-
mar
People across the country observe 21 February to remember the (a)________
They organize various (b) to celebrate the day. At dawn they, putting on b
badges, attend the (c) barefoot and sing mouring songs which remind them of
(d) sacrifice of the martyrs. The shaheed Minar gets flooded with (e)
flowers.
Answers
Explanation:
\textbf{Given:} \: \frac{ {3}^{ - 5} \times {2}^{ - 6} \times {5}^{4} }{ {6}^{ - 4} \times {10}^{2} }Given:
6
−4
×10
2
3
−5
×2
−6
×5
4
\textbf{To Find: } \text{Simplify}To Find: Simplify
\fbox{Solve}
Solve
= \frac{ {3}^{ - 5} \times {2}^{ - 6} \times {5}^{4} }{ {6}^{ - 4} \times {10}^{2} }=
6
−4
×10
2
3
−5
×2
−6
×5
4
= \frac{ {3}^{ - 5} \times {2}^{ - 6} \times {5}^{4} }{(2 \times 3 {)}^{4} (2 \times 5 {)}^{2} }=
(2×3)
4
(2×5)
2
3
−5
×2
−6
×5
4
= \frac{ {3}^{ - 5} \times {2}^{ - 6} \times {5}^{4} }{ {2}^{ - 4} \times {3}^{ - 4} \times {2}^{2} \times {5}^{4} }=
2
−4
×3
−4
×2
2
×5
4
3
−5
×2
−6
×5
4
= \frac{ {5}^{4} - 2}{ {2}^{ - 4 + 2 + 6} \times {3}^{ - 4 + 5} }=
2
−4+2+6
×3
−4+5
5
4
−2
\textbf{Formula = } \: \left[ {a}^{m} \div {a}^{n} = {a}^{m - n} \right]Formula = [a
m
÷a
n
=a
m−n
]
= \frac{ {5}^{2} }{ {2}^{4} \times 3}=
2
4
×3
5
2
⇒ \frac{25}{16 \times 3} = \frac{25}{48}⇒
16×3
25
=
48
25
\begin{gathered} \\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙﷻ}\end{gathered}