Question

please help me to solve this. but if you don't know answer then don't answer anything else. if you dare to do this then you will be reported.¯\_(ツ)_/¯​

Answer 1

Solutions :

1) Use the law of exponents & write answer in positive exponents.

(a) $\Big(\dfrac{1}{3}\Big)^{-4}$

$\implies \Big(\dfrac{3}{1}\Big)^{4}$

$\implies 3^4$

$\implies 81$

(b)$\Big(\dfrac{-2}{5}\Big)^{-6}$

$\implies \Big(\dfrac{-5}{2}\Big)^{6}$

$\implies \dfrac{15625}{64}$

(c) $(-3)^{2}\times (-3)^{3}$

$\implies (-3)^{2+3}$

$\implies (-3)^{5}$

(d) $\Big(\dfrac{5}{4}\Big)^4\times \Big(\dfrac{5}{4}\Big)^{-3}$

$\implies \Big(\dfrac{5}{4}\Big)^{4-3}$

$\implies \dfrac{5}{4}$

(e) $\Big(\dfrac{5}{13}\Big)^{-4}\times \Big(\dfrac{5}{13}\Big)^{-2}$

$\implies \Big(\dfrac{5}{13}\Big)^{-4-2}$

$\implies \Big(\dfrac{5}{13})^{-6}$

$\implies \Big(\dfrac{13}{5})^{6}$

(f) $(2)^2 \times (2)^3\times (2)^4$

$\implies (2)^{2+3+4}$

$\implies (2)^{9}$

(g) $(4)^{3}\: \div\:4^{5}$

$\implies 4^{3-5}$

$\implies \Big(\dfrac{1}{4}\Big)^{2}$

(h) $(6)^{2}\:\div\:(6)^{-1}$

$\implies 6^{(2-(-1))}$

$\implies 6^{3}$

(i) $\Big(\dfrac{-1}{5}\Big)^{3} \: \div\: \Big(\dfrac{-1}{5}\Big)^{2}$

$\implies \Big(\dfrac{-1}{5}\Big)^{3-2}$

$\implies \Big(\dfrac{-1}{5}\Big)^{-1}$

$\implies \Big(\dfrac{-5}{1}\Big)^{1}$

$\implies -5$

(j) $\Big[\Big(\dfrac{-1}{3}\Big)^4\Big]^{-2}$

$\implies \Big(\dfrac{-1}{3}\Big)^{-8}$

$\implies \Big(-3\Big)^{8}$

(k) $(5^2)^{0}$

$\implies (5)^{2\times0}$

$\implies (5)^{0}$

$\implies 1$

(l) 9³ ÷ 729

= 9³ ÷ 9³

= 9³¯³

= 9¹

= 9

(3) Simplify the following :

(a) z² × z³

$\implies z^{2+3}$

$\implies z^{5}$

(b) $z^0$

$\implies 1$

(c) $(z^5)^{-2}$

$\implies z^{(5\times-2)}$

$\implies z^{-10}$

(d) $z^5\: \div\: z^6$

$\implies z^{(5-6)}$

$\implies z^{-1}$

$\implies \dfrac{1}{z}$

(e) $5^z\:\times\: 2^z$

$\implies (5\times2)^z$

$\implies 10^z$

(f) $3^{z}\: \div\: 4^{z}$

$\implies \Big(\dfrac{3}{4}\Big)^z$

(g) $\dfrac{z^3}{z^5}\:\times z^2$

$\implies z^{3-5+2}$

$\implies z^{5-5}$

$\implies z^{0}$

$\implies 1$

(h) $4 \times z^2$

$\implies 2^2 \times z^2$

$\implies (2z)^2$

For Answer 2 & 4 refer attachment

Answer 2

Answer:

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