Question

Necesito la respuesta y debo entregarlo hoy ;-;

Answer 1

1)

$P(x) = {x}^{3} - {2x}^{2} - 4x + 2$

...

$P(0) = {(0)}^{3} - 2 {(0)}^{2} - 4(0) + 2 \\ P(0) = 2$

2)

a)

$= - {5x}^{2} + x - 3 + {x}^{2} + 2 {x}^{4} + 2 \\ = {2x}^{4} - {4x}^{2} + x - 1$

b)

$= {2x}^{3} - x + 1 - {x}^{2} - {2x}^{4} - 2 \\ = - {2x}^{4} + {2x}^{3} - {x}^{2} - x - 1$

3)

$= 2 - ( \frac{3}{2} - ( \frac{9}{4} + \frac{1}{5} )) \\ = 2 - ( \frac{3}{2} - ( \frac{45 + 4}{20} )) \\ 2 - ( \frac{3}{2} - \frac{49}{20} ) \\ = 2 - ( \frac{30 - 49}{20} ) \\ = 2 - ( - \frac{19}{20} ) \\ = 2 + \frac{19}{20} \\ = \frac{40 + 19}{20} \\ = \frac{59}{20}$

4)

$9x - {( \frac{1}{2} )}^{ - 1} = x + 6 \\ 9x - 2 = x + 6 \\ 9x - x = 6 + 2 \\ 8x = 8 \\ x = 1$

5)

a)

${13}^{2} = {5}^{2} + {x}^{2} \\ 169 = 25 + {x}^{2} \\ {x}^{2} = 169 - 25 \\ {x}^{2} = 144 \\ x = 12 \: cm$

b)

$A = \frac{5 \times 12}{2} \\ A = 5 \times 6 \\ A = 30 \: {cm}^{2}$

c)

$P = 5 + 12 + 13 \\ P = 30 \: cm$

6)

$E(x) = {3x}^{2} + x - m$

$E(2) = 3{(2)}^{2} + (2) - m \\ E(2) = 12 + 2 - m \\ E(2) = 14 - m$

Por dato:

$E(2) = 6$

Reemplazando:

$6 = 14 - m \\ m = 14 - 6 \\ m = 8$

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