Matemática, perguntado por marysanchezpablo, 2 meses atrás

me ajudem por favor

Anexos:

Soluções para a tarefa

Respondido por paulakern38

Resposta:

a) A (-2, 5), B (3,7) e C (5,11)

$\left[\begin{array}{ccc}-2&5&1\\3&7&1\\5&11&1\end{array}\right] = 0\\\\\left[\begin{array}{ccc}-2&5&1\\3&7&1\\5&11&1\end{array}\right] = \left[\begin{array}{ccc}-2&5\\3&7\\5&11\end{array}\right]$

-2 . 7 . 1 = -14 1 . 7 . 5 = 35

5 . 1 . 5 = 25 -2 . 1 . 11 = -22

1 . 3 . 11 = 33 5 . 3 . 1 = 15

(-14 + 25 + 33) - (35 - 22 +15)

44 - 28 = 16 Não está alinhado

b) A (0, 4), B (-6,2) e C (8,10)

$\left[\begin{array}{ccc}0&4&1\\-6&2&1\\8&10&1\end{array}\right] = 0\\\\\left[\begin{array}{ccc}0&4&1\\-6&2&1\\8&10&1\end{array}\right] = \left[\begin{array}{ccc}0&4\\-6&2\\8&10\end{array}\right]$

0 . 2 . 1 = 0 1 . 2 . 8 = 16

4 . 1 . 8 = 32 0 . 1 . 10 = 0

1 . -6 . 10 = - 60 4 . -6 . 1 = - 24

(0 + 32 - 60) - (16 + 0 - 24)

- 28 - (- 8) = -28 + 8 = -20 Não está alinhado

c) A (1,5), B ( 3,4) e C (2,2)

$\left[\begin{array}{ccc}1&5&1\\3&4&1\\2&2&1\end{array}\right] = 0\\\\\left[\begin{array}{ccc}1&5&1\\3&4&1\\2&2&1\end{array}\right] = \left[\begin{array}{ccc}1&5\\3&4\\2&2\end{array}\right]$

1 . 4 . 1 = 4 1 . 4 . 2 = 8

5 . 1 . 2 = 10 1 . 1 . 2 = 2

1 . 3 . 2 = 6 5 . 3 . 1 = 15

(4 + 10 + 6) - (8 + 2 + 15)

20 - 25 = - 5 Não está alinhado.

d) A (-4, -3), B (-1, 1) e C (2, 5)

$\left[\begin{array}{ccc}-4&-3&1\\-1&1&1\\2&5&1\end{array}\right] = 0\\\\\left[\begin{array}{ccc}-4&-3&1\\-1&1&1\\2&5&1\end{array}\right] = \left[\begin{array}{ccc}-4&-3\\-1&1\\2&5\end{array}\right]$

-4 . 1 . 1 = - 4 1 . 1 . 2 = 2

- 3 . 1 . 2 = - 6 - 4 . 1 . 5 = - 20

1 . - 1 . 5 = - 5 - 3 . - 1 . 1 = 3

(- 4 - 6 - 5) - ( 2 - 20 + 3)

- 15 - (- 15)

- 15 + 15 = 0 Está alinhado

e) A (-1,02), B (1, 3) e C (2, 5)

$\left[\begin{array}{ccc}-1&2&1\\1&1&3\\3&4&1\end{array}\right] = 0\\\\\left[\begin{array}{ccc}-1&2&1\\1&1&3\\3&4&1\end{array}\right] = \left[\begin{array}{ccc}-1&2\\1&3\\3&4\end{array}\right]$