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3 years ago

any whole number containing 3 or more digits can be ÷ by 4 if the last two digits form a number that is divisible by 4 why?

Mathematics

1 answer:

NARA [144]3 years ago

8 0

Answer:

100 is always divisible by 4. Therefore it doesn't matter which number is up front.

the two other digits are important because there are 2-digit-numbers that are not divisible by four. "14" is not divisible by four, but "24" is.

34 isn't divisible, 44 is, 54 isn't.

So both digits will need to form a number that is divisible by 4, or it will not work.

Step-by-step explanation:

n/A

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Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

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3 years ago

The triangles shown below must be congruent.

leonid [27]

These triangles are congruent by AAS condition hence the statement is true

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3 years ago

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Write a single transformation that maps ABC onto A' B' C'

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Answer:

Either of ...

(x, y) ⇒ (-x, -y)
Rotation 180° about the origin

Step-by-step explanation:

There are at least two ways to express the transformation that maps each coordinate to its opposite.

1. reflection across the origin: (x, y) ⇒ (-x,-y)

2. rotation 180° (either direction) about the origin.

Take your pick.

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3 years ago

Prove that (n+5)^2 - (n+3)^2 is a multiple of 4 for all positive integer values of n.

Drupady [299]

Answer:

(n+5)² - (n+3)² =

= (n² + 10n + 25) - (n² + 6n + 9)

= n² + 10n + 25 - n² - 6n - 9

= 4n + 16

= 4(n + 4) ⋮ 4

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3 years ago

I need help please???

Solnce55 [7]

Answer:

A(-5,-1) B(-1,1) C(3,4) D(2,-4) E(0,-6)

Step-by-step explanation:

(x,y)

x being the horizontal line (going across)

y being the vertical line (up and down)

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3 years ago

any whole number containing 3 or more digits can be ÷ by 4 if the last two digits form a number that is divisible by 4 why?​

any whole number containing 3 or more digits can be ÷ by 4 if the last two digits form a number that is divisible by 4 why?