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2 years ago

15

Classify the real number.

\sqrt{15} " align="absmiddle" class="latex-formula">

1 answer:

lilavasa [31]2 years ago

5 0

Answer:

<h2><em><u>Irrational</u></em><em><u> </u></em><em><u> </u></em></h2>

Step-by-step explanation:

<em><u>Firstly</u></em><em><u>, </u></em>

According to rational and irrational,

$\sqrt{15} \: is \: irrational$

<em><u>Since</u></em><em><u>,</u></em>

Natural numbers, Whole Numbers and Integers all come under <em><u>Rational</u></em><em><u> </u></em><em><u>number</u></em><em><u>.</u></em>

<em><u>Hence</u></em><em><u>,</u></em>

<em><u> $\sqrt{15}$ </u></em>

<em><u>Is</u></em><em><u> </u></em><em><u>an</u></em><em><u> </u></em><em><u>irrational</u></em><em><u> </u></em><em><u>number</u></em><em><u>. </u></em>

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<img src="https://tex.z-dn.net/?f=f%28n%29%3D3%5En" id="TexFormula1" title="f(n)=3^n" alt="f(n)=3^n" align="absmiddle" class="la

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$24(f(n))$

Step-by-step explanation:

$f(n+3) = 3^{n+3}\\f(n+1) = 3^{n+1}$

so

$f(n+3) - f(n+1) = 3^{n+3} - 3^{n+1}$

$= 27(3^n) - 3(3^n)$

$= 24(3^n)$

$= 24(f(n))$

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