Question

16. (URCA/2019.1) O valor da expressão abaixo.
$\sqrt{ {27}^{1.333...} } + 4sen(330^{o} ) + {6}^{ log_{6}(13) }$

A) 17

B) 18

C) 19

D) 20

E) 24​

Answer 1

Resposta:

1,333...=1+1/3 =4/3

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sen 330=sen(360-30)=sen360*cos30-sen30*cos360= 0-1/2=-1/2

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y =6^(log₆ 13)

log y = log 6^(log₆ 13)

log y =(log₆ 13) * log 6

log y =(log 13/log 6) * log 6  =log 13

log y =log 13  ==> y=13

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√(27^(4/3)) + 4* (-1/2)  + 13

√(3³^(4/3)) + 4* (-1/2)  + 13

√(3⁴) + 4* (-1/2)  + 13

=3² -2 +13

=9 -2+13 = 20

Letra D

Answer 2

Resposta:

Alternativa d)

Explicação passo-a-passo:

Vamos resolver por partes:

√27^{1,3333.....}

1,33333.....=1+3/9=1+1/3=(3+1)/3=4/3

27=3^{3}

√27^{1,3333.....}^=√3^{3.4/3}=√3^{4}=3^{1/2.4}=3^{2}=9

4sen(330°)= -4sen(30°)= -4.1/2= -2

6^{log₆13}=x

log₆6^{log₆13}=log₆x

log₆13.log₆6=log₆x

log₆13=log₆x

x=13

√27^{1,3333.....}+4sen(330°)+6^{log₆13}=9-2+13=20