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PSYCHO15rus [73]

3 years ago

12

Prove that n^2 > 2n + 1 for all natural numbers n >= 3

1 answer:

mafiozo [28]3 years ago

7 0

${n}^{2} - 2n - 1 > 0$
$n = 3 > > 9 - 6 - 1 = 2 > 0$
suppose the following is true

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1252 divided by 48 equals 26 r 4

Ksenya-84 [330]

I assume that in this question, you are asked to evaluate the value(s) of r. To answer that question, dividing 1252 by 48 gives us 26.0833 or 26. Mathematically expressing the solution,
26 = 26r^4
Then, divide the equation by 26, giving us 1 = r^4. Evaluate r. This will give us answers of +-1.

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4 years ago

It took my question down twice here's a real math question I guess

aliina [53]

Answer:

$n-3$

Interval notation: $(-\infty, -10)\cup(-3,\infty)$

Step-by-step explanation:

<u>First inequality:</u>

<u /> $n+8$

Therefore, this inequality restricts:

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<u>Second inequality:</u>

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Therefore, with both of these restrictions together, we have:

$\fbox{$n \in \mathrm{R}; n-3$}\\\mathrm{or\:}\fbox{$n-3$}\\\mathrm{or\:}\fbox{$(-\infty, -10)\cup(-3,\infty)$}$ .

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1 + 5 = 6
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I hope this helps you

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