Respuesta :
[tex]\bf \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{\qquad k\qquad }}}}{}&\cfrac{}{\qquad x\qquad }
&&y=\cfrac{{{ k}}}{x}
\end{array}
\\ \quad \\
\textit{we know that}
\begin{cases}
(1.6, 6)\\
-------\\
x=1.6\\
y=6
\end{cases}\implies y=\cfrac{k}{x}\implies (6)=\cfrac{k}{(1.6)}[/tex]
solve for "k", to find the "constant of variation",
and plug it back in the y = k/x, for the equation.
now, about (8,y)
namely, when x = 8, what's "y"?
well, just set x = 8, in y =k/x to get "y"
solve for "k", to find the "constant of variation",
and plug it back in the y = k/x, for the equation.
now, about (8,y)
namely, when x = 8, what's "y"?
well, just set x = 8, in y =k/x to get "y"
Answer:
The missing value i.e. the value of y is:
[tex]y=1.2[/tex]
Step-by-step explanation:
We know that two variables x and y are said to be in inverse variation if there exist a constant k such that:
[tex]y=\dfrac{k}{x}\\\\i.e.\\\\k=xy[/tex]
We are given that:
The pair of points is on the graph of an inverse variation.
The points are:
(1.6, 6) and (8, y)
i.e.
[tex]k=6\times 1.6\\\\i.e.\\\\k=9.6[/tex]
Also,
[tex]k=8\times y\\\\i.e.\\\\y=\dfrac{k}{8}\\\\i.e.\\\\y=\dfrac{9.6}{8}\\\\i.e.\\\\y=1.2[/tex]
Hence, the missing value is: 1.2
