Is #c= e/-4 # a direct variation, inverse variation, or neither?

Question

Nathan Axel · Accepted Answer

See Below.

I'm considering #e# as a variable. As We have #c = e/-4#, We can Write this as #c/e = -1/4# [Dividing both sides by #e#] As #c/e# is equal to a constant, we can say #c prop e# or, #c# and #e# are in direct variation. But, This results in a confusing situation. If we take #c = 1#, we get, #e = -4.# But if we put #c = 2#, we get, #e = -8#. So, Actually, If we increase #c#, e is decreasing, and again, with the same ratio, #-1/4#. So, Defintion of Variation and The Results we are getting are contradicting each other. But still, The graph of the relation shows it is direct. graph{y = (x/(-4)) [-10, 10, -5, 5]} I don't think this is a variation. But This is my opinion. I need consultations.

Abigail Allen · Answer

A constant thus neither

#e# is a constant #e~~2.71828.... ->2.72# to 2 decimal places Thus #e/(-4)# is also a constant

Eli Assouline · Answer

undefined The equation c = e/-4 is an inverse variation.