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MrRissso [65]
2 years ago
6

Marj needs to round $345.67$, and she wants the result to be as large as possible. She has to round to the nearest thousand, the

nearest hundred, the nearest ten, the nearest integer, the nearest tenth, or the nearest hundredth. What is the largest possible rounded value she can attain?
Mathematics
1 answer:
Andre45 [30]2 years ago
4 0

Answer:

$346.00

Step-by-step explanation:

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d1i1m1o1n [39]

Answer:

∠A = 88°

∠B = 92°

∠C = 88°

∠D = 92°

Step-by-step explanation:

∠A + ∠B = 180°

(2x + 4) + (3x - 34) = 180

reduce:

5x - 30 = 180

5x = 210

x = 42

∠A = 2(42) + 4 = 88°

∠B = 3(42) -34 = 92°

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∠D = ∠B = 92°

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3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

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:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

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

brainly.com/question/28196765

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1 year ago
Do you guys know what is after you become an expert??
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Answer:

Ace is the next level after expert.

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