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3 years ago

What is the difference of the polynomials? (5x - 3) - (2x + 7).

Mathematics

1 answer:

alukav5142 [94]3 years ago

7 0

Answer:

The answer is

Step-by-step explanation:

(5x - 3) - (2x + 7).

<u>Remove the brackets</u>

That's

5x - 3 - 2x - 7

<u>Group like terms and simplify</u>

That's

5x - 2x - 3 - 7

We have the final answer as

Hope this helps you

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amm1812

Answer:

To Prove: $9e^x$ is equal to the sum of its Maclaurin series.

Step-by-step explanation:

If $f(x) = 9e^x$ , then $f ^{(n + 1)(x)} =9e^x$ for all n. If d is any positive number and |x| ≤ d, then $|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.$

So Taylor's Inequality, with a = 0 and M = $9e^d$ , says that $|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.$

Notice that the same constant $M = 9e^d$ works for every value of n.

But, since $lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$$ ,

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It follows from the Squeeze Theorem that $lim_{n\to\infty} |R_n(x)|=0$ and therefore $lim_{n\to\infty} R_n(x)=0$ for all values of x.

$THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|$

By this theorem above, $9e^x$ is equal to the sum of its Maclaurin series, that is,

$9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}$ for all x.

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3 years ago

What is the difference of the polynomials? (5x - 3) - (2x + 7).​

What is the difference of the polynomials? (5x - 3) - (2x + 7).