Question

oi socorro alguem ai me ajudah​

Answer 1

1)  Bom vamos calcular as potências normalmente de cada alternativa

   a)$2^{4}$ = 16

   b) (2/5)² = 4/25

   c) (-5)³ = -125

   d) 2,7³ = 19,683

   e)(-1/4)³ = -1/64

   f) $15^{0}$ = 1

   g)$(3/11)^{1}$ = 3/11

2) a) $(-4)^{5}.(-4)^{3} = (-4)^{5+3} = (-4)^{8}$

    b) $(1/2)^{8} / (1/2)^{6} = (1/2)^{8-6} = (1/2)^{2}$

    c) $(x^{3}) ^{4} = x^{3.4} = x^{12}$

    d) $5^{2} .5^{7} = 5^{9}$

    e) $[(2/3)^{2} ]^{7} = (2/3)^{7.2} = (2/3)^{14}$

     f)$(-3)^{9} / (-3)^{6} = (-3)^{9-6} = (-3)^{3}$

Espero ter ajudado!

Answer 2

$1)a) \: 2 {}^{4} = 2.2.2.2 = 16 \\ \\ b) \: ( \frac{2}{5} ) {}^{2} = ( \frac{2}{5} . \frac{2}{5} ) = \frac{4}{25}$

$c) \: ( - 5) {}^{3} = ( - 5). ( - 5).( - 5) = \\ 25.( - 5) = - 125 \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$d) \: 2.7 {}^{3} = 2.7 \times 2.7 \times 2.7 = 19.683$

$e) \: ( \frac{ - 1}{4} ) {}^{3} = ( \frac{ - 1}{4} ).( \frac{ - 1}{4} ).( \frac{ - 1}{4} ) = \\ \\ \frac{1}{16}.( \frac{ - 1}{4} ) = \frac{ - 1}{64 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$f) \: 15 {}^{0} = 1$

$g) \: ( \frac{3}{11} ) {}^{1} = \frac{3}{11}$

$2)a) \: ( - 4) {}^{5} .( - 4) {}^{3} = ( - 4) {}^{5 + 3} = ( - 4) {}^{8}$

$b) \: ( \frac{1}{2} ) {}^{8} \div ( \frac{1}{2} ) {}^{6} = ( \frac{1}{2} ) {}^{8 + 6} \\ \\ ( \frac{1}{2}) {}^{14} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$c) \: (5 {}^{3}) {}^{4} = 5 {}^{3.4} = 5 {}^{12}$

$d) \: 5 {}^{2}.5 {}^{7} = 5 {}^{2 + 7} = 5 {}^{9}$

$e) \: |( \frac{2}{3}) {}^{3} | {}^{7} = ( \frac{2}{3}) {}^{3.7} = ( \frac{2}{3} ) {}^{21}$

$f) \: ( - 3) {}^{9} \div ( - 3) {}^{6} = ( - 3) {}^{9 - 6} = ( - 3) {}^{3}$

ESPERO TER AJUDADO!