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if h varies inversely as i and h=3 when i=8, find h when i=6​

Sagot :

INVERSE VARIATION

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Solve:

[tex] \implies \sf \large h = \frac{k}{i} \\ [/tex]

[tex]\large \tt \red{given} \begin{cases} \sf \: h = 3 \\ \sf \: i = 8 \end{cases} \\ \\ [/tex]

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

[tex] \implies \sf \large 3 \times 8 = \frac{k}{ \cancel8} \times \cancel8 \\ [/tex]

[tex] \implies \sf \large \therefore \: k = 24[/tex]

[tex] \: [/tex]

[tex]\large \tt \red{given(2)} \begin{cases} \sf \: k = 24 \\ \sf \: i = 6 \end{cases} \\ \\ [/tex]

[tex] \implies \sf \large h = \frac{24}{6} \\ [/tex]

[tex] \implies \sf \large \therefore \orange{ h = 4}[/tex]

[tex] \: [/tex]

Final Answer:

[tex] \large\implies \sf \huge \orange{ h = 4}[/tex]

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