Question

me ajudem. eu viagei e perdi essa materia. Por favor eu quero as contas.
as contas numero 5 6 12 13 16 e 19.

Answer 1

Nesses exercícios, você usará a fórmula de Bháskara

$\boxed{x = \frac{-b\ +\ ou\ -\ \sqrt{\Delta}}{2a}}$

Para encontrar o Delta.

$\boxed{\Delta = b^2-4(a)(c)}$

5) $x^2+11=12x\\\\ x^2-12x+11= 0\\\\ \Delta = (-12)^2-4(1)(11)\\\\ \Delta = 144-44\\\\ \boxed{\Delta =100}$

$x = \frac{-(-12)\ +\ ou\ -\ \sqrt{100}}{2*1}\\\\ x = \frac{12\ +\ ou\ -\ 10}{2}\\\\ x^i = \frac{12+10}{2} = \frac{22}{2} = 11\\\\ x^i^i = \frac{12-10}{2} = \frac{2}{2} = 1$

Ele quer a média aritmética. Para isso, vamos somar e dividir por 2.

$M = \frac{11+1}{2}\\\\ M = \frac{12}{2}\\\\\ \boxed{M=6}$

6) $5x^2-31x+6=0\\\\ \Delta = (-31)^2-4(5)(6)\\\\ \Delta = 961-120\\\\ \boxed{\Delta = 841}$

$x = \frac{31\ +\ ou\ -\ 29}{10}\\\\ x^i = \frac{31+29}{10} = \frac{60}{10} = 6\\\\ x^i^i = \frac{31-29}{10} = \frac{2}{10} = \frac{1}{5}$

Ele quer a soma dos termos da fração, que são 1 e 5, por isso 1+5 é igual a 6.

12) $S = \frac{-b}{a}\\\\ P = \frac{c}{a}$

$S = -\frac{5}{2}\\\\ P = -\frac{3}{2}\\\\ \frac{S}{P} = \frac{\frac{-5}{2}}{\frac{-3}{2}} = \frac{10}{6} = \frac{5}{3}$

Alternativa A.

13) $\sqrt{2x^2-4x+9} = 2x-3\\\\ (\sqrt{2x^2-4x+9})^2 = (2x-3)^2\\\\ 2x^2-4x+9 = 4x^2-12x+9\\\\ 2x^2-8x = 0\\\\ 2x(x-4) = 0\\\\ \boxed{x=0}\\\\ x-4=0\\\\ \boxed{x=4}$

c) entre 3 e 5

16) $x^2+3xy=0\\\\ x(x+3y) = 0\\\\ x_1 = 0\\\\ x-y = 2\\\\ 0-y = 2\\\\ \boxed{y_1 = -2}$

$x+3y = 0\\\\ x = -3y\\\\\ x-y = 2\\\\ -3y-y = 2\\\\ -4y = 2\\\\ \boxed{y=-\frac{1}{2}}$

$y_1+y_2\\\\ -2+(-\frac{1}{2})\\\\ -2-\frac{1}{2}\\\\ y_1+y_2 = \boxed{-\frac{5}{2}}$

19) $x-y = 1\\ x^2+y^2 = 8,5\\\\ x= y+1\\\\ (y+1)^2+y^2 = 8,5\\\\ y^2+2y+1+y^2 = 8,5\\\\ 2y^2+2y-7,5 = 0\\\\ \Delta = 4+60\\\\ \boxed{\Delta = 64}\\\\ y^1 = \frac{3}{2}\\\ y^2 = -\frac{5}{2}$

$x = y+1\\\\ x = 1,5+1\\\\ x = 2,5\\\\\ x+y\\\\ 2,5+1,5\\\\ \boxed{4}$