[tex]\large\bold{COMBINED\: VARIATION}[/tex]
[tex]\footnotesize\textsf{m varies directly as h and inversely as the square of d}[/tex]
[tex]\bold{EQUATION:}[/tex] [tex]\boxed{\tt m=\frac{kh}{d^2}}[/tex] where k is the constant of the variation.
[tex]\bold{GIVEN:}[/tex]
[tex]\bold{UNKNOWN:}[/tex]
- constant of the variation k
- m if h=4 and d=20
[tex]\bold{SOLUTION:}[/tex]
First, we have to find the value of k
[tex] \large \tt m = \frac{kh}{ {d}^{2} } \\ \\ \large \tt 200 = \frac{k(2)}{ {(10)}^{2} } \\ \\ \large \tt 200 = \frac{2k}{100} \\ \\ \large \tt (100)(200) = \bigg( \frac{2k}{ \cancel{100}} \bigg)( \cancel{100}) \\ \\ \large \tt 20 \: 000 = 2k \\ \\ \large \tt \frac{20 \: 000}{2} = \frac{2k}{2} \\ \\ \large \boxed{\tt k = 10 \: 000}[/tex]
Now, we will substitute the value of k to find the value of m if h=2 and d=20.
[tex]\begin{array}{l} \Large \tt m = \frac{kh}{ {d}^{2} } \\ \\ \Large \tt m = \frac{(10 \: 000)(4)}{ {(20)}^{2} } \\ \\ \Large \tt m = \frac{40 \: 000}{400} \\ \\ \Large \red{ \boxed{ \tt m = 100}}\end{array}[/tex]
[tex]\\[/tex]
[tex]\therefore\boxed{\textsf{m=100 when h=2 and d=20.}}[/tex]
[tex]\\ \\[/tex]
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