Suppose y varies directly with x. If y = 6 when x = 2, find x when y = 12. 4 3

1 answer:

6 0

$\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{we also know that } \begin{cases} y=6\\ x=2 \end{cases}\implies 6=k(2)\implies \cfrac{6}{2}=k\implies 3=k \\\\\\ therefore\qquad \boxed{y=3x} \\\\\\ \textit{when y = 12, what is \underline{x}?}\qquad 12=3x\implies \cfrac{12}{3}=x\implies 4=x$

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Patrick raced round a 440 metre circular track and stopped suddenly after 900 metres . How far was she from the starting point a

castortr0y [4]

Answer:

20 meters

Step-by-step explanation:

The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.

So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.

The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be $900-880=20$ meters from the starting point.