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Byju's Answer
Standard VII
Mathematics
Exponents with Unlike Bases and Same Exponent
Evaluate: 1...
Question
Evaluate:
(
17
)
2
−
(
16
)
2
A
0
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B
1
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C
33
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D
1089
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Solution
The correct option is
B
33
(
17
)
2
−
(
16
)
2
=
(
17
−
16
)
(
17
+
16
)
=
(
1
)
(
33
)
=
33
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0
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Exponents with Unlike Bases and Same Exponent
Standard VII Mathematics
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