If x is inversely proportional to y, and x = 60 when y = 0.5, how do you find x when y = 12?

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Answer 1 · 2016-03-31T16:02:54

#y=30/x#

At #y=12"; "x= 2 1/2 -= 5/2#

Let #k# be a constant. Then:

#y=kxx1/x = k/x#

It is given that:

#y=k/x" "->" "0.5=k/60#

Multiply both sides by 60 giving

#(0.5xx60) = kxx (60/60)#

But #60/60=1" and "0.5xx60=30#

#=>k=30#

So the relationship is: #y=30/x#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For #y=12# we have

#12=30/x#

#x=30/12 = 2 6/12 -> 2 1/2#