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Inter Conversion between Standard and Normal Forms

Simplify:i 13...

Question

Simplify:
(i) $\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3}$
(ii) $(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}$
(iii) ${\{{(\frac{1}{2})}^{- 1} \times (- 4)^{- 1}\}}^{- 1}$
(iv) ${[{\{{(\frac{- 1}{4})}^{2}\}}^{- 2}]}^{- 1}$
(v) ${\{{(\frac{2}{3})}^{2}\}}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times 6^{- 1}$

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Solution

$(i)left(left(frac{1}{3} right )^{-3}- left(frac{1}{2} right )^{-3}right )div left(frac{1}{4} right )^{-3}=left(frac{1}{(1/3)^3}- frac{1}{left (1/2 right )^3}right )divfrac{1}{(1/4)^{3}}$ ---> (a⁻ⁿ = 1/(aⁿ))
$=left(frac{1}{(1/27)}- frac{1}{left (1/8 right )}right )divfrac{1}{(1/64)}$
$=left(frac{27}{1}- frac{8}{1}right )div64$
$=left(19right )timesfrac{1}{64}$
$=frac{19}{64}$

$(ii)left(3^2-2^2right )times left(frac{2}{3} right )^{-3}=left(9-4 right )timesfrac{1}{(2/3)^3}$ ---> (a⁻ⁿ = 1/(aⁿ))
$=5timesfrac{1}{8/27}$
$=5timesfrac{27}{8}$
$=frac{135}{8}$

$(iii)left(left(frac{1}{2}right )^{-1}times left(-4 right )^{-1}right)^{-1}=left (left(frac{1}{1/2}right )times left(frac{1}{-4} right ) right )^{-1}$ ---> (a⁻¹ = 1/a)
$=left (2times left (frac{1}{-4} right ) right )^{-1}$
$=left (frac{1}{-2} right )^{-1}$
$=frac{1}{1/(-2)}$ ---> (a⁻¹ = 1/a)
=-2

$(iv)left(left(left(frac{-1}{4} right )^2 right )^{-2} right )^{-1}=left (left (frac{(-1)^2}{4^2} right )^{-2} right )^{-1}$ --> ((a/b)ⁿ = aⁿ/(bⁿ))
$=left (left (frac{1}{16} right )^{-2} right )^{-1}$ ---> (a⁻ⁿ = 1/(aⁿ))
$=left (left (frac{1}{(1/16)^2} right ) right )^{-1}$
$=left (frac{1}{(1/256)} right )^{-1}$
$=256^{-1}$ ---> (a⁻¹ = 1/a)
$=frac{1}{256}$

$(v)left(left(frac{2}{3} right )^2 right )^3timesleft(frac{1}{3} right )^{-4}times3^{-1}times6^{-1} =left (frac{2^2}{3^2} right )^3timesfrac{1}{(1/3)^4}timesfrac{1}{3}timesfrac{1}{6}$ ---> ((a/b)ⁿ = aⁿ/(bⁿ)) and (a⁻ⁿ = 1/(aⁿ))
$=left (frac{4}{9} right )^3timesfrac{1}{(1/81)}timesfrac{1}{3}timesfrac{1}{6}$
$=frac{4^3}{9^3}times81timesfrac{1}{18}$ ---> ((a/b)ⁿ = aⁿ/(bⁿ))
$=frac{64}{729}times81timesfrac{1}{18}$
$=frac{64}{9}timesfrac{1}{18}$
$=64timesfrac{1}{162}$
$=frac{64}{162}$
$=frac{32}{81}$

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Similar questions

Simplify:

$(i) {{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}} \div {(\frac{1}{4})}^{- 3} (i i) (3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3} (i i i) {(\frac{1}{2}) \times (- 4)^{- 1}}^{- 1} (i v) {[{{(\frac{- 1}{4})}^{2}}^{- 2}]}^{- 1} (v) {{(\frac{2}{3})}^{2}}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times 6^{- 1}$

Q. Simplify:
(i)

(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}

(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}

[{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}] \div {(\frac{1}{4})}^{- 3}

(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}

Subtract :

(i) $\frac{3}{4}$ from $\frac{1}{3}$ (ii) $\frac{- 5}{6}$ from $\frac{1}{3}$ (iii) $\frac{- 8}{9}$ from $\frac{- 3}{5}$

(iv) $\frac{- 9}{7}$ from -1 (v) $\frac{- 18}{11}$ from 1 (vi) $\frac{- 13}{9}$ from 0

(vii) $\frac{- 32}{13}$ from $\frac{- 6}{5}$ (viii) -7 from $\frac{- 4}{7}$

Q. Simplify:
(i)

{\{4^{- 1} \times 3^{- 1}\}}^{2}

{\{5^{- 1} \div 6^{- 1}\}}^{3}

{(2^{- 1} + 3^{- 1})}^{- 1}

{\{3^{- 1} \times 4^{- 1}\}}^{- 1} \times 5^{- 1}

(4^{- 1} - 5^{- 1}) \div 3^{- 1}

1^{3} + 1^{2} + 1 + 2^{3} + 2^{2} + 2 + 3^{3} + 3^{2} + 3 + . . . 3 n

=

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