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if y varies directly as the cube of x and inversely as the square of z, and y=8 when x=2 and z=3. find y when x=3 and z=6​

Sagot :

COMBINED VARIATION

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» Solve:

[tex] \: : \implies \sf \large y = \frac{k {x}^{3} }{ {z}^{2} } \\ [/tex]

[tex]\large \tt \red{given} \begin{cases} \sf \: y = 8 \\ \sf \: x = 2 \\ \sf \: z = 3\end{cases}[/tex]

» Substitute to find the constant (k)

[tex]\implies \sf \large 8 = \frac{k {2}^{3} }{ {3}^{2} } \\ [/tex]

[tex]\implies \sf \large 8 = \frac{k(8)}{ 9 } \\ [/tex]

[tex]\implies \sf \large \cancel8 \times \frac{9}{ \cancel8} = \frac{k(8)}{ 9 } \times \frac{9}{8} \\ [/tex]

[tex]\implies \sf \large 9 = \frac{k( \cancel{72})}{ \cancel{72}} \\ [/tex]

[tex]\implies \sf \large 9 = k[/tex]

[tex]\implies \sf \large k = 9[/tex]

» Now find the value of (y) using the second given.

[tex]\large \tt \red{given} \begin{cases} \sf \: k = 9 \\ \sf \: x = 3 \\ \sf \: z = 6\end{cases}[/tex]

[tex] \implies \sf \large y = \frac{9( {3}^{3} )}{ {6}^{2} } \\ [/tex]

[tex] \implies \sf \large y = \frac{9 (27)}{36} \\ [/tex]

[tex] \implies \sf \large y = \frac{9 \div 9(27)}{36 \div 9} \\ [/tex]

[tex] \implies \sf \large y = \frac{27}{4} \\ [/tex]

[tex] \implies \sf \large y = 6 \frac{3}{4} [/tex]

[tex] \: [/tex]

Final Answer:

[tex] \tt \huge» \: \purple{6 \frac{3}{4} }[/tex]

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