Question

f={(√2;3),(9;5)(π;3),(0;-3,(-√2;1),(-π;5)}



alguien que me ayude es urgente ​

Answer 1

Respuesta:

$f=-171\pi ^{0.3}\sqrt{\pi }$

Explicación paso a paso:

$f=\left(\sqrt{2}\cdot \:3\right)\left(9.5\right)\left(\pi ^{0.3}\right)\left(0-3\left(-\sqrt{2}\cdot \:1\right)\left(-\sqrt{\pi }\right)\right)$

$f=\sqrt{2}\cdot \:3\cdot \:9.5\pi ^{0.3}\left(0-3\sqrt{2}\cdot \:1\cdot \sqrt{\pi }\right)$

$f=3\sqrt{2}\cdot \:9.5\pi ^{0.3}\left(-3\sqrt{2}\sqrt{\pi }+0\right)$

$f=3\sqrt{2}\cdot \:9.5\pi ^{0.3}\left(-3\sqrt{2}\sqrt{\pi }\right)$

$f=-\sqrt{2}\cdot \:3\cdot \:9.5\pi ^{0.3}\cdot \:3\sqrt{2}\sqrt{\pi }$

$f=-\sqrt{2}\sqrt{2}\cdot \:85.5\pi ^{0.3}\sqrt{\pi }$

$f=-2\cdot \:85.5\pi ^{0.3}\sqrt{\pi }$

$f=-171\pi ^{0.3}\sqrt{\pi }$