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3 years ago

9

Approximate the distance between (0, -5) and (4, 1) to the nearest tenth, if necessary.

1 answer:

frutty [35]3 years ago

8 0

Answer:

7.7

Step-by-step explanation:

To solve this question we must use Pythagoras (a²+b²=c²) Let those two dots be A and B respectively.Imagine a line is drawn that pases through both dots,

Imagine that the line given is the hypotenuse (the longest side of a right-angled triangle) of a right-angled triangle and the other sides as a line from A going to the right, and a line from B going downwards, making the final vertexes for the triangle at coordinates: 0,-5 4,1 and 4,-5 (the vertex we made). now we find the lengths for the sides: 4 and 6 so we use these to plug them into pythagoras's equation:

4²+6²=c²

24+36=c²

√60=c

c=7.74....

But all we need is until the tenths so the answer is 7.7 (as the hundredths digit, 4, rounds down)

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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.

6 0

3 years ago

after 2/5 of the kittens were given away, there was k kittens and p puppies left in the pet store. write and algebraic expressio

Rufina [12.5K]

Answer:

Algebraic expression in terms of k

k= T- p

Step-by-step explanation:

As

no of kittens left = k

no of puppies left =p

If total left after giving away kittens =T

T = p + k

In terms of k the expression will be

k= T- p

Keywords: Algebra

Learn more about algebra at:

#learnwithBrainly

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