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alexandr1967 [171]

3 years ago

15

WILL MARK YOU BRAINLIEST

1 answer:

Shalnov [3]3 years ago

4 0

Answer:

the answer is 3/2

Step-by-step explanation:

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Can you ever use a calculator to determine if a number is rational or irrational

Tcecarenko [31]

By definition and according to Google:

"Irrational numbers<span> cannot be represented as terminating or repeating decimals."

So the answer to your question would have to be 'no' as calculators were designed to give us approximate answers when dealing with irrational numbers.

It would be hard to guess whether a large number produced by a calculator is irrational or not given that fact that many rational numbers can be incredibly long.
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4 years ago

Convert the decimal number to the base stated<br><br> 113 to base 6

Oduvanchick [21]

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mafiozo [28]

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4 years ago

I love you man<br><br><br> SAY IT BACK<br><br> SAY IT BAAACK

iren2701 [21]

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Step-by-step explanation:

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3 years ago

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Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1]

Alekssandra [29.7K]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

3 years ago