Question

Preciso de ajuda, please

Answer 1

Explicação passo-a-passo:

( Equação Exponencial )

a )

$\sf ( \frac{1}{5} ) {}^{x} = \frac{1}{625}$

$\sf {5}^{ - x} = {5}^{ - 4}$

$\sf - x = - 4$

$\boxed{\boxed{\sf x = 4}}$

b )

$\sf {9}^{x} = \frac{1}{27}$

$\sf {3}^{2x} = {3}^{ - 3}$

$\sf 2x = - 3$

$\boxed{\boxed{\sf x = - \frac{ 3}{2} }}$

c )

$\sf {49}^{x + 1} = \sqrt[3]{7}$

$\sf {7}^{2x + 2} = {7}^{ \frac{1}{3} }$

$\sf 2x + 2 = \frac{1}{3}$

$\sf 2x = \frac{1}{3} - 2$

$\sf 2x = \frac{1 - 6}{3}$

$\sf 2x = - \frac{5}{3}$

Dívida ambos por 2

$\boxed{\boxed{\sf x = - \frac{ 5}{6} }}$

( Razões Trigonométricas )

a )

=>Valor de X

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>tan30</strong><strong> = \frac{x}{9 \sqrt{3} }$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong> \frac{ \sqrt{3} }{3} = \frac{x}{9 \sqrt{3} }$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> \frac{ \sqrt{3} }{3} = \frac{ \sqrt{3} x}{27}$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>27 \sqrt{3} = 3 \sqrt{3} x$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = \frac{27 \sqrt{3} }{3 \sqrt{3} }$

$\boxed{\boxed{</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = 9}}$

=>Valor de Y

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>sen30</strong><strong> = \frac{9}{y}$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> \frac{1}{2} = \frac{9}{y}$

$\boxed{\boxed{</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>y</strong><strong> = 18}}$

b )

=>Valor de X

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>tan60</strong><strong> = \frac{12}{x}$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> \sqrt{3} = \frac{12}{x}$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> \sqrt{3} x = 12$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = \frac{12}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = \frac{12 \sqrt{3} }{3}$

$\boxed{\boxed{</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = 4 \sqrt{3} }}$

=>Valor de Y

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>cos60</strong><strong> = \frac{4 \sqrt{3} }{y}$

$</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> \frac{1}{2} = \frac{4 \sqrt{3} }{y}$

$\boxed{\boxed{</strong><strong>\</strong><strong>s</strong><strong>f</strong><strong> </strong><strong>y</strong><strong> = 8 \sqrt{3}}}$

Espero ter ajudado um pouco !!!