If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

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If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

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Answer 1 · 2016-05-25T09:46:53

$y = \frac{c}{z^{2}} \to y = \frac{500}{6^{2}} \to y = 13 \frac{8}{9} \to \frac{125}{9}$ $y=c/z^2" "->" "y=500/6^2" "->" "y=13 8/9 -> 125/9$

$y = k x \to y = \frac{2}{5} \times 3 \to y = \frac{6}{5}$ $y=kx " "->" " y=2/5xx3" "->" "y=6/5$

Explanation:

As both $x$ $x$ and $z$ $z$ are related to $y$ $y$ we can express them as follows:

Given: $y α x α \frac{1}{z^{2}}$ $" "color(red)( y)" "color(blue)(alpha)" "color(red)( x)" "color(blue)(alpha)" "color(red)( 1/z^2)$

Where $α$ $alpha$ means proportional to.

Let $k and c$ $k" and "c$ be constants of variation. Then we have:

$y = k x = \frac{c}{z^{2}}$ $y=k x=c/z^2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the values of k and c$ $color(blue)("Determine the values of "k" and "c)$

Base condition is that at $y = 20; x = 50; z = 5$ $y=20 ; x=50 ; z=5$

$To determine k$ $color(brown)("To determine "k)$
$\Rightarrow y = k x \to 20 = k (50)$ $=>y=kx" "->" "20=k(50)$

Divide both sides by 50

$\frac{20}{50} = k \times \frac{50}{50}$ $20/50=kxx50/50" "$

but $\frac{50}{50} = 1$ $50/50=1$

$k = \frac{20}{50} = \frac{2}{5}$ $color(brown)(k=20/50=2/5$

'...........................................................
$To determine c$ $color(brown)("To determine "c)$

$\Rightarrow y = \frac{c}{z^{2}} \to 20 = \frac{c}{5^{2}}$ $=>y=c/z^2" "->" "20=c/(5^2)$

Multiply both sides by 25

$c = 20 \times 25 = 500$ $color(brown)(c=20xx25= 500)$

'.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of y at x = 3$ $color(blue)("Determine the value of "y" at "x=3)$

$y = k x \to y = \frac{2}{5} \times 3$ $y=kx " "->" " y=2/5xx3$

$y = \frac{6}{5}$ $color(blue)( y=6/5)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of y at z = 6$ $color(blue)("Determine the value of "y" at "z=6)$

$y = \frac{c}{z^{2}} \to y = \frac{500}{6^{2}}$ $y=c/z^2" "->" "y=500/6^2$

$y = 13 \frac{8}{9} \to \frac{125}{9}$ $y=13 8/9 -> 125/9$

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If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

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