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erma4kov [3.2K]
3 years ago
13

Lisa, an experienced shipping clerk can fill an order in 9 hours. Felipe a clerk needs 11 hours to do the same job. Working toge

ther how long will it take them to fill the order.
Mathematics
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer:

4  hours and  57  minutes.

Step-by-step explanation:

The least common multiple of  9  and  11  is  99 .  

In  99  hours, Sue could fill  11 orders (99 divided by 11)

Felipe could fill 9  orders ( 99  divided by 11 ), making a total of  

9 + 11 = 20  orders if they both work.

So for both of them working to fill one order would take:  

99 /20  hours.

To express in hours and minutes:

99 /20 = 80 /20 + 19 /20 = 4 +  3 ⋅ 19 /3 ⋅ 20 = 4 +  57 /60

That's  4  hours and  57  minutes, since a sixtieth of an hour is one minute.

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Write the Taylor Series for f(x) = sin(x)center
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Answer:

Taylor series of sin(x) centered at x = 0.

Step-by-step explanation:

Taylor series expansion:

\sum_{n=0}^N f^n(a)\displaystyle\frac{(x-a)^n}{n!}

Here, f(x) = sin (x) and a = 0

f(x) = \sin x, f(0) = 0\\f'(x) = \cos x, f'(0) = 1\\f''(x) = -\sin x, f''(0) = 0\\f'''(x) = -\cos x, f'''(0) = -1\\f^4(x) = \sin x, f^4(0) = 0\\f^5(x) = \cos x, f^5(0) = 0

Putting all the values and expanding, we get,

f(x) = f(a) + \displaystyle\frac{f'(a)(x-a)}{1!} + \displaystyle\frac{f''(a)(x-a)^2}{2!} + \displaystyle\frac{f'''(a)(x-a)^3}{3!} + \displaystyle\frac{f^4(a)(x-a)^4}{4!} + \displaystyle\frac{f^5(a)(x-a)^5}{5!} + ...\\\\= \sin 0 + \displaystyle\frac{x}{1!} +  \displaystyle\frac{(0)x^2}{2!} + \displaystyle\frac{(-1)x^3}{3!} + \displaystyle\frac{(0)x^4}{4!} + \displaystyle\frac{(1)x^5}{5!} + ...

Solving, we get

\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} - \displaystyle\frac{x^7}{7!} + ...\\\\\sin x = x - \displaystyle\frac{x^3}{6} + \displaystyle\frac{x^5}{120} - \displaystyle\frac{x^7}{5040} + ...

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