✏️VARIATION
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Problem: If x varies directly as y³ and inversely as √z. if x = 8 when y = 2 and z = 4 find.
Solution: Write a combine variations in which k is the constant of the variation.
- [tex] \begin{gathered}x = \frac{k {y}^{3} }{ \sqrt{z} } \end{gathered}[/tex]
- Find the constant.
- [tex] \begin{gathered}8 = \frac{k {(2)}^{3} }{ \sqrt{4} } \end{gathered}[/tex]
- [tex] \begin{gathered}8 = \frac{k(8) }{2 } \end{gathered}[/tex]
- [tex] \begin{gathered}8 \cdot2 = \frac{k(8) }{ \cancel2 } \cdot \cancel2\end{gathered}[/tex]
- [tex]8 \cdot2 = k(8)[/tex]
- [tex] \begin{gathered} \frac{16}{8} = \frac{k \cancel{(8) }}{ \cancel8 } \end{gathered}[/tex]
4. Find x when y = 3 and z = 9.
- [tex] \begin{gathered}x = \frac{2 {(3)}^{3} }{ \sqrt{9} } \end{gathered}[/tex]
- [tex] \begin{gathered}x = \frac{2(27)}{3 } \end{gathered}[/tex]
- [tex] \begin{gathered}x = \frac{54}{3 } \end{gathered}[/tex]
- Therefore, the value of x is:
- [tex] \large \boxed{ \sf \green{ \: 18 \: }}[/tex]
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5. y when x = 4 and z = 16.
- [tex] \begin{gathered}4 = \frac{2y^{3} }{ \sqrt{16} } \end{gathered}[/tex]
- [tex] \begin{gathered}4 = \frac{2y^{3} }{4} \end{gathered}[/tex]
- [tex] \begin{gathered}4 \cdot4= \frac{8y }{ \cancel4} \cdot \cancel4 \end{gathered}[/tex]
- [tex] \begin{gathered} \frac{16}{8} = \frac{ \cancel{ 8}y}{ \cancel8 } \end{gathered}[/tex]
- Therefore, the value of y is:
- [tex] \large \boxed{ \sf \green{ \: \: 2\: \: }}[/tex]
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