Question

answer it correctly please​

Answer 1

Answer:

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$\dashrightarrow\sf \bigg\lgroup\dfrac{1}{4}\bigg\rgroup^{\large-2}-3\bigg\lgroup8\bigg\rgroup^{\dfrac{2}{3}}\times4^{\large0}+\bigg[\dfrac{9}{16}\bigg]^{\dfrac{-1}{2}}\\\\\\\dashrightarrow\sf (4)^2 - 3(2)^{\bigg\lgroup\large3 \times \dfrac{2}{3}\bigg\rgroup} \times 1 + \bigg[\dfrac{16}{9}\bigg]^{\dfrac{1}{2}}\\\\\\\dashrightarrow\sf 16 - 3(2)^2 \times 1 +\bigg[\dfrac{4}{3}\bigg]^{\bigg\lgroup\large2 \times \dfrac{1}{2}\bigg\rgroup}\\\\\\\dashrightarrow\sf 16 - 3 \times 4 \times 1 + \dfrac{4}{3}\\\\\\\dashrightarrow\sf 16 - 12 + \dfrac{4}{3}\\\\\\\dashrightarrow\sf 4 + 1\frac{1}{3}\\\\\\\dashrightarrow \underline{\boxed{\sf 5\frac{1}{3}}}$

How to approach exponential questions?

• First remember all the exponential rules given below.
• If power is negative, then to change it into positive. We reciprocate base number.
• If power is zero, it equals to 1
• If power is in fraction, try to find power of base which can cancel denominator. Like that question can be easy.

$\rule{200}{1}$

$\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}$