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vfiekz [6]
3 years ago
9

What is 4.5609+3.098

Mathematics
1 answer:
Nikolay [14]3 years ago
4 0

Answer:

7.6589

Step-by-step explanation:

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30t - 20=70 two step equations
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Answer:

Hi!

To solve this we can begin by adding 20 to both sides

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3 years ago
Read 2 more answers
(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
Fofino [41]

Answer:

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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Answer:

I don't know

Step-by-step explanation:

Sorry cause i have no answer too

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