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seropon [69]
3 years ago
5

What is the modulus of the complex number z=e^itheta - 1 ?

Mathematics
1 answer:
timama [110]3 years ago
8 0

Answer:

eahti-1

Step-by-step explanation:

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72

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The next two numbers are 0.24 0.36 Your just adding 0.06. 
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Henry's family members choose numbers to assign
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Help me with these please​
olchik [2.2K]

Step-by-step explanation:

(1) y = x e^(x²)

Take derivative with respect to x:

dy/dx = x (e^(x²) 2x) + e^(x²)

dy/dx = 2x² e^(x²) + e^(x²)

dy/dx = (2x² + 1) e^(x²)

Take derivative with respect to x again:

d²y/dx² = (2x² + 1) (e^(x²) 2x) + (4x) e^(x²)

d²y/dx² = (4x³ + 2x) e^(x²) + 4x e^(x²)

d²y/dx² = (4x³ + 6x) e^(x²)

Substitute:

d²y/dx² − 2x dy/dx − 4y

= (4x³ + 6x) e^(x²) − 2x (2x² + 1) e^(x²) − 4x e^(x²)

= 4x³ + 6x − 2x (2x² + 1) − 4x

= 4x³ + 6x − 4x³ − 2x − 4x

= 0

(2) y = sin⁻¹(√x)

sin y = √x

sin²y = x

Take derivative with respect to x:

2 sin y cos y dy/dx = 1

sin(2y) dy/dx = 1

dy/dx = csc(2y)

Take derivative with respect to x again:

d²y/dx² = -csc(2y) cot(2y) 2 dy/dx

d²y/dx² = -2 csc²(2y) cot(2y)

Substitute:

2x (1 − x) d²y/dx² + (1 − 2x) dy/dx

= 2 sin²y (1 − sin²y) (-2 csc²(2y) cot(2y)) + (1 − 2 sin²y) csc(2y)

Use power reduction formula:

= (1 − cos(2y)) (1 − ½ (1 − cos(2y))) (-2 csc²(2y) cot(2y)) + (1 − (1 − cos(2y))) csc(2y)

= (1 − cos(2y)) (1 − ½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cos(2y) csc(2y)

= (1 − cos(2y)) (½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cot(2y)

= (cos(2y) − 1) (1 + cos(2y)) csc²(2y) cot(2y) + cot(2y)

= (cos²(2y) − 1) csc²(2y) cot(2y) + cot(2y)

= -sin²(2y) csc²(2y) cot(2y) + cot(2y)

= -cot(2y) + cot(2y)

= 0

8 0
3 years ago
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