\( \left. \begin{array} { l }{ 5 \times 397 = 5 \times ( 400 - \ldots ) } \\ { = ( 5 \times \ldots , ( 5 \times \_ ) } \\ { = \square } \\ { = } \end{array} \right. \)

QUESTÃO : 04 As coordenadas do vértice da parábola que representa a função: \( y=x^{2}-6 x+8 \), são: \( \begin{array}{lllll}\text { a) }(-1,4) & \text { b) }(3,-1) & \text { c) }(-1,1) & \text { d) }(0,1) & \text { e) }(1,3)\end{array} \)

Multiply the powers: \(7^2 \times 7^3\)

\( \left(2 x^{2}-3 x+9\right)(x-5) \) (2) \( \left(x^{2}-2 x+1\right)\left(x^{2}+3 x+4\right) \) \( (3-2 y)\left(2 x^{2}-3 x y+4 y^{2}\right) \) \( (3)\left(3 x^{2}-x+2\right)\left(x^{2}-4 x+5\right) \) \( (3)+3 x+2)\left(x^{2}-x+1\right) \)

\( \left(2 x^{2}-3 x+9\right)(x-5) \) (2) \( \left(x^{2}-2 x+1\right)\left(x^{2}+3 x+4\right) \) (3) \( (x-2 y)\left(2 x^{2}-3 x+4 y^{2}\right) \) (4) \( \left(3 x^{2}-x+2\right)\left(x^{2}-4 x+5\right) \) (3) \( \left(x^{2}+3 x+2\right)\left(x^{2}-x+1\right) \)

Let \( d \) represent distance, \( r \) rate, and \( t \) time. Write the equation. \( d=\square \)