Question

please help me please please please​

Answer 1

Answer:

The answer is -

Step-by-step explanation:

1.

a. -4 X -4 X -4 = $(-4)^{3}$ [As the number is multiplied by itself three times]

b. P/Q X P/Q = $(\frac{P}{Q})^{2}$ [As the number is multiplied by itself three times]

2.

= $[(\frac{4}{5})^{2}]^{-3}$

Follow the laws of exponents

= $(\frac{4}{5})^{2 X -3}$ [Using law of exponent, $[(\frac{m}{n})^{a}]^{b}$ =  $(\frac{m}{n})^{a X b}$)

=  $(\frac{4}{5})^{-6}$

= $(\frac{5}{4})^{6}$ [To make the exponent positive, did reciprocal of the base]

3.

a. $(1^{2} + 2^{2} + 3^{2})^{2}$

As we know

$1^{2}$ = 1

$2^{2}$ = 4

$3^{2}$ = 9

So,

= $(1^{2} + 2^{2} + 3^{2})^{2}$

= $(1 + 4 + 9)^{2}$

= $(14)^{2}$

b. $(\frac{1}{4})^{-1} X (\frac{1}{3})^{-1}$

Follow the laws of exponents

= $(\frac{1}{4})^{-1} X (\frac{1}{3})^{-1}$

= $(\frac{1}{4} X \frac{1}{3})^{-1}$ [Using law of exponent, $a^{m} X b^{m}$ = $(a X b)^{m}$]

= $(\frac{1}{12})^{-1}$

= $(12)^{1}$ [To make the exponent positive, did reciprocal of the base]

4.

a. $5^{0}$ = 1 [As any number raised to the power 0 is always 1(result)]

b. $3^{3} X 3^{2}$

Follow the laws of exponents

= $3^{3} X 3^{2}$

= $(3)^{3 + 2}$ [Using law of exponent, $a^{m} X a^{n} = a^{m + n}$]

= $(3)^{5}$

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