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    Calculus A Complete Course
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    Exam 4: Transcendental Functions
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    Solve the Initial Value Problem + 5
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Solve the Initial Value Problem + 5

Question 126

Question 126

Multiple Choice

Solve the initial value problem Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  + 5 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  + 2y = 0; y(0) = 2, Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  (0) = -5.


A) y = 2 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  - 3t Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t
B) y = 2 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  - 5t Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t
C) y = 2 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  - 7t Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t
D) y = -2 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  - 3t Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t
E) y = 2 Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t  - 3t Solve the initial value problem   + 5   + 2y = 0; y(0)  = 2,   (0)  = -5. A)  y = 2   - 3t   B)  y = 2   - 5t   C)  y = 2   - 7t   D)  y = -2   - 3t   E)  y = 2   - 3t

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