The Student Room Group

Proportionality question

How do you do this type of question?
Reply 1
Original post by Peanut247
How do you do this type of question?


Do you know how to rewrite a statement like "x is directly proportional to the square of z" in terms of indices and some unknown constant(s)?
Reply 2
Original post by davros
Do you know how to rewrite a statement like "x is directly proportional to the square of z" in terms of indices and some unknown constant(s)?


x [directly proportional to] k(z^2)
Original post by Peanut247
x [directly proportional to] k(z^2)


x=kz2x = kz^2
Reply 4
Original post by Peanut247
x [directly proportional to] k(z^2)


Not quite - see TenOfThem's correction!

Now you should be able to write something similar for the case of inverse proportion :smile:
Reply 5
Original post by davros
Not quite - see TenOfThem's correction!

Now you should be able to write something similar for the case of inverse proportion :smile:



y=1/(z^3)
Original post by Peanut247
y=1/(z^3)


y=kz3y = \dfrac{k}{z^3}
Reply 7
Original post by Peanut247
y=1/(z^3)


Again, see TenOfThem's correction - you missed out the constant!!

Now you just need to combine these things in various ways :smile:
Reply 8
Original post by davros
Again, see TenOfThem's correction - you missed out the constant!!

Now you just need to combine these things in various ways :smile:


This was the bit I got stuck on.. I tried making them both equal z but just flopped!
Reply 9
Original post by TenOfThem
y=kz3y = \dfrac{k}{z^3}

Thank you :P
Original post by Peanut247
This was the bit I got stuck on.. I tried making them both equal z but just flopped!


Try making them both z6z^6
Reply 11
Original post by TenOfThem
Try making them both z6z^6


y=k/z^3 becomes 1/y^2 = kz^6

x=kz^2 becomes x^3 = kz^6

So x^3 = 1/y^2 making the answer D? :colondollar::colondollar:
Original post by Peanut247
y=k/z^3 becomes 1/y^2 = k^2z^6

x=kz^2 becomes x^3 = k^3z^6

So x^3 = k/y^2 making the answer D? :colondollar::colondollar:


But, yes, D

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