Find the constant of variation in the inverse relation below. X | Y 2 | 5 4 | 2.5 5 | 2 6 | 5/3 A. 1/10 B. 1/3 C. 3 D. 10
@sleepyjess @Compassionate @dan815 Please HELP!
@TheSmartOne
http://media.education2020.com/evresources/2003-20-04-00-00_files/i0150000.jpg
\(\large { \begin{array}{llll} x&y \\\hline\\ 2&5\\ 4&2.5\\ 5&2\\ 6&\frac{5}{3} \end{array} \\ \quad \\ \begin{array}{llllll} \textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\ \textit{something}&=&\cfrac{{\color{brown}{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\ y&=&\cfrac{{\color{brown}{\textit{n}}}}{}&\cfrac{}{x} \\\hline\\ &&y=\cfrac{{\color{brown}{ n}}}{x} \end{array} \\ \quad \\ \textit{we know that when }x=5\qquad y=2\implies 2=\cfrac{{\color{brown}{ n}}}{5} }\) solve for "n", or "constant of variation"
My result is 0.4=N , i divide both side by 5.
@jdoe0001
ok let us do that
How?
or better yet.. let us check if n = 0.4 then say \(\bf x=5\qquad y=\cfrac{n}{5}\implies y=\cfrac{0.4}{5}\implies y=0.08 \\ \quad \\ 0.08\ne 2\) recall that when x = 5, y = 2 but using 0.4 for "n", didn't really give us 2 for "y"
I'm confuse :(
well.. you said n = 0.4 but using 0.4 for the "n" value in the inverse variation didn't yield 2 if we plug x = 5 we should get y = 2 but instaed we've got 0.08 so \(\Large n\ne 0.4\)
\(\bf \textit{we know that when }x=5\qquad y=2\implies 2=\cfrac{{\color{brown}{ n}}}{5}\) so... what do you think is "n"?
1/10?
1/10? well.. if you cross-multiply \(\bf 2=\cfrac{{\color{brown}{ n}}}{5}\) what does that give for "n"? hint: multiply both sides by 5
10
there you have it then \(\bf \textit{we know that when }x=5\qquad y=2\implies 2=\cfrac{{\color{brown}{ n}}}{5}\implies 10={\color{brown}{ n}}\) "n" or the "constant of inverse variation" is 10 :)
Thanks for the help :))
yw
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