Respuesta :
[tex]\bf \qquad \textit{ inverse proportional variation}\\\\\\
\begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}[/tex]
"The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth"
namely "w" varies inversely to [tex]\bf d^2[/tex] or [tex]\bf w=\cfrac{k}{d^2}[/tex]
then
"A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. "
namely when w = 192, d = 3900
so [tex]\bf w=\cfrac{k}{d^2}\qquad \begin{cases} w=192\\ d=3900 \end{cases}\implies 192=\cfrac{k}{3900^2} \\\\\\ \textit{solve for "k", to find the}\\ \textit{"constant of variation"}\\ \textit{once you've found it, plug it back in at }w=\cfrac{k}{d^2}[/tex]
"The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth"
namely "w" varies inversely to [tex]\bf d^2[/tex] or [tex]\bf w=\cfrac{k}{d^2}[/tex]
then
"A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. "
namely when w = 192, d = 3900
so [tex]\bf w=\cfrac{k}{d^2}\qquad \begin{cases} w=192\\ d=3900 \end{cases}\implies 192=\cfrac{k}{3900^2} \\\\\\ \textit{solve for "k", to find the}\\ \textit{"constant of variation"}\\ \textit{once you've found it, plug it back in at }w=\cfrac{k}{d^2}[/tex]
Answer:
Option D is correct.
Step-by-step explanation:
Edg said so...
