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Let's focus on the given equation. The C represents the cost, while the x must be the number of tacos, which is what we're solving for. Since we're given the boundary that C must be 300, then the solution would be:
300 = x² - 40x + 610
x² - 40x + 310 = 0
Apply the quadratic formula where a = 1, b = -40 and c = 310. The roots of x are:
x = 29.49≈30
x = 10.51≈11
<em>Thus, she needs to sell either 30 or 11 tacos.</em>
300 = x² - 40x + 610
x² - 40x + 310 = 0
Apply the quadratic formula where a = 1, b = -40 and c = 310. The roots of x are:
x = 29.49≈30
x = 10.51≈11
<em>Thus, she needs to sell either 30 or 11 tacos.</em>
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
Find out more information about nth taylor polynomial here
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