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Sveta_85 [38]
3 years ago
10

What is 6 tenths 20 hundredths 4 ones and 50 hundreds

Mathematics
1 answer:
Elina [12.6K]3 years ago
3 0

Answer:

The answer is 25064.

Step-by-step explanation:

<u>There is a mistake in the question so the correct is below:</u>

What is 6 tenths 20 thousandths 4 ones and 50 hundreds.

Now, to solve it:

So, we arrange it in sequence:

20 thousandths 50 hundreds 6 tenths and 4 ones.

Now,

20 thousandths = 20 <u>000</u>

50 hundreds      = 50 <u>00</u>

6 tenths              = 6 <u>0</u>

4 ones                = 4

Thus, 20 thousandths 50 hundreds 6 tenths and 4 ones is :

20,000 + 5000 + 60 + 4.

= 25064.

Therefore, the answer is 25064.

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2 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
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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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Step-by-step explanation:

root 75 is also 5 root 3 (after simplifying)

when that is converted into a decimal we get

8.660254038...

Rounding to the nearest hundred of that number we focus on these two numbers

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answer:

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step-by-step explanation:

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