the amount of gasoline used by car varies jointly as the distance traveled ang the square root of the speed. suppose a car used 25 liter will on a 100 kilometers trip ar 100 km/he. about how many liters will it use on a 150 kilometers trip at 64 km/hr?
JOINT VARIATION
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» Assign variables to each following:
- Let (g) be the amount of gasoline.
- Let (d) be the distance travelled.
- Let (s) be the speed.
[tex] \: : \implies \sf \large g = kd \sqrt{s} [/tex]
[tex]\large \tt \red{given} \begin{cases} \sf \: g = 25 \\ \sf \: d = 100 \\ \sf \: s = 100\end{cases}[/tex]
» Substitute to find the constant (k).
[tex]\implies \sf \large 25 = k(100) (\sqrt{100})[/tex]
[tex]\implies \sf \large 25 = k(100)(10)[/tex]
[tex]\implies \sf \large 25 = k(1000)[/tex]
[tex]\implies \sf \large \frac{25}{1000} = \frac{k( \cancel{1000})}{ \cancel{1000}} \\ [/tex]
[tex]\implies \sf \large \frac{1}{40} =k \\ [/tex]
[tex]\implies \sf \large k = \frac{1}{40} \\ [/tex]
» Now find how many liters (g) will be used if the distance (d) is 150 km and the speed (s) is 64 kmph.
[tex]\large \tt \red{given} \begin{cases} \sf \: k = \frac{1}{40} \\ \sf \: d = 150 \\ \sf \: s = 64\end{cases}[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (150)( \sqrt{64} )[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (150)( 8 )[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (1200)[/tex]
[tex] \implies \sf \large g = \frac{1200}{40} \\ [/tex]
[tex] \implies \sf \large g = 30 \\ \\ [/tex]
Final Answer:
[tex] \tt \huge» \: \purple{30\, L \: of \: gasoline}[/tex]
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