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What is the solution set of the inequality \[\left ( \frac{\pi}{2} \right )^{(x-1)^2}\leq \left ( \frac{2}{\pi} \right )^{x^2-5x-5}?\]

Note by John Marvin Macaraeg 5 years, 10 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

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(pi/2)^(x^2-2x+1) < (pi/2)^(5x+5-x^2) We apply the natural Log funtion (x^2-2x+1) * ln(pi/2) < (5x+5-x^2) * ln(pi/2) x^2-2x+1<5x+5-x^2 2x^2-7x-4<0 2(x-4)(x+1/2)<0 [-1/2,4] I forgot the equal sign

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a little more space b/w the letters will make this soln. perfect.....:)

Sorry for that, I am new to tying here :)

@Ahmed Taha – no problem :)

Thanks guys! :)

[-1/2,4] You should be able to work out how I got this. HINT: what can you do since the bases are reciprocals?

yeah...answer...[-1/2,4]...agreed...:)

is this a question in a Philippine Math Olympiad?

Yes! :) On the QS.

x<=-2

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewest(pi/2)^(x^2-2x+1) < (pi/2)^(5x+5-x^2) We apply the natural Log funtion (x^2-2x+1) * ln(pi/2) < (5x+5-x^2) * ln(pi/2) x^2-2x+1<5x+5-x^2 2x^2-7x-4<0 2(x-4)(x+1/2)<0 [-1/2,4] I forgot the equal sign

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a little more space b/w the letters will make this soln. perfect.....:)

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Sorry for that, I am new to tying here :)

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Thanks guys! :)

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[-1/2,4] You should be able to work out how I got this. HINT: what can you do since the bases are reciprocals?

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yeah...answer...[-1/2,4]...agreed...:)

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is this a question in a Philippine Math Olympiad?

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Yes! :) On the QS.

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x<=-2

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