Advanced Math Archive: Questions from November 23, 2022
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- Solve the IVP using the Laplace transform \[ y^{\prime \prime}-y=\left\{\begin{array}{lr} e^{2 t} & , 0 \leq t2 answers -
- Solve the IVP using the Laplace transform \[ y^{\prime \prime}+9 y=\left\{\begin{array}{lr} \cos (t) & , 0 \leq t2 answers -
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1. Differentiate the given functions: a) \( y=2 x^{7}+3 \cos x-3^{x}+7 \) b) \( y=x^{4} \cos x+\frac{x^{2}+3}{5 e^{x}+1} \) c) \( y=\tan \sqrt{x^{2}+9}+\frac{3}{x^{5}}+\sin ^{5} \frac{x}{7}-x^{7} \ln2 answers -
Find \( f^{\prime \prime \prime}(z) \) of \( f(z)=19 z^{3} e^{z} \) \[ f^{\prime \prime \prime}(z)= \]2 answers -
1. Simplify the following: (i) \( (a b)^{2}\left(a^{-1} b^{-1}\right)^{2} \) (ii) \( 3^{n} \times 9^{n} \) (iii) \( 5 a^{4} b^{3}\left(a^{2} b^{5}\right)^{2} \) (iv) \( 9\left(a^{6} \div a^{3}\right)2 answers -
a) Show that from the 2-point Lagrange interpolation is equivalent to the Newton interpolation, that is: where a and b are the interpolation coefficients of Newton's method b) Show that the quadra
a) Demuestre que a partir de la interpolación de Lagrange de 2 puntos es equivalente a la interpolación de Newton, es decir: \[ f(x)=\frac{x-x_{1}}{x_{0}-x_{1}} f\left(x_{0}\right)+\frac{x_{0}-x}{x_0 answers -
Given the first-order and linear ODE dx/dt =1+xt2 under the initial condition x(0) = 0 dt a) Solve the ODE and show that the solution is given by Appropriately use the method of integration of sim
4. Dada la EDO de primer orden y lineal \( \frac{d x}{d t}=1+x t^{2} \) bajo la condición incial \( x(0)=0 \) a) Resuelva la EDO y muestre que la solución está dada por \[ x(t)=e^{t^{3} / 3} \int_{2 answers -
2 answers
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Parte 3: 30 puntos Aplicaciones: Una compañia de venta de especias para cocinar determina que el costo de fabricar máquinas para empacar pimienta es dada por la función... \[ C(x)=0.001 x^{2}-0.020 answers -
Un volante cuyo momento de inercia es \( 4 \mathrm{~kg} \cdot \mathrm{m}^{2} \) está inicialmente en reposo. Si se le aplica repentinamente un par de \( 5 \mathrm{~N} \cdot \mathrm{m} \) (en sentido2 answers -
2 answers
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Solve using Cramer's Rule. 2x − z = 10 3x − y + 3z = 0 4x + 2y + 3z = −6 (x, y, z) =
Solve using Cramer's Rule. \[ \left\{\begin{aligned} 2 x-z=& 10 \\ 3 x-y+3 z=& 0 \\ 4 x+2 y+3 z=&-6 \end{aligned}\right. \]2 answers