What's is 24/60 as a decimal

Mathematics

1 answer:

Levart [38]2 years ago

6 0

Answer:

0.4

Step-by-step explanation:

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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:

5 0

2 years ago

Nasser invested $550 at 6% simple interest for 5 years. What was his investment worth after 5

kirill [66]

Answer:

B. $715

Step-by-step explanation:

By multiplying your starting value, in this case $550, and your (simple) interest, 6%, or by 1.06 as 0.06 being your interest value and the 1.00 accommodating your starting value you will multiply, $550 × 1.06 = $583. Repeat these steps for however many years you are account for, in this case, five years. Giving you a total of $715

4 0

3 years ago

Can any one help me i really need it

Vilka [71]

34 try the app Cymath it’s great for math

5 0

2 years ago

Read 2 more answers

CORRECT ANSWER PLEASE!!!!! 150 points

juin [17]

Answer: -7

Step-by-step explanation: