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qpHalcy0nBest ResponseYou've already chosen the best response.0
What's going on with y as x keeps getting bigger?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Why growth?\[0.05^2 <0.05^1 \]That proves that it's decay itself.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
If the second power is less than the first, then we have an exponential decay. Get it?
 one year ago

qpHalcy0nBest ResponseYou've already chosen the best response.0
Yes, I believe the OP means raised to the power of x. Not linear in x.
 one year ago

HashirBest ResponseYou've already chosen the best response.0
if its a raised power then may be its decay
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
@Hashir The question is\[5 \times 0.5^x\]
 one year ago

qpHalcy0nBest ResponseYou've already chosen the best response.0
Well, if its raised to x, its DEFINITELY decay, not maybe.
 one year ago

greener6Best ResponseYou've already chosen the best response.0
so if the number in the parentheses is lower then 1 its decay
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
When people copy and paste questions, they don't type out ^ before powers.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[y(x) = 5(0.5)^x\] lets try some points \[y(0) = 5(0.5)^0=5\] \[y(1) = 5(0.5)^1=5\times0.5=2.5\] \[y(2) = 5(0.5)^2=5\times0.25=1.25\] as x goes up y decreases , decay
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
dw:1343887585554:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
dw:1343887677903:dw
 one year ago
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