: Consider the following system of linear equations. ⎩⎪⎪⎨⎪⎪⎧−2x1+3x2+2x3= 3 −x1−…

Answered: Attempt 1: 6 attempts remaining.…

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.55P

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Instructions: USE KMAPS or LC to reduce the equations and or Generate the equations by inspection using SOPs'.
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Below is a circuit diagram. Create a set of equations by using mesh analysis. The value of Rnsid is equal to 625. Express the final answer as rectangular coordinates.
Find a system of differential equations and initial conditions for the currents in the network given in the schematic diagram. Assume that all initial currents are zero. Solve for the currents in each branch of the network. What is the system of equations? Is A. O B. O C. N dI₂ 4-41₁-8=0, 1₁ + 1₂ + q' = 0 and -2- dt 1₁ (0)=0, 1₂ (0)=0, 13(0) = 0 dl₁ dt I₁ (0) = 0, 1₂ (0) = 0, 13(0) = 0 dl₁ dq dt dt 1₁ (0)=0, 1₂ (0)=0, 13 (0) = 0 dq where q is the charge on the capacitor. dt dI₂ 4+41₁ +8=0, I₂ + dt I₁ (0)=0, 1₂ (0) = 0, 13 (0) = 0 dq 4-4- -81₂ = 0, −1₁+1₂ + q' = = 0 and -2 - 12q + 8I₂ = 0 dt dq 4+4- +81₂ = 0, I₂ + +₁ = 0 and 2 dt dq 13 (Type exact answers.) What is the current in each branch of the network? 4₁ = | 1₂ = dq dt +1₁ = 0 and 2- dq dt dI₂ -=0 dt - 12q+8- +12q +81₂ = 0 +12q +8I₂ = 0 4 V 492 ww 1₁ 8 H ell 1₂ 292 www 13
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for the given circuit find p & q for the load as shown
Hello, I am confused on this question, the answer is attached as well but I am having difficulty with converting the complex numbers and the reasoning behind these steps. Thanks.
The subplot function in MATLAB/OCTAVE is used to O put a title for the plot. O put label (information) to an x-axis. O draw data in continuous format. O none of them. To enter data in a matrix form in MATLAB/OCTAVE, each row should be ended by O comma(,) O space O colon O semi colon() The output data of MATLAB/OCTAVE program is displayed in window. O edit O figure O command O Graphic
For the circuit below, fill in the values for the mesh equations. Between the vaiables there will be a space for a + or - sign but the first entry or the voltage value will have to be entered with a negative sign if necessary, just enter the number if it's positive. Just enter the values for the mesh equations. Notice that there is a space for the + or - sign between 1 & 12 and 12 & 13. 22k2 4.7k2 10k2 1.0kQ 47k2 25V 15V kO*12 kOʻ13 = kO*11 kO"12 kO*13 = kO 12 kOʻ13 =
can you explain to me why the convolution of these two sinc fucntions which are rect functions equal a rect function? I thought when you convolute a rectangle and a rectangle with two different lengths, it makes a trapezoid shape. The answer says it is rectangle funtion so my code is corrcect but i dont understand why this is true. I need a step by step solution and explanation. here is the problem: Consider the signal x(t) = 2 sinc(6t + 3).the output y (t) = (x *h )(t) when x(t) ispassed through a filter with impulse response h (t) = 4sinc(2t -1). the matlab code was just to help me visuallize
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