Question

Resolva a inequaçao: 6(4×+2) >9+7(4×-5).

Calcule a distançia entre os pontos A(0, 4) e B(15 ,-4). Ajudemmmm??????

Answer 1

$\\ 6(4^x + 2) \ \textgreater \ 9 + 7(4^x-5) \\ \\ 6*4^x +6*2\ \textgreater \ 9+7*4^x-7*5 \\ \\ 6*4^x+12 \ \textgreater \ 9-35+7*4^x \\ \\ 6*4^x-7*4^x \ \textgreater \ -26-12 \\ \\ -1*4^x\ \textgreater \ -36 *(-1) \\ \\ 4^x \ \textless \ 36 \\ \\ 4^x \ \textless \ 6^2 \\ \\ log(4^x) < log(6^2) \\ \\ x*log(4) \ \textless \ 2*log(6) \\ \\ x \ \textless \ 2* \frac{log(6)}{log(4)} \\ \\ x \ \textless \ 2,5849625007$

B)

Formula para a distancia de um ponto qualquer:

$\\ d(A,B) = \sqrt{(x-xo)^2+(y-yo)^2} \\ \\ d(A,B) = \sqrt{(15-0)^2+(-4-4)^2} \\ \\ d(A,B) = \sqrt{15^2 + (-8)^2} \\ \\ d(A,B) = \sqrt{225+64} = \sqrt{289} = \sqrt{17^2} = 17$