If x varies directly with y and x=3.5 when y=14 find x when y=18

1 answer:

4 0

$\bf \qquad \qquad \textit{direct proportional variation}
\\\\
\textit{\underline{x} varies directly with \underline{y}}\qquad \qquad x=ky\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\textit{we also know that }
\begin{cases}
x=3.5\\
y=14
\end{cases}\implies 3.5=k14\implies \cfrac{3.5}{14}=k
\\\\\\
\cfrac{1}{4}=k\qquad therefore\qquad \boxed{x=\cfrac{1}{4}y}
\\\\\\
\textit{when y = 18, what is \underline{x}?}\qquad x=\cfrac{1}{4}(18)$

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The graph shows information about the speed of a vehicle during the final 50 seconds of a journey. At the start of the 50 seconds the speed is k metres per second. The distance travelled during the 50 seconds is 1.35 kilometres.

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The graph is attached. The total time = 50 seconds, total distance = 1.35 km = 1350 m

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b) From the graph, the total distance covered is the area of the graph. The graph is made up of a rectangle and triangle, the area of the graph is equal to the sum of area of rectangle and area of triangle.

$Total \ distance=Total\ area=Area\ of \ rectangle+Area\ of \ triangle\\Total \ distance=(length *breadth)+\frac{1}{2}base*height\\1350 \ m=(40*k)+0.5*10*k\\1350=40k+5k\\45k=1350\\k=1350/45\\k=30\ m/s$ c) i) The gradient in the last 10 seconds is the ratio of change in speed to change in time