Question

CLASS 7 MATHS

CH-13- EXPONENTS AND POWERS

FIND THE RECIPROCAL OF

A) (-2/3)^{4}\\

B) (-2/3)^{3} × (3/4)^{4}\\

Answer 1

A ) : ( - 2 / 3 )⁴

From the properties of exponents and powers, we know :

Any negative number with an even number in its power is equal to the sfame number with positive sign ( instead of negative ).


Therefore,

= > ( - 2 / 3 )^4

= > ( 2 / 3 )^4

= > ( 2 x 2 x 2 x 2 ) / ( 3 x 3 x 3 x 3 )

= > 16 / 81


Then, reciprocal of 16 / 81 = 81 / 16


Hence,
Reciprocal of ( - 2 / 3 )^4 is 81 / 16.



B ) : ( - 2 / 3 )^3 + ( 3 / 4 ).

From the properties of exponents and powers, we know :

Any negative number with an even number in its power is equal to the sfame number with positive sign ( instead of negative ).

But in this case, ( - 2 / 3 )^3 is with a negative number in its power. So, ( - 2 / 3 )^3 will ( - 2 / 3 )^3 only.


= > ( - 2 / 3 )^3 x ( 3 / 4 )^4

= > [ ( - 2 x - 2 x - 2 ) / ( 3 x 3 x 3 ) ] x [ ( 3 x 3 x 3 x 3 ) / ( 4 x 4 x 4 x 4 )

= > - 3 / 32


Hence,
Reciprocal of ( - 3 / 32 ) is - 32 / 3.
$\:$

Answer 2

$\huge \bf \green{Hey \: there !! }$

A.) Find the reciprocal of ( -2/3 )⁴ .

$\sf = { (\frac{ - 2}{3}) }^{4} . \\ \\ \sf = \frac{ - 2 \times - 2 \times - 2 \times - 2}{3 \times 3 \times 3 \times 3 \times 3} . \\ \\ \sf = \frac{16}{81} . \\ \\ \sf the \: reciprocal \: is \: \large \boxed{ \pink{ = \frac{81}{16} .}}$

B.) Find the reciprocal of ( -2/3 )³ × ( 3/4 )⁴ .

$\sf = {( \frac{ - 2}{3} )}^{3} \times { (\frac{3}{4} )}^{4} . \\ \\ \sf = \frac{ - 2 \times - 2 \times - 2 \times \cancel3 \times \cancel3 \times \cancel3 \times 3}{ \cancel3 \times \cancel3 \times \cancel3 \times 4 \times 4 \times 4 \times 4} . \\ \\ \sf = \frac{ - 2 \times - 2 \times - 2 \times 3 }{4 \times 4 \times 4 \times 4} . \\ \\ \sf = \frac{ - 3}{2 \times 2 \times 2 \times 4} . \\ \\ \sf = \frac{ - 3}{32} . \\ \\ \sf the \: reciprocal \: is \: \huge \boxed{ \pink{ \sf = \frac{ - 32}{3} .}}$

✔✔ Hence, it is solved ✅✅.

THANKS

#BeBrainly.