If #y# varies directly as the cube of #x#, what is the value of #y# in the ordered pairs #(4,16)# and #(2,y)#?

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Answer 1 · 2016-07-03T03:11:22

#y=2#

We can interpret "y varies directly as the cube of x" as following formula:

#y=kx^3#, where k = some constant that we need to find.

We are given one order pair #(4,16)# from which we can find the value of k.

So, #16=k*4^3#

#16=k*64#

#k=16/64 = 1/4#

Now, we can use our original formula #y=kx^3# and pluging #1/4# for #k#. That would give us #y=1/4x^3#.

To find #y# in #(2,y)#, we just need to plugin 2 for #x#.

#y=1/4*2^3#

#y=1/4*8#

#y=2#