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3 years ago

20%5C%3A%20inversely%20%5C%5C%20as%20%5C%3A%20their%20%5C%3A%20sum%20%5C%3A%20if%20%5C%3A%20x%20-%203%20%5C%3A%20when%20%5C%3A%20u%20%3D%203%20%5C%3A%20%20%5C%5C%20and%20%5C%3A%20v%20%3D%201%20%5C%3A%20what%20%5C%3A%20is%20%5C%3A%20the%20%5C%3A%20value%20%5C%3A%20of%20%5C%3A%20x%20%5C%3A%20if%20%5C%5C%20u%20%3D%203%20%5C%3A%20and%20%5C%3A%20v%20%3D%203" id="TexFormula1" title="x \: varies \: as \: product \: u \: v \: and \: inversely \\ as \: their \: sum \: if \: x - 3 \: when \: u = 3 \: \\ and \: v = 1 \: what \: is \: the \: value \: of \: x \: if \\ u = 3 \: and \: v = 3" alt="x \: varies \: as \: product \: u \: v \: and \: inversely \\ as \: their \: sum \: if \: x - 3 \: when \: u = 3 \: \\ and \: v = 1 \: what \: is \: the \: value \: of \: x \: if \\ u = 3 \: and \: v = 3" align="absmiddle" class="latex-formula">
Pls help me with this

Mathematics

1 answer:

EleoNora [17]3 years ago

4 0

[Note that you don't have to evaluate k to solve the problem]

$x = \frac{kuv}{(u+v)}$

That means that

$x \frac{(u+v)}{(uv)} = k, a \: constant$

So, you want $x$ such that

$x \frac{(3+3)}{(3 \times 3)} = 3 \frac{(3+1)}{(3 \times 1)}$

$x=6$

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