B
3. In DABCD, side BC < side AD
side BC || side AD and if
side BA = side CD
then prove that ZABC = ZDCB.
A
Answers
Answer:
Question -
({(729)}^{ \frac{ - 5}{3} } ) ^{ \frac{ - 1}{2} } = {(729)}^{ \frac{ - 5}{3} \times \frac{ - 1}{2} }((729)
3
−5
)
2
−1
=(729)
3
−5
×
2
−1
Answer -
LHS -
({(729)}^{ \frac{ - 5}{3} })^{ \frac{ - 1}{2} }((729)
3
−5
)
2
−1
we \: know \: ({a}^{m} ) ^{n} = {a}^{mn}weknow(a
m
)
n
=a
mn
(729) ^{ \frac{5}{6} }(729)
6
5
729 = {3}^{6}729=3
6
({3}^{6}) ^{ \frac{5}{6} }(3
6
)
6
5
{3}^{5} = 2433
5
=243
_____________________________________
RHS -
(729) ^{ \frac{ - 5}{3} \times \frac{ - 1}{2} }(729)
3
−5
×
2
−1
(729) ^{ \frac{5}{6} }(729)
6
5
729 = {3}^{6}729=3
6
( {3}^{6} ) ^{ \frac{5}{6} }(3
6
)
6
5
{3}^{5} = 2433
5
=243
Hence :
Verified : LHS = RHS .
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