Question

Plz don't spam...........plz solve with full steps........... They r three different sums

Answer 1

Answer:

3) 32

4) 1

5) (7/9)³

Step-by-step explanation:

$3) \: \: \dfrac{{5}^{ - 2} \times {2}^{3}}{ {10}^{ - 2} }$

$= {5}^{ - 2} \times {2}^{3} \times {10}^{2}$

$= {5}^{ - 2} \times {2}^{3} \times {(5 \times 2)}^{2}$

$= {5}^{ - 2} \times {2}^{3} \times {5}^{2} \times {2}^{2}$

$= {5}^{ (- 2 + 2)} \times {2}^{(3 + 2)}$

$= {5}^{0} \times {2}^{5}$

$= 1 \times 32 = 32$

Therefore our required answer is 32.

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4) As we see there is power of 0 in whole it's mean the final answer will be 1 .

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$5) {( \dfrac{9}{7})}^{ - 6} \times {( \dfrac{7}{9}) }^{ -3}$

${( \dfrac{7}{9})}^{6} \times {( \dfrac{7}{9} )}^{ - 3}$

${( \dfrac{7}{9})}^{6 - 3}$

${(\dfrac{7}{9})}^{3}$

Hence, Our required answer is (7/9)³

${\boxed{\red{\textsf{Some Important Laws of Indices}}}}$

${a}^{n}.{a}^{m}={a}^{(n + m)}$

${a}^{-1}=\dfrac{1}{a}$

$\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}$

${({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}$

${a}^{\frac{1}{x}}=\sqrt[x]{a}$

$a^0 = 1$

$[\text{Where all variables are real and greater than 0}]$

Answer 2

Step-by-step explanation:

the answer is attached ............... not very clear