\[ y(n)=3 x(n-1)+\frac{1}{x(n)}+4 x(n) \] Checu for (1) Unearty (2) Static (3) Time Variant (4) Camal
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The given system is: \[ y(n) = 3x(n-1) + \frac{1}{x(n)} + 4x(n) \] - **Superposition**: If \( x_1(n) \) and \( x_2(n) \) are inputs, then for linearity, \( y(n) \) should satisfy: \[ y(n) = 3(x_1(n-1) + x_2(n-1)) + \frac{1}{x_1(n) + x_2(n)} + 4(x_1(n) + Show more…
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Find the value of x if f(x) = 4x^3 - 26x^2 + 6x - 8. To find the value of x, we need to solve the equation f(x) = 0. 4x^3 - 26x^2 + 6x - 8 = 0 We can factor out a common factor of 2 from each term: 2(2x^3 - 13x^2 + 3x - 4) = 0 Now we can use synthetic division to find the roots of the equation. By trying different values of x, we find that x = 2 is a root of the equation. Using synthetic division, we divide 2x^3 - 13x^2 + 3x - 4 by (x - 2): 2 | 2 -13 3 -4 - 4 -18 -30 -------------- 2 -17 -15 -34 The result is 2x^2 - 17x - 15 - 34/(x - 2). Now we can factor the quadratic equation 2x^2 - 17x - 15 = 0: (2x + 3)(x - 5) = 0 Setting each factor equal to zero, we find two possible values for x: 2x + 3 = 0 --> x = -3/2 x - 5 = 0 --> x = 5 Therefore, the values of x that satisfy the equation f(x) = 0 are x = -3/2 and x = 5.
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dy/dx - 2y/x = x^-1y^-1 , y(1) = 3
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