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The length of a rectangle is 5t+6 and its height is t^6 , where t is time in seconds and the dimensions are in inches. Find the rate of change of area, A, with respect to time

Respuesta :

[tex]\bf \textit{area of a rectangle}\\\\ A=l\cdot w\qquad \begin{cases} l=length\\ w=width\\ -----\\ l=5t+6\\ w=t^6 \end{cases}\implies A(t)=(5t+6)t^6 \\\\\\ A(t)=5t^7+6t^6\implies \cfrac{dA}{dt}=35t^6+36t^5[/tex]

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