Maths M.C.Q. [1 Marks Each]

$[100 - 99^\circ ] \times 100$
$($as we know a to the power zero equals to $1$ so $99$ to the power zero equals to $1)$
$= (100 - 1) \times 100$
$= 100 \times 99 = 9900.$
Hence, option $(iii)$ is correct.

$\Big(\frac{7}{12}\Big)^{-4}\times\Big(\frac{7}{12}\Big)^{3\text{x}}=\Big(\frac{7}{12}\Big)^5$
$\Rightarrow\Big(\frac{7}{12}\Big)^{3\text{x}-4}$
$=\Big(\frac{7}{12}\Big)^5$
$=3\text{x}-4=5$
$=3\text{x}=5+4=9$
$\Rightarrow\text{x}=\frac{9}{3}$
$=3$

$5.05\times10^6$
$=5.05\times1000000=5050000$

The question given to us is:
$\frac{6^{\text{m}}}{6^{-3}}=6^{5}$
This can be re-written as:
$6^{\text{m}-(-3)}=6^{5}\big(\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}\big)$
Thus, we get:
$6^{\text{m+3}}=6^{5}$
Now, comparing the exponents in the above equation, we will get:
$\text{m}+3=5$
$\therefore\text{m}=5-3=2$
Thus, the last option, option IV, $m = 2$, is the correct option.

Reciprocal of $\Big(\frac{-3}{4}\Big)^\circ=\Big(\frac{4}{3}\Big)^\circ$
(as we know a to the power zero equals to $1)$
$=\Big(\frac{4}{3}\Big)^\circ=1$

For multiplicative inverse, a is called multiplicative inverse of b, if $a \times b = 1.$
Put b = $\Big(-\frac{5}{9}\Big)^{-99}$
$\Rightarrow\text{a}\times\Big(\frac{-5}{9}\Big)^{-99}=1$
$\Rightarrow\text{a}=\frac{1}{\frac{-5}{9}}^{-99}$
$\Rightarrow\text{a}=\Big(-\frac{5}{9}\Big)^{99}\ \Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$

$\text{m}=8$
$=\frac{5^2-5^{-3}}{5^2}=5^3$
$=\frac{5^{[8+(-3)]}}{5^2}$
$=\frac{5^5}{5^2}$
$=5^3=5^3$
Hence Prove.