Question

Me podrían ayudar porfabor es para aprobar la materia porfabor ​

Answer 1

Respuesta:

Explicación paso a paso:

Explicación paso a paso:

$5.-x^{3}-0.9x^{2}+0.5x-0.24:x-0.4=\\\\=\begin{matrix}0.4&:&1&-0.9&+0.5&-0.24\\&&&+0.4&-0.2&+0.12\\\hline&&1&-0.5&+0.3&-0.12\end{matrix}\\q(x)=x^{2}-0.5x+0.3\\r(x)=-0.12$

$\texttt{Factorizar}\\-\frac{15}{16}x^{4}y^{2}+\frac{3}{4}x^{2}y^{4}+\frac{27}{4}x^{4}y^{5}=-\frac{3}{4}x^{2}y^{2}\left(\frac{5}{4}x^{2}-y^{2}-9x^{2}y^{3}\right)\\\\6.\frac{1}{2}a^{2}x-2ax^{2}+ax-\frac{1}{2}ab+2bx-b\\\\\texttt{Sacando los factores comunes y factorizado:}\\\frac{1}{2}ax(a-4x+2)-\frac{1}{2}b(a-4x+2)=(a-4x+2)\frac{1}{2}(ax-b)=\\\\=\frac{1}{2}(a-4x+2)(ax-b)\\\\7.\frac{4}{25}m^{2}-\frac{18}{25}mn+0.81n^{2}=\frac{4}{25}m^{2}-\frac{18}{25}mn+\frac{81}{100}n^{2}\\\sqrt{\frac{4}{25}m^{2}}=\frac{2}{5}m\\\sqrt{\frac{81}{100}n^{2}}=\frac{9}{10}n\\2\left(\sqrt{\frac{4}{25}m^{2}}\right)\left(\sqrt{\frac{81}{100}n^{2}}\right)=2*\frac{2}{5}*\frac{9}{10}=\frac{18}{25}mn\\\texttt{eso implica que es un trinomio cuadrado perfecto,por lo tanto:}\\\\\frac{4}{25}m^{2}-\frac{18}{25}mn+\frac{81}{100}n^{2}=\left(\frac{2}{5}m-\frac{9}{10}n\right)^{2}\\\\8.-t^{2}+5t-24\\\texttt{Completando el cuadrado:}\\t^{2}+5t=24\\t^{2}+5t+\left(\frac{5}{2}\right)^{2}=24+\left(\frac{5}{2}\right)^{2}\\t^{2}+5t+\frac{25}{4}=24+\frac{25}{4}\\(t+\frac{5}{2})^{2}=\frac{121}{4}\\(t+\frac{5}{2})^{2}-\frac{121}{4}=0$[/tex]