Question

-Considere a função F(x)=4X/3+1/5, responda:

a) Qual o ponto de interseção com o eixo das abscissas(x)?
b) Qual o ponto de interseção com o eixo das ordenadas(y)?
c)Calcule o valor de (F(1))/(F(2)) + (F(3))/(F(4)).

Answer 1

Resposta:

a.) Quando a reta intersectar com o eixo das abscissas, y terá valor zero. Assim, temos:

$f(x) = \frac{4x}{3}+\frac{1}{5}  \\\\0 =\frac{4x}{3}+\frac{1}{5}\\\\\frac{4x}{3} = -\frac{1}{5}\\\\\\20x = -3\\x = -\frac{3}{20}$

b.) Quando a reta intersectar com o eixo das ordenadas, x terá valor zero. Portanto:

$y = \frac{4x}{3}+\frac{1}{5}\\\\y = \frac{4*0}{3}+\frac{1}{5}\\\\y = \frac{1}{5}$

c.) $F(1) = \frac{4}{3} +\frac{1}{5} \\\\F(1) = \frac{20+3}{15} \\\\F(1) = \frac{23}{15} \\\\\\F(2) = \frac{8}{3} +\frac{1}{5}\\\\F(2) = \frac{40+3}{15} \\\\F(2) = \frac{43}{15} \\\\\\F(3) = \frac{12}{3} +\frac{1}{5} \\\\F(3) = \frac{60+3}{15} \\\\F(3) = \frac{63}{15} \\\\\\F(4) = \frac{16}{3} +\frac{1}{5} \\\\F(4) = \frac{80+3}{15} \\\\F(4) = \frac{83}{15}$

$\\\frac{F(1)}{F(2)} = \frac{23}{15} * \frac{15}{43} = \frac{23}{43} \\\\\frac{F(3)}{F(4)} = \frac{63}{15} * \frac{15}{83} = \frac{63}{83} \\\\\frac{F(1)}{F(2)} + \frac{F(3)}{F(4)} = \frac{23}{43} + \frac{63}{83} = \frac{1909+2709}{3569} = \frac{4618}{3569}$