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=c/z^2" "->" "y=500/6^2" "->" "y=13 8/9 -> 125/9#

#y=kx " "->" " y=2/5xx3" "->" "y=6/5#

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

Given:#" "color(red)( y)" "color(blue)(alpha)" "color(red)( x)" "color(blue)(alpha)" "color(red)( 1/z^2)#

Where #alpha# means proportional to.

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

#y=k x=c/z^2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the values of "k" and "c)#

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

#color(brown)("To determine "k)#
#=>y=kx" "->" "20=k(50)#

Divide both sides by 50

#20/50=kxx50/50" "#

but #50/50=1#

#color(brown)(k=20/50=2/5#

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

#=>y=c/z^2" "->" "20=c/(5^2)#

Multiply both sides by 25

#color(brown)(c=20xx25= 500)#

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

#y=kx " "->" " y=2/5xx3#

#color(blue)( y=6/5)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "y" at "z=6)#

#y=c/z^2" "->" "y=500/6^2#

#y=13 8/9 -> 125/9#