The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:
Since, c = 5 so,
Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
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You basically just have to do . . .
257/54
4.759 . . .
≈ 4.76 hours
Answer: 4.76 hours
Answer:
Step-by-step explanation:
The GCF of these monomials is : u^3v^2
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Formula
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a² + b² = c²
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Find the missing leg.
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15² + b² = 18²
b² = 18² - 15²
b² = 99
b = √99
b = 9.95 cm
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Answer: 9.95 cm
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I hope this helps you
40=8.5=5.2^3
24=8.3=3.2^3
