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MrRa [10]
3 years ago
12

The sum of 2 consecutive integers divided by 4 is 189.5

Mathematics
1 answer:
Arada [10]3 years ago
4 0

Answer:

That is <em>not possible</em>. The sum of consecutive integers must be odd, so the quotient will end in 0.25 or 0.75.

Step-by-step explanation:

Let n and n+1 represent two consecutive integers. Their sum will be ...

... sum = n + (n+1) = 2n+1

Dividing by 4 gives

... sum/4 = (2n+1)/4 = (1/2)n + 1/4

The term (1/2)n will have a decimal fraction part of either 0.0 or 0.5, depending on whether the first integer (n) is even or odd (respectively). Adding 1/4 to that will give a decimal fraction part of either 0.25 or 0.75.

The quotient given in the problem statement ends in 0.5, which does not match either 0.25 or 0.75.

_____

The sum of the <em>consecutive </em><u><em>even</em></u><em> integers</em> 378 and 380 will have a quotient of 189.5 when divided by 4.

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\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



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\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

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Add the answer from Step 1 to the numerator

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Step 3

Write answer from Step 2 over the denominator

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