[R:75r]

258 De canna Provincie⁶⁹¹

Canna Provincie, que est palmi 8, venditur pro libris \({\phantom{1}7~~\phantom{1}5 \over 12~~20}\) 3, et queratur

787 £ pal. \({\phantom{1}7~~\phantom{1}5 \over 12~~20}\) 3 8 pensa est 4 per 7 139 \({1~~4~~7~~\phantom{1}7~~11 \over 4~~8~~9~~12~~20}\) 1 \({1 \over 9} {3 \over 4} 3\)

⁶⁹² quantum valeant palmi \({1 \over 9}\) \({3 \over 4}\) 3. Describes questionem, et multiplicabis \({1 \over 9}\) \({3 \over 4}\) 3 per \({\phantom{1}7~~\phantom{1}5 \over 12~~20}\) 3 et divides per 8, ordinans \({\phantom{1}1~~\phantom{1}0 \over 12~~20}\) in capite virgule divisionis, ideo quia locus in quo ponenda est summa est sub libris, scilicet sub \({\phantom{1}7~~\phantom{1}5 \over 12~~20}\) 3⁶⁹³: exibit libra⁶⁹⁴ \({1~~4~~7~~\phantom{1}7~~11 \over 4~~8~~9~~12~~20}\) 1, ut in questione ostenditur⁶⁹⁵. 259 Et scias quia quot soldos valuerit ipsa canna, tot denarios cum totidem mediis denariis valebit palmus. Verbi gratia: cum canna valet soldos 14, palmus valet denarios 14 cum totidem obulis, hoc est denarios 21.

260 ⁶⁹⁶ Item eadem canna valet libre \({11 \over 20}\) 5, et queritur quantum valeant

111 3 \({11 \over 20}\) 5 1 pensa est 6 per 7 439 \({3~~\phantom{1}9~~\phantom{1}2 \over 8~~12~~20}\) 76 \({3~~5 \over 4~~8}\) 13

canne 13 et palmi \({3 \over 4}\) 5, hoc est canne \({3~~5 \over 4~~8}\) 13, hoc est palmi \({3 \over 4}\) 109. Describes questionem sic, et multiplicabis \({11 \over 20}\) 5 per \({3~~5 \over 4~~8}\) 13 et divides per 1, hoc est quod⁶⁹⁷ multiplicabis 5 per suam virgulam; erunt 111, et 13 per suam virgulam; erunt 439, que multiplicabis per 111: erunt 48729 que divides per ruptos, scilicet per \({1~~0~~\phantom{1}0 \over 4~~8~~20}\). 261 Sed cum locus in quo ponenda est summa sit sub libris, scilicet sub \({11 \over 20}\) 5, oportet nos habere \({\phantom{1}1~~\phantom{1}0 \over 12~~20}\). Sed non possumus habere \({1 \over 12}\), quia minuit nobis \({1 \over 3}\) ex ea⁶⁹⁸. Pones 3 super 1 in questione, et multiplicabis 48729 per 3 et divides summam per eadem 3 et per \({1~~0~~\phantom{1}0 \over 4~~8~~20}\), hoc est⁶⁹⁹ per \({1~~\phantom{1}0~~\phantom{1}0 \over 8~~12~~20}\): exibunt libre \({3~~\phantom{1}9~~\phantom{1}2 \over 8~~12~~20}\) 76.