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2 years ago

-6 divided by 2/3 Help me

Mathematics

1 answer:

Lina20 [59]2 years ago

8 0

Its -9
because in your calculator you would put -6/ by 2/3

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use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r

Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

If we ask that $f$ has continuous $(n+1)$ th derivative we can apply the mean value theorem for integrals. Then, there exists $c$ between $a$ and $x$ such that

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x$

Hence,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .$

Thus,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

and the Taylor theorem with Lagrange remainder is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ .

Step-by-step explanation:

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3 years ago

What value of x will make the equation below true?<br> 14−(x+7)=3

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14-x-7=3
7-x=3
-x=-4
x=4

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2 years ago

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r-ruslan [8.4K]

Answer:

(0,1)

Step-by-step explanation:

I put it in desmos :)

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2 years ago

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Determine whether quantities vary directly or inversely and find the constant of variation. A teacher grades 25 students essays

Lesechka [4]

IDK the constant and quantities...
First, find how many hours it takes for him to grade 1 essay, 0.16. Then multiply that by 35 to get 5.6.
I hope this is right -_-'

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2 years ago

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2789 dimensions because it’s simple u multiple them

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