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a.y=(1/2)x
b.y=(1/4)x
c.y=(1/2)2x
d.y=2x

2007-03-30 08:30:25 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

For exponential growth, you need the x as an exponent.

From the way you have it written, none of them show x as an exponent. So, I'd say NONE.

2007-03-30 08:35:09 · answer #1 · answered by Mathematica 7 · 0 0

If you meant this:

a.y=(1/2)^x
b.y=(1/4)^x
c.y=(1/2)^(2x)
d.y=2^x

then only d is exponential growth. You don't need a negative exponent to be decay, you can also have a base between 0 and 1, which the other 3 choices have, making them all decay. If this isn't what you meant, then fix your question and we can stop guessing and get it right :-)

2007-03-30 09:02:03 · answer #2 · answered by Kathleen K 7 · 0 0

i'm afraid you're all incorrect. A, B and D are linear applications. i'm assuming that for the time of C you recommend (a million/2)^(2x) and not what you wrote (in any different case C is likewise linear). If that's what you meant, than C is exponential, yet no longer advance. It truthfully decreases exponentially.

2016-12-08 14:39:40 · answer #3 · answered by ? 4 · 0 0

I'd say c since I think you meant the 2x to be an exponent. The rest are all linear.

2007-03-30 08:38:24 · answer #4 · answered by ironduke8159 7 · 0 0

if by "c" you mean

y = (1/2)^(2x)

then C. is the answer.

y = ax^r is exponential growth
y = ax^(-r) is exponential decay.

2007-03-30 08:36:43 · answer #5 · answered by Sherman81 6 · 0 0

prety sure none of them are

2007-03-30 08:35:58 · answer #6 · answered by w1ckeds1ck312121 3 · 0 0

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