Respuesta :
the area of a parallelogram is pretty straighforward, A = bh, base * height.
now, let's first off convert those mixed fractions to "improper", and then plug them in,
[tex]\bf \stackrel{mixed}{140\frac{5}{8}}\implies \cfrac{140\cdot 8+5}{8}\implies \stackrel{improper}{\cfrac{1125}{8}} \\\\\\ \stackrel{mixed}{11\frac{1}{4}}\implies \cfrac{11\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{45}{4}}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{area of a parallelogram}\\\\ A=bh\qquad \begin{cases} A=\frac{1125}{8}\\\\ h=\frac{45}{4} \end{cases}\implies \cfrac{1125}{8}=b\left( \cfrac{45}{4} \right)\implies \cfrac{\quad\frac{1125}{8} \quad }{\frac{45}{4}}=b \\\\\\ \cfrac{1125}{8}\cdot \cfrac{4}{45}=b\implies \cfrac{25}{2}\cdot \cfrac{1}{1}=b\implies \cfrac{25}{2}=b\implies 12\frac{1}{2}=b[/tex]
now, let's first off convert those mixed fractions to "improper", and then plug them in,
[tex]\bf \stackrel{mixed}{140\frac{5}{8}}\implies \cfrac{140\cdot 8+5}{8}\implies \stackrel{improper}{\cfrac{1125}{8}} \\\\\\ \stackrel{mixed}{11\frac{1}{4}}\implies \cfrac{11\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{45}{4}}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{area of a parallelogram}\\\\ A=bh\qquad \begin{cases} A=\frac{1125}{8}\\\\ h=\frac{45}{4} \end{cases}\implies \cfrac{1125}{8}=b\left( \cfrac{45}{4} \right)\implies \cfrac{\quad\frac{1125}{8} \quad }{\frac{45}{4}}=b \\\\\\ \cfrac{1125}{8}\cdot \cfrac{4}{45}=b\implies \cfrac{25}{2}\cdot \cfrac{1}{1}=b\implies \cfrac{25}{2}=b\implies 12\frac{1}{2}=b[/tex]
