Advanced Math Archive: Questions from June 04, 2022
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CORRECT ANSWER (please show steps): -π/16
47. JJJ y dv y dV, where D = {(x, y, z): x² + y² + z² ≤ 1, x ≥ 0, y ≥ 0, z ≤ 0}1 answer -
CORRECT ANSWER (please show steps): 2/3
45. J y dV, where D = {(x, y, z): x² + y² ≤ 1, x ≥ 0, y ≥ 0,0 ≤ ≈ ≤ 2}1 answer -
use matlab U. se the gradient method to find the maximum of the function f(x,y)=36y−(3x2+9y2)−24x−67 with initial point x0=(−9,−9) and λ=0.09 . (The number λ is also known as the step size
(1 point) Usa el método del gradiente para encontrar el máximo de la función f(x, y) = 36y - (3x² +9y²) - 24x - 67 con punto inicial xo = (-9,-9) y λ = 0.09. (El número X también se conoce com1 answer -
use matblat The gradient method will be used to find the minimum value of the function f(x,y)=(x2+y2−4x−18y+175)2 The iterations are started at the point (x0,y0)=(2.3,2) and λ=0.003 is used. (The
(1 point) Se utilizará el método del gradiente para hallar el valor mínimo de la función f(x, y) = (x² + y² − 4x − 18y + 175)² Se inician las iteraciones en el punto (xo, yo) = (2.3, 2) y s1 answer -
Prove that the 3rd Frenet's equation has this form, using fields dot product. This way we have Frenet's equations: T' = kN N' = -kT + \tau B B' = -\tau N Remark: Curvature gives the reciprocal of the
Asignación: Demuestre que la tercera ecuación de Frenet tiene esta forma, usando las propiedades del producto punto de los campos. De esta forma tenemos las ecuaciones de Frenet: T' = KN N' = −KT+1 answer -
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Consider the differential equation y′′ + 2y′ + 2y = f (t), such that y(0) = 0 and y′(0) = 5. The function f (t) is given by: Let Y (s) = L {y(t)}. Determine the value of Y (0.2).
1. [2 puntos] Considere la ecuación diferencial y” +2y' + 2y = f(t), tal que y(0) = 0 y y'(0) = 5. La función f(t) está dada por: f(t) 6 2 1 5 t Sea Y(s) = L {y(t)}. Determine el valor de Y (0.2)1 answer -
By solving the equation, we obtain:
2. [2 puntos] Al resolver la ecuación t A) f(t) = 1 - e⁹t 1 B) f(t) 1 + e⁹t t C) f(t) 1+e⁹t 1 D) f(t) 1 - e⁹t = = S [10 = f(u)du = t - S "f(u)du se obtiene: еди,1 answer -
Consider the integral equation: By applying the Laplace transform to both sides of the above equation, it is obtained that the nu- merator of the function F (s) is of the form (a2s2 + a1s + a0) (s2 +
3. [2 puntos] Considere la ecuación integral: -20t f(t)- 10e = 16t - -C sen(t – u)f(u)du Al aplicar la transformada de Laplace a ambos lados de la ecuación anterior, se obtiene que el nu- merador1 answer -
By applying the Convolution Theorem to compute Determine the value of a + b
a 4. [1 punto] Al aplicar el Teorema de Convolución para calcular L-¹. ¹ {x²+Q²*2²+ b} 1 /* [sen (7t - 2u) + sen(12u - 7t)] du Determine el valor de a + b. se obtiene:1 answer -
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#23 : solve in format: y(x)=c1^x + c2^x2 #5, 13 solve in format: y(x)=(c1 + c2x) e^x # 8 solve in format y(x) = e^ax(cl cosb + c2 sinbx) Thank you !
3. y" + 6y + 8.96y = 0 4. y" + 4y + (772² + 4)y = 0 5. y" + 2ny' + m²y = 0 6. 10y" 32y + 25.6y = 0 7. y" + 4.5y = 0 - 8. y"+y+3.25y = 0 9. y" + 1.8y' - 2.08y = 0 10. 100y" +240y' + (1967² 11. 4y"4y2 answers -
The gradient method will be used to find an optimal point of the function f(x,y)=(0.4x−sin(0.8y))3 The iterations are started at the point (x0,y0)=(0.4,−0.3) and λ=0.04 is used. The first iterati
(1 point) Se utilizará el método del gradiente para hallar un punto óptimo de la función f(x, y) = (0.4x - sin(0.8y))³ Se inician las iteraciones en el punto (co, yo) = (0.4,-0.3) y se ut1 answer -
(1 point) Use the gradient ascent method to optimize the function f(x,y)=28y−(3x2+19y2+6xy+18x)+14 with initial point x0=(−5,−7) and λ=0.05. The first two points of the iteration are x1=( , ) x
(1 point) Se utilizará el método del gradiente para hallar un punto óptimo de la función f(x, y) = (0.4x - sin(0.8y))³ Se inician las iteraciones en el punto (co, yo) = (0.4,-0.3) y se ut1 answer -
(1 point) The function f(x)=4(x+18)(x+12)(x−7)(x−16) has a local minimum between x=7 and x=16. Use the following algorithm to estimate that minimum. 1. Given an interval [a,b] where there is a uni
(1 point) La función f(x) = 4(x + 18)(x + 12) (x − 7)(x − 16) tiene un minimo local entre x = 7yx = 16. Usa el siguiente algoritmo para estimar dicho minimo. 1. Dado un intervalo [a, b] donde exi1 answer -
The function f(x)=−3(x+11)(x+4)(x−10)(x−20) has a local bound (maximum or minimum) between x=10 and x=20. Use the following algorithm to estimate that extremum. 1. Given three initial values x0,
(1 point) La función f(x) = −3(x+11)(x + 4)(x − 10)(x − 20) tiene un extremo (máximo o mínimo) local entre x = 10 y x = 20. Usa el siguiente algoritmo para estimar dicho extremo. 1. Dados tre1 answer -
The gradient method will be used to find a maximum point of the function f(x,y,z)=4xyz−5xy2z+4 The iterations are started at the point (x0,y0,z0)=(1.8,1.6,1.5) and λ=0.003 is used. (The number λ i
(1 point) Se utilizará el método del gradiente para hallar un punto máximo de la función f(x, y, z) = 4xyz - 5xy²z + 4 - 0.003. (El número λ también se Se inician las iteraciones en el punto (1 answer -
(1 point) The gradient method will be used to find the minimum value of the function f(x,y)=(x2+y2−20x−18y+241)2 The iterations are started at the point (x0,y0)=(2.1,2.1) and λ=0.009 is used. (Th
(1 point) Se utilizará el método del gradiente para hallar el valor mínimo de la función ƒ(x, y) = (x² + y² – 20x − 18y + 241)² Se inician las iteraciones en el punto (xo, Yo) = (2.1, 2.1)1 answer