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Sagot :
The liter of oil tin contained
[tex] = 10 \ \frac{3}{5} [/tex]
The liter of oil used and leak
[tex] = 1 \frac{2}{7} + 7 \ \frac{1}{10} [/tex]
[tex] = \frac{9}{7} + \frac{71}{10}[/tex]
[tex] = \frac{9 \times 10}{7 \times 10} + \frac{71 \times 7}{10 \times 7} [/tex]
[tex] = \frac{90}{70} + \frac{497}{70} [/tex]
[tex] = \frac{587}{70} [/tex]
The liter of oil left in the tin
[tex] = \frac{53}{5} - \frac{587}{70} [/tex]
[tex] = \frac{53 \times 14}{5 \times 14} - \frac{587 \times 1}{70 \times 1} [/tex]
[tex] = \frac{742 - 587}{70} [/tex]
[tex] = \frac{155 \div 5}{70 \div 5} [/tex]
[tex] = \frac{31}{14} [/tex]
[tex] = 2 \frac{3}{14} [/tex]
~nightmare 5474~
Answer:
2.2 l of oil was left.
Step-by-step explanation:
[tex]Total=10\frac{3}{5} =10.6l\\Used=7\frac{1}{10} =7.1l\\Leaked=1\frac{2}{7} =1.28l\\[/tex]
Total used=7.1l+1.28l=8.38l
[tex]Left=10.6l-8.38l=2.22l[/tex]
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