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elixir [45]
2 years ago
15

Whats greater brianliest + MAX ponits or 4 x 2160 4 x 2610

Mathematics
2 answers:
astra-53 [7]2 years ago
5 0

It is equal. The same either way. :)

Paraphin [41]2 years ago
4 0

Hey there!

PROBLEM #1. 4 * 2,160

4 * 2,160
= 2,160 * 4

= 2,160 + 2,160 + 2,160 + 2,160

= 4,320 + 4,320

= 8,640
Therefore, the answer should be: 8,640

Problem #2. 4 * 2,610
4 * 2,610
= 2,610 * 4

= 2,610 + 2,610 + 2,610 + 2,610

= 5,220 + 5,220

= 10,440

Therefore, the answer should be:

10,440


Good luck on your assignment & enjoy your day!



~Amphitrite1040:)

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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
3 years ago
The square root of 12 is irrational.<br><br> True<br> False
dalvyx [7]
True.

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This continues indefinitely and can not be represented as a ratio of two integers.

So it is not rational, that is irrational.
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3 years ago
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dlinn [17]

Answer:

0, 3

Step-by-step explanation:

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3 years ago
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If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.
zmey [24]

Answer:

a=3 and b=5.

Step-by-step explanation:

So I believe the problem is this:

\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}

where we are asked to find values for a and b such that the equation holds for any x in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)

2(4x+3)=a(x+7)+b(x-3)

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

2(4\cdot -7+3)=a(-7+7)+b(-7-3)

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-50=-10b

Divide both sides by -10:

\frac{-50}{-10}=b

5=b

Now we have:

2(4x+3)=a(x+7)+b(x-3) with b=5

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

2(4\cdot 3+3)=a(3+7)+b(3-3)

2(12+3)=a(10)+b(0)

2(15)=10a+0

30=10a

Divide both sides by 10:

\frac{30}{10}=a

3=a

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mario62 [17]

Answer:

Absolute error = 39

Percent Error = 6.10%

Step-by-step explanation:

Actual value = 639 pounds

Observed Value = 600 pounds

The formula to find absolute error is:

Absolute Error = Actual value - Observed Value

                        = 639 - 600

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The formula to find Percent error is:

Percent Error = (Actual value - Observed Value / Actual value)*100

                       = (639 - 600/639)*100

                       = 6.10%

                 

7 0
3 years ago
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