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

2 Answers
Mar 18, 2018

Answer:

See Below.

Explanation:

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.

Mar 18, 2018

Answer:

A constant thus neither

Explanation:

#e# is a constant #e~~2.71828.... ->2.72# to 2 decimal places

Thus #e/(-4)# is also a constant