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2 years ago

9

Are irrational number? justify the reaason !!!! 5√32

1 answer:

sleet_krkn [62]2 years ago

3 0

Answer:

The given number is irrational.

Step-by-step explanation:

It is known that $\sqrt{2}$ is irrational.

Given number can be converted as:

$5\sqrt{32} =$
$5\sqrt{16*2} =$
$5\sqrt{16}*\sqrt{2} =$
$5*4*\sqrt{2} =$
$20\sqrt{2}$

Since $\sqrt{2}$ is irrational, $20\sqrt{2}$ is also irrational (<em>the product of an irrational number and a rational number is irrational</em>), so the given number is irrational.

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Dmitrij [34]

The answer is A.
I know because sqrt(8) is between 2 and 3.

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JulsSmile [24]

$x^3=\boxed{x}\cdot x^2$ , and

$\boxed{x}(x^2+2x+1)=x^3+2x^2+x$

Subtract this from $x^3$ to get a remainder of

$x^3-(x^3+2x^2+x)=-2x^2-x$

$-2x^2=\boxed{-2}\cdot x^2$ , and

$\boxed{-2}(x^2+2x+1)=-2x^2-4x-2$

Subtract this from the previous remainder to get a new remainder of

$(-2x^2-x)-(-2x^2-4x-2)=3x+2$

$3x$ does not divide $x^2$ , so we stop here.

What we've done is to write

$\dfrac{x^3}{x^2+2x+1}=x-\dfrac{2x^2+x}{x^2+2x+1}$

then

$\dfrac{x^3}{x^2+2x+1}=x-2+\dfrac{3x+2}{x^2+2x+1}$

and we stop here because the remainder term $(3x+2)$ has a degree less than the degree of the denominator.

Alternatively, we can be a bit tricky and notice that

$x^2+2x+1=(x+1)^2$

Now,

$(x+1)^3=x^3+3x^2+3x+1$

so that

$\dfrac{x^3}{(x+1)^2}=\dfrac{(x+1)^3-(3x^2+3x+1)}{(x+1)^2}$

We can divide the first term by $(x+1)^2$ easily to get

$\dfrac{x^3}{(x+1)^2}=x+1-\dfrac{3x^2+3x+1}{(x+1)^2}$

Next,

$(x+1)^2=x^2+2x+1$

so that

$\dfrac{x^3}{(x+1)^2}=x+1-\dfrac{3((x+1)^2-(2x+1))}{(x+1)^2}-\dfrac{3x+1}{(x+1)^2}$

$\dfrac{x^3}{(x+1)^2}=x+1-3+\dfrac{6x+3}{(x+1)^2}-\dfrac{3x+1}{(x+1)^2}$

$\dfrac{x^3}{(x+1)^2}=x-2+\dfrac{3x+2}{(x+1)^2}$

which is the same result as before.

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Answer:

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Step-by-step explanation:

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image attached!

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