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Firlakuza [10]
3 years ago
14

Which is the solution set of a for -3(-x a) 7<5 when x=-1 ?

Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer: CAN I DO YOU PLS

Step-by-step explanation:

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Heather has a bag containing 4 marbles. 1 red, 1 blue, and 2 green. She draws 1 marble out of the bag, replaces it. And then dra
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Answer: 32

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3 years ago
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EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If
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$,

We have lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0

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
Hello can someone do this please
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Answer:

y=-2x+4

Step-by-step explanation:

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What is the solution to this inequality?
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