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2 years ago

9

Guadalupe can do a job in 15 hours. If she works with lan, together they can get the job done in 10 hours. How long would it tak

e lan to do the job alone?

1 answer:

krok68 [10]2 years ago

7 0

Answer:

30 hours

Step-by-step explanation:

The equation we use is 1/15 + 1/x = 1/10. X being the amount of time it takes Ian alone. X works out to 30 hours which is the answer

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3 years ago

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(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

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a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$

$e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...$

Thus, the first three terms of the above series are:

$\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b)

The series for $e^x + 3 \ cos \ x$ is $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}$

let consider the series; $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}$

$|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty$

Thus it converges for all value of x

Let also consider the series $\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}$

It also converges for all values of x

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3 years ago