The Amount Of Gasoline Used By Car Varies Jointly As The Distance Traveled A…

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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