ALGUEM QUE MANJE DE CALCULO PRA EXPLICAR?.... Pergunta de ideia deFawnUniverso

December 2019 1 36 Report

ALGUEM QUE MANJE DE CALCULO PRA EXPLICAR?

Lista de comentários

andresccp Equação da reta:
$\boxed{\boxed{y= \frac{y_a-y_b}{x_a-x_b} *(x-x_a)+y_a}}$

a)
a=(1,-3) , b=(5,7) ... reta que passa por a e b

$y= \frac{-3-7}{1-5}*(x-1)+(-3)\\\\y= \frac{5}{2}(x-1)-3\\\\y= \frac{5(x-1)}{2}- \frac{3*2}{2} \\\\y= \frac{5x-5-6}{2} \\\boxed{y= \frac{5x-11}{2} }$

b) reta que passa por a=(-5,10) , b=(7,-10)

$y= \frac{10-(-10)}{-5-7} (x-(-5))+10\\\\y= \frac{-5}{3}(x+5)+10\\\\y= \frac{-5(x+5)+30}{3} \\\\y= \frac{-5x-25+30}{3} \\\\ \boxed{y= \frac{-5x+5}{3} }$

c) isando o y

$x+(y-1)^2=0\\\\ (y-1)^2=-x\\\\y-1=\pm \sqrt{-x} \\\\y= \pm \sqrt{-x}+1 \\\\ \text{parte superior } \to y=\sqrt{-x}+1\\\\\text{parte inferior} \to y=-\sqrt{-x}+1$

d) isolando o y

$x^2+(y-2)^2=4\\\\(y-2)^2=4-x^2\\\\(y-2)= \pm\sqrt{4-x^2} \\\\y=\pm \sqrt{4-x^2}+2\\\\ \text{parte superior }\to y= \sqrt{4-x^2}+2\\\\\text{parte inferior }\to y= -\sqrt{4-x^2}+2$

1 votes Thanks 1

FawnUniverso MUITO OBRIGADA