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\[11\div 33\]
+
−
.
ln
>
<
×
÷
/
log
≥
≤
(
)
log
x
=
%
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11#33
11#33
11\div 33
1
Use the algorithm method.
.
3
3
3
3
3
3
3
3
|
1
1
.
9
.
9
1
.
1
0
9
9
1
1
0
9
9
1
1
0
9
9
1
1
0
9
9
1
1
0
9
9
1
1
2
Therefore,
11
÷
33
≈
0.333333
11\div 33\approx0.333333
1
1
÷
3
3
≈
0
.
3
3
3
3
3
3
.
0.333333
0.333333
0
.
3
3
3
3
3
3
Done
0.33333333333333
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