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pantera1 [17]
3 years ago
9

One case of books weighs 35 pounds. One case of magazines weighs 23 pounds.A book store wants to ship 72 cases of books and 94 c

ases of magazines to another store. What is the total weight of the shipment
Mathematics
1 answer:
RSB [31]3 years ago
6 0
Shipment = 72 cases of books + 94 cases of magazines
1 case of books = 35 pounds
1 case of magazines = 23 pounds

Shipment = 72 × 35 + 94 × 23 pounds

72 × 35 = 2520
94 × 23 = 2162

Shipment = 2520 + 2162 pounds

Shipment = 4682 pounds
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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
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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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-Dominant- [34]

Answer:

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Step-by-step explanation:

$8 \frac{\tan^2(\theta)}{\sec(\theta)} \csc^2(\theta)=8 \csc(\theta)$

<u>Note that </u>

\tan^2(\theta)\csc^2(\theta)=\sec^2(\theta)

<u>You can confirm it: </u>

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<u>Therefore</u>

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