-5 + n is another way to write the expression.
Double is double, so you can solve
.. 2 = 1*e^(.08t)
and get the correct value of t.
Taking logs,
.. ln(2) = .08t
.. ln(2)/.08 = t ≈ 8.7 . . . . years
Lizzie has 18 dimes and 12 quarters
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
We know that,
value of 1 dime = $ 0.10
value of 1 quarter = $ 0.25
Given that LIzzie has 30 coins
number of dimes + number of quarters = 30
d + q = 30 ---- eqn 1
Also given that the coins total $ 4.80
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 4.80

0.1d + 0.25q = 4.8 ------ eqn 2
Let us solve eqn 1 and eqn 2
From eqn 1,
d = 30 - q ---- eqn 3
Substitute eqn 3 in eqn 2
0.1(30 - q) + 0.25q = 4.8
3 - 0.1q + 0.25q = 4.8
0.15q = 1.8
<h3>q = 12</h3>
From eqn 3,
d = 30 - 12
<h3>d = 18</h3>
Thus she has 18 dimes and 12 quarters
__ 5 __, __
the first would be 2
after 5 is 8
after 8 is 11
Hope this helps!!!