Calculus Archive: Questions from April 10, 2022
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-1 y = x sin-(x) + V1 – 22 Select the correct response: -1 - y' = sin(x) y' = 0 y' = 1 y = -x sin-|(x)1 answer -
Use the Chain Rule to find dz/dt z = x2 + y2 + xy, x = sin t, y = e' cos(x + 4y), x = 514, y = 1/1 z = V1 + x2 + y2, x = ln t, y = cos t1 answer -
3 answers
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evaluate the integral 48, 52 plz
48. sec’y cos"(tan y) dy ) 50. V1 – cos 40 do *** Vi | * 52. x sec x tan x dx1 answer -
1 answer
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1 answer
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Help With #2 Please
2. y = 1-54 Use the guidelines of this section to sketch the curve. 1. y = x3 + 3x2 2x 12x2 + 18x 3. y = x4 – 4x 4. y = x+ – 8x2 + 8 5. y = x(x – 4)3 6. y = x3 – 5x 7. y = 5x5 – 3x3 + 16x 8.1 answer -
Help with #4 Please
2. y = 1-54 Use the guidelines of this section to sketch the curve. 1. y = x3 + 3x2 2x 12x2 + 18x 3. y = x4 – 4x 4. y = x+ – 8x2 + 8 5. y = x(x – 4)3 6. y = x3 – 5x 7. y = 5x5 – 3x3 + 16x 8.1 answer -
Help with #16 Please
2. y = 1-54 Use the guidelines of this section to sketch the curve. 1. y = x3 + 3x2 2x 12x2 + 18x 3. y = x4 – 4x 4. y = x+ – 8x2 + 8 5. y = x(x – 4)3 6. y = x3 – 5x 7. y = 5x5 – 3x3 + 16x 8.1 answer -
Help with #20 Please
2. y = 1-54 Use the guidelines of this section to sketch the curve. 1. y = x3 + 3x2 2x 12x2 + 18x 3. y = x4 – 4x 4. y = x+ – 8x2 + 8 5. y = x(x – 4)3 6. y = x3 – 5x 7. y = 5x5 – 3x3 + 16x 8.1 answer -
0 answers
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2,3,7
- 6 Solving Initial Value Problems. Solve the initial problems in Exercise Group 6.2.5.1-8 using the Laplace transform. 1. " – 2y - 3y 3e2, y() - 1.7(0) = 0 2. y" -y-2y = 422. y(0) = -1,7(0) = 1 du1 answer -
Evaluate SS | Fax f(x, y, z) DV for the function f and region W specified. f(x, y, z) = ex + y + z; W: 0 SXS 3,0 s y sx, 0 SZ S5 X1 answer -
1 answer
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Find all the second partial derivatives. f(x, y) = x6y8 + 7x8y fxx(x, y) = fxy(x, y) fyx(x, y) = fyy(x, y) = Submit Answer1 answer -
0 answers
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Find all the second partial derivatives. f(x, y) = x7y6 + 6x4y fxx(x, y) = fxy(x, y) = fyx(x, y) = fyy(x, y) =1 answer -
I need #8
7.5 EXERCISES In Problems 1-14, solve the given initial value problem using the method of Laplace transforms. 1. y" – 2y' + 5y = 0; y(0) = 2, y'(0) = 4 - = 2. y" - y' – 2y = 0; y(0) = -2, y'(0) =1 answer -
Situation: An object is launched from a mound with a height of 4 feet at a speed of 48 feet per second. A) determine the position function that describes the displacement of the object. B)
Instrucciones: Analalice los datos provistos por el siguiente ejercicio para determinar la función posición (1) = -g1² de un objeto en caida libre. Luego presente todo el proceso que le llevara a l0 answers -
Evaluate the triple integral. SITE y dv, where E = {(x, y, z)| 0 5 X 5 7,0 s y s x,x - y s zsx+y} (sxs1 answer -
45-56 Use logarithmic differentiation to find the derivative of the function. 45. y = (x2 + 2)²(x4 + 4) + e-*cos²x 46. y = x' + x + 1 х 47 y 48. y= Vxex-*(x + 1)2/3 x+ + 1 49. y = x* 51. y =rsin. =1 answer