`y=0.35(2.3)^x`

If x represents the number of years, we have to determine the two values of y in consecutive terms to determine if the value of y increases or decreases after a year.

So, if x=1,

`y=0.35(2.3)^1 = 0.35*2.3=0.805`

And if x=2,

`y=0.35(2.3)^=0.35*5.29=1.8515`

Since the value of y increases after...

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`y=0.35(2.3)^x`

If x represents the number of years, we have to determine the two values of y in consecutive terms to determine if the value of y increases or decreases after a year.

So, if x=1,

`y=0.35(2.3)^1 = 0.35*2.3=0.805`

And if x=2,

`y=0.35(2.3)^=0.35*5.29=1.8515`

Since the value of y increases after a year, then it is the percent increase that we have to solve.

To do so, use the formula,

`% Increase = (Amount of Increase)/(Origi n al Amount)*100 `

`% Increase=(1.8515-0.805)/0.805 * 100=1.0465/0.805*100=1.3*100`

`% Increase=130%`

**Hence, the annual percent increase is 130%.**