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>
Answer:
see explanation
Step-by-step explanation:
given
f'(x) = 12cos x - 4sin x, then
f'() = 12cos() - 4sin()
[ cos() = -cos(), sin( = - sin() ]
= -12cos() + 4sin()
= - 12 × + 4 ×
= - 6 + 2 ≈ - 2.536 ( 3 dec. places )
Answer:
obviously the answer is 72
The model that represents 0.30+0.03 is A.
5 hope this helps sorry if wrong