Question

Pueden resolver esto

Answer 1

$b)\frac{15}{2}x^4y^{-2}\left(0.13x^5y^4\right)$
$\mathrm{Quitar\:los\:parentesis}:\quad \left(a\right)=a$
$=\frac{15}{2}x^4y^{-2}\cdot \:0.13x^5y^4$
$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^b\cdot \:a^c=a^{b+c}$
$y^{-2}y^4=\:y^{-2+4}=\:y^2$
$=0.13x^9y^2\frac{15}{2}$
$\mathrm{Multiplicar\:fracciones}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$
$=0.13\cdot \frac{15x^9y^2}{2}$
$=\frac{15x^9y^2\cdot \:0.13}{2}$
$\mathrm{Multiplicar\:los\:numeros:}\:15\cdot \:0.13=1.95$
$=\frac{1.95x^9y^2}{2}$
$\mathrm{Dividir:}\:\frac{1.95}{2}=0.975$
$=0.975x^9y^2$

$d)\frac{5}{4}a^2x\left(\frac{2}{5}a^2x+3ax-8\right)$
$=\frac{5a^2x}{4}\left(\frac{2a^2x}{5}+3ax-8\right)$
$=\frac{5a^2x\left(\frac{2a^2x}{5}+3ax-8\right)}{4}$
$=\frac{5a^2x\frac{2a^2x+15ax-40}{5}}{4}$
$=\frac{a^2x\left(2a^2x+15ax-40\right)}{4}$