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arsen [322]
3 years ago
15

Please give me four more

Mathematics
1 answer:
Harman [31]3 years ago
6 0
Just factor 48  factors. It has many.
2 * 2 * 12
3 * 2 *  8
4 * 4 * 3
1 * 2 * 24

And the factors given above can have the height and the smallest of the other factors interchanged.

So 3 * 8 * 2 is not the same thing as 3 * 2 * 8 although your teacher may argue that they are the same. All you have to do is put a different face on the bottom to make the height and width different.

Other teachers would say they were different. I have given you what you requested, so I think you are done with this question.
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To \; find \; the \; maximum \; number \; of \; items \; possible \; in each \; packet,\\we \; need \; to \; find \; the \; Greatest \; Common \; Factor\\ \; of \; 527 \; Pencils, \; 646 \; erasers \; and \; 748 \; sharpeners.\\\\We \; need \; to \; list \; the \; Factors \; of \; each \; number \; as \; given \; below\\ \; and \; identify \; the \; greatest \; common \; factor.

The \; factors \; of \; 527 \; are:\\1, \; 17, \; 31, \; 527\\\\The \; factors \; of \; 646 \; are:\\1, \; 2, \; 17, \; 19, \; 34, \; 38, \; 323, \; 646\\\\The \; factors \; of \; 748 \; are:\\1, \; 2, \; 4, \; 11, \; 17, \; 22, \; 34, \; 44, \; 68, \; 187, \; 374, \; 748\\\\Then \; the \; greatest \; common \; factor \; is \; 17.

Conclusion: \\There \; will \; 31 \; Pencils, \; 38 \; Erasers,\\ \; and \; 44 \; Sharpeners \; in \; each \; of \; SEVENTEEN \; packet.

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Step-by-step explanation:

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