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ddd [48]
2 years ago
7

1

Mathematics
1 answer:
Artemon [7]2 years ago
3 0

Answer:

x = 4

Step-by-step explanation:

2x − y = 11      ⇒    -y = - 2x +11     ⇒      y = 2x - 11

x + 3y = -5

x + 3(2x - 11) = -5

x + 6x - 33 = -5

      +33       +33  

7x = 28

÷7     ÷7

x = 4

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Write the slope intercept form that satisfies the following conditions
Olin [163]

y = 3x - 13

Step-by-step explanation:

The line will have the same slope as the given line, which is m = 3. Therefore, the equation of the line in its slope-intercept form is

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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
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
3 years ago
James and Tyson are buds. The sum of their ages is 97. The difference of their ages is 19. Find their ages.
sweet [91]

Answer:

James is 58 and Tyson is 39 years old.

Step-by-step explanation:

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x - y = 19    +

--------------

2x = 116

x = 58

--------------

y = 97 - 58

y = 39

7 0
2 years ago
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