P(t) = 40(2)^(kt) <span>when t=10, (1990), N = 55 </span> <span>55 = 40(2)^(10k) </span> <span>1.25 = 2^(10k) </span> <span>take the ln of both sides, hope you remember your log rules </span> <span>10k = ln 1.25/ln 2 </span> <span>10k = .32193 </span> <span>k = .032193 </span>
<span>so P(t) = 40(2)^(.032193t) </span>
<span>in 2000, t = 20 </span> <span>P(20) = 40(2)^(.032193(20)) </span> <span>= 62.5 million </span>
<span>for the formula </span> <span>P(t) = a(2)^(t/d), d = the doubling time </span> <span>so changing .032193t to t/d </span> <span>= .032193t </span> <span>= t/31.06 </span>
<span>so the doubling time is 31.06 </span>
<span>another way would be to set </span> <span>80 = 40(2)^(.032193t) </span> <span>2 = (2)^(.032193t) </span> <span>.032193t = ln 2/ln 2 = 1 </span> <span>t = 31.06</span>
Looking at <span>4x3 + x2 – 8x – 2, I see immediately that x^2 can be factored out of the first 2 terms => 4x^2(x+1) and -2 out of the last 2 terms => -2(4x+1).
Notice how the factor (4x+1) shows up twice.
We can factor this (4x+1) out, obtaining (x^2 - 2) (4x+1). These are the desired factors, found "by grouping."
To get y by itself subtract 2x from both sides. The left side becomes 3y=12-2x. Then divide by 3 on both sides to get y by itself. Which you get y=4-(2/3)x