Set h to 640 and solve for t:
640 = -490t^2 + 1120t
Subtract 640 from both sides:
-490t^2 + 1120t - 640 = 0
The formula to solve a quadratic equation is:
x = -b -/+ sqrtroot (b^2-4ac)/(2a) where a = -490, b = 1120 and c = -640
Solve:
x = -1120 -/+ sqrtroot (1120^2-4(-490)(-640) )/ 2(-490)
x = 8/7 = 1.1428 = 1.14
Time was 1.14 seconds
<span>437 = (21 + x)(21 - x), then x = 2, -2</span>
Exponential growth of the form:
F=Ar^t
F=100(1.22)^t
F(5)=100(1.22)^5
F(5)=270.27
F(5)=270 to the nearest frog...
....
1.22=r^12
r=1.22^(1/12)
F=100(1.22^(1/12)^t)
F=100(1.22^(t/12)) now it will produce monthly populations...
Say we did the same as before but instead of 5 years you have 60 months...
F(60)=100(1.22^(60/12))
F(60)=270 to the nearest frog... :P
Answer:
(0,3) and (5,0)
Step-by-step explanation:
Check the forward differences of the sequence.
If 
, then let 
be the sequence of first-order differences of 
. That is, for n ≥ 1,
so that 
.
Let 
be the sequence of differences of ,
and we see that this is a constant sequence, 
. In other words, is an arithmetic sequence with common difference between terms of 2. That is,
and we can solve for 
in terms of 
:
and so on down to
We solve for 
in the same way.
Then
and so on down to
