Question

Plz solve this asap.... _/\_

Answer 1

a^-3 = 1/a^3

use this property to solve all of em :)

Answer 2

☞︎︎︎ Q1) Simplify :

$\: \: \: \: \: \: \:$

$\mathtt{a) \: \: 3 {}^{ - 3} } \times { - 7}^{ - 3} = {?}^{}$

$\: \: \: \: \: \: \:$

☞︎︎︎ Solution (with explanation :

$\: \: \: \: \: \: \:$

➪ reciprocal the base as the power is negative

$\: \: \: \: \: \: \:$

$\mathtt{ \implies \: { \dfrac{1}{3} }^{3} \times { \dfrac{ - 1}{7} }^{3} }$

$\: \: \: \: \: \: \:$

❥︎ We know,

$\: \: \: \: \: \: \:$

• 1³ = 1

$\: \: \: \: \: \: \:$

• -1³ = -1

$\: \: \: \: \: \: \:$

• 3³ = 27

$\: \: \: \: \: \: \:$

• 7³ = 343

$\: \: \: \: \: \: \:$

$\mathtt{ \implies \: { \dfrac{1}{27} }^{} \times { \dfrac{ - 1}{343} }^{} }$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies \: {\dfrac{ - 1}{9261} }}$

$\: \: \: \: \: \: \:$

✰ We see the value is very large-

$\: \: \: \: \: \: \:$

$\therefore \: \dfrac{ - 1}{21}$

$\: \: \: \: \: \: \:$

✞︎ As,

$\: \: \: \: \: \: \:$

• 21³ = 9261

$\: \: \: \: \: \: \:$

༒︎ We could write it as ,

$\: \: \: \: \: \: \:$

$\large \looparrowright \boxed{ \: { - 21}^{ - 3} }$

$\: \: \: \: \: \: \:$

☕︎ Or,

$\: \: \: \: \: \: \:$

$\mathtt{a) \: \: 3 {}^{ - 3} } \times { - 7}^{ - 3} = {?}^{}$

$\: \: \: \: \: \: \:$

❥︎ Take the powers common

$\: \: \: \: \: \: \:$

$\implies(3 \times - 7 {)}^{ - 3}$

$\: \: \: \: \: \: \:$

$\large \looparrowright{ \boxed{ = - 21 {}^{ - 3} }}$

$\: \: \: \: \: \: \:$

_____________________________________________

$\: \: \: \: \: \: \:$

$\mathtt{b) \: \left( \dfrac{1}{2} \right)^{ - 2} } + \left( \dfrac{1}{3} \right)^{ - 2} + \left( \dfrac{1}{4} \right)^{ - 4}$

$\: \: \: \: \: \: \:$

➪ reciprocal the base as the power is negative

$\: \: \: \: \: \: \:$

$\implies \mathtt{ {2}^{2} } + {3}^{2} + {4}^{4}$

$\: \: \: \: \: \: \:$

❥︎ We know,

$\: \: \: \: \: \: \:$

• 2² = 4

$\: \: \: \: \: \: \:$

• 3² = 9

$\: \: \: \: \: \: \:$

• 4⁴ = 256

$\: \: \: \: \: \: \:$

$\implies \mathtt{ {4}^{} } + {9}^{} + {256}^{}$

$\: \: \: \: \: \: \:$

$\mathtt{ = \: 260 + 9}$

$\: \: \: \: \: \: \:$

$\large \looparrowright{ \boxed{ = 269 {}^{ } }}$

$\: \: \: \: \: \: \:$

_____________________________________________

$\: \: \: \: \: \: \:$

$\mathtt{c) \: ( {5}^{ - 1} \times {2}^{ - 1} ) \times {6}^{ - 1} }$

$\: \: \: \: \: \: \:$

➪ reciprocal the base as the power is negative

$\: \: \: \: \: \: \:$

$\mathtt{ \implies \: \left( \dfrac{1}{5} \right)^{ } } \times \left( \dfrac{1}{2} \right)^{ } \times \left( \dfrac{1}{6} \right)^{}$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies \: \left( \dfrac{1}{10} \right)} \times \left( \dfrac{1}{6} \right)^{}$

$\: \: \: \: \: \: \:$

$\large \looparrowright \boxed{ = \mathtt{\left( \dfrac{1}{60} \right)}}$

$\: \: \: \: \: \: \:$

_____________________________________________

$\: \: \: \: \: \: \:$

$\mathtt{d) \: \: \dfrac{25 \times {p}^{ - 4} }{ {5}^{ - 3 } \times 10 \times {p}^{ - 6} } }$

$\: \: \: \: \: \: \:$

➪ reciprocal the base as the power is negative

$\: \: \: \: \: \: \:$

$\mathtt{ \implies\: \: \dfrac{25 \times {p}^{ - 4}} { \dfrac{1 \times 10 \times {1}^{}}{ {5}^{3} \times {p}^{6} }}}$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies\: \: \dfrac{25 \times {1}^{}} { \dfrac{1 \times 10 \times {1}^{} \times {p}^{4} }{ {5} \times 5 \times 5 \times {p}^{6} }}}$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies\: \: \dfrac{25 } { \dfrac{1 \times 2 \times {p}^{4} }{ {5} \times 5 \times {p}^{6} }}}$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies\: \: \dfrac{25 } { \dfrac{1 \times 2 \times {p}^{4}}{ 25 \times {p}^{6} }}}$

$\: \: \: \: \: \: \:$

$\mathtt{ \implies\: \: \dfrac{25} { \dfrac{ 2 \times {p}^{4}}{ 25 {p}^{6} }}}$

$\: \: \: \: \: \: \:$

✞︎ Denominator of fraction will reciprocal

$\: \: \: \: \: \: \:$

$\implies \mathtt{ \dfrac{25 \times 25 {p}^{ 6} }{2 \times {p}^{4} } }$

$\: \: \: \: \: \: \:$

➪ In multiplication, when the basis are same then the powers are added

$\: \: \: \: \: \: \:$

$\implies \mathtt{ \dfrac{25 \times 25 {p}^{ 6 + 4} }{2 } }$

$\: \: \: \: \: \: \:$

$\implies \mathtt{ \dfrac{25 \times 25 {p}^{ 10} }{2 } }$

$\: \: \: \: \: \: \:$

$\large \looparrowright \boxed{ = \mathtt{ \dfrac{ 625 {p}^{ 10} }{2 } }}$

$\: \: \: \: \: \: \:$

_____________________________________________