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Stoichiometry: C3H8  +  5O2  -->  3 CO2  +  4 H2O PV  =  nRT ,  V = nRT/P, (38.8)*10^3 g  C3H8  =  879.9 mole C3H8 reacts with                                 5*(879.9) mole O2  = 4399.7 mole O2                     V =  [4399.7 mole O2 * (.082 L atm/mole °K) * (293 °K)]/1.01...

# Find the Laplace transform? (answer)

f(t) = e^-(t-x)  sin x dx, Using Euler’s identity, e^im = cos m + i sin m, integrate using the exponential, and take the imaginary part (Im) after integration. f(t) = e^-(t-x) * (e^ix) dx = e^(-t + x + ix) = e^-t * (e^(1 + i)x) dx int f(t) = int [0, t] e^-t * (e^(1 + i)x) dx = e^-t...

# Find the inverse Laplace transform? (answer)

e^(-10s)/(s(s^2+3s+2)) Factor out e^(-10s) for partial fraction decomposition. 1/(s(s^2+3s+2))  =   1/ s(s+1)(s+2)  =   A/s  +  B/(s+1)  +  C/(s+2) Using the Heaviside coverup method. when s = 0,   A= 1/ (  )(0 + 1)(0 + 2)  =  1/2 when s = -1, B= 1/(-1)(  )(-1 + 2)  =  -1 When...

# Find the inverse Laplace transform? (answer)

Another method, that involves less writing is the Oliver Heaviside coverup method.  Google it and it's well explained.  Just another instrument in your armamentarium for solving partial fraction decomposition.

# Solve the differential equation? (answer)

dx/dy  +    2x  -   (1/y)e^(-2y)  =  0 dx/dy  +   2x   =   (1/y)e^(-2y) x'      +     2x   =   (1/y)e^(-2y) I.F. = e^2y x  =   (e^-2y) ∫ dy/y x  =   (e^-2y) [ln y  +  c] The trick is to integrate wrt y rather than x

# complex numbers equation derivation (answer)

The absolute value, or modulus of the number z = a + bi is defined by |z| = √(a² + b²) |z1 + z2|  = |(a1 + a2) + (b1 + b2)| = √[(a1 + a2)² + (b1 + b2)²] |z1 + z2|² = (a1 + a2)² + (b1 + b2)²

y'+(1/2)y=2cos(t) vy' + 1/2 vy  =  2 v cos t                 Let  1/2  v  = dv/dt = v' (vy)'  =   2 v cos t                                       1/2 dt   = dv/v vy     =  2 int v cos t dt                             1/2 t  +  c   = ln v  y  =  (1/v)* 2  int v cos t dt...

# Complete the identity: csc(t)(sin(t) + cos(t)) = ? (answer)

You have a typo here.

x-z=arctan(yz) 1 - (dz/dx) =  (y(dz/dx)/(1 + (yz)^2) 1  =  (dz/dx)((1 +  y/(1 + (yz)^2))) dz/dx  =  (1 + (yz)^2)/(1 + y + (yz)^2)

# Find the area of the domain D which is in the first quadrant? (answer)

You must paramaterize.  It is easiest if you draw a picture.   [(1/3)^(1/2), 1 ] int (3x - 1/x)dx  +  [1, (3)^(1/2)] int (3/x  -  x)dx   (3/2 x^2 - ln x) [(1/3)^(1/2), 1 ]  +  (3 ln x  -  (x^2/2) [1, (3)^(1/2)]   Performing the arithmetic gives ln 3   A = ln 3

sin x = (e^ix – e^-ix)/2i Let sin x = -4 + sqrt2/2 = k = (e^ix – e^-ix)/2i 2ik = (e^ix – e^-ix) (e^ix – e^-ix) –2ik = 0 e^2ix – 2ike^ix –1 = 0 Let u = e^ix u^2 – 2iku – 1 = 0, u = (2ik ± (((-2ik)^2 – 4(1)(-1))^(1/2))/2 (-2.586i ± 1.64)/2 u=  e^ix   ≈  -.4735i, -2...

f = ye^x(x)+cos(x).             Subscripts are  tedious.  Let f(xn) be the nth derivative wrt  x. f(x1) = y(xe^x + e^x) - sin x f(x2) = y(xe^x + 2e^x) - cos x f(x3) = y(xe^x + 3e^x) + sin x f(x3)(y1) = (xe^x + 3e^x) f(x3)(y2) = 0 All derivatives after this are zero....

# Find the normal vector to the plane? (answer)

3 vectors defining the plane are ,<2, 0, 0>, <0, 3, 0>, <0, 0, 1> x1 - x2= <2, -3, 0> x1 - x3= <2, 0, -1>  i         j          k 2        -3         0                    = 3i + 2j + 6k 2         0        -1

# A tepee with a dirt floor in the shape of a right cone has a slant height of (answer)

dA = 2 pi r dh, the limits of integration in terms of h [0, 22.8]      =  2 pi (12.5 - (12.5/22.8)h) dh  A   = 2 pi (12.5h  - (1/2)(12.5/22.8)h^2)   [0, 22.8]       = 2pi (12.5)(22.8)(1 - 1/2)       = 285 pi ft^2

# ind all solutions for Sin^2x - 2sinx - 3 = 0 on the interval [0 , 2pi ) (answer)

This will factor: (sin x - 3)(sin x + 1),  sin x = -1, x = 3pi/2 The solution sin x = 3 has no real solution.

# Use Stokes' Theorem to evaluate the surface integral? (answer)

F= From the paramaterization,  F(r(t)),  <  0,  -cos t,  (cos t) ln (sin t)^2>, (everything multiplied by z vanishes) r'(t) = i  (- sin t) dt   +  j  (cos t) dt + k (0) Int [ F(r(t))*r'(t)]  dt  = Int [ i (0)(- sin t)  +  j (-cos t)(cos t)  +  k (cos t)...

(h - k)/a  =  b^(cx + d) ln ((h - k)/a) =  (cx + d) ln b (ln ((h - k)/a)/ln b) = cx + d  (ln ((h - k)/a)/ln b)  -  d  =  cx  ((ln ((h - k)/a)/ln b) - d)/c  =  x  (((ln ((h - k)/a)))/ln b) - d)/c  =  x