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Byju's Answer
Standard VII
Mathematics
Exponents with Unlike Bases and Same Exponent
Simplify: [ ...
Question
Simplify:
[
4
0
+
4
2
−
2
5
]
×
3
−
2
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Solution
=
[
4
0
+
4
2
−
2
5
]
×
3
−
2
=
[
1
+
16
−
32
]
×
1
9
=
(
17
−
32
)
×
1
9
=
−
15
×
1
9
=
−
5
3
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0
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gives
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Exponents with Unlike Bases and Same Exponent
Standard VII Mathematics
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