Question

College algebra online is not for me Can someone help me understand these questions

Answer 1

Answer: (1a) 20,158.7            (1b)  131,072,000

              (2) F(t) = 150e²ⁿ   use t instead of n

              (3a) 8015                  (3b) 8,606,040,719,360

Step-by-step explanation:

$P(t) = P_oe^{kt}\\\\\bullet 2P_0=4000\\\bullet P_o=2000\\\bullet t=30\\\\\\4000=2000e^{30k}\\.\quad 2=e^{30k}\\ln(2)=30k\\\\\dfrac{ln(2)}{30}=k\qquad \longrightarrow \qquad P(t)=2000e^{\frac{ln(2)}{30}t}\\\\\\\\P(100)=2000e^{\frac{ln(2)}{30}(100)}\\.\qquad \quad =\boxed{20,158.7}\\\\\\P(6\cdot 60)=2000e^{\frac{ln(2)}{30}(8\cdot 60)}\\.\qquad \qquad = 2000 e^{16ln(2)}\\.\qquad \qquad =\boxed{131,072,000}$

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$F(t)=P_0e^{kt}\\\\\bullet P_0=150\\\bullet k=0.2\\\\F(t)=\boxed{150e^{0.2t}}$

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$P(t) = P_oe^{kt}\\\bullet P(t)=2P_o\\\bullet P_o=P_o\\\bullet t=10\\\\\underline{\text{Find k:}}\\2P_o=P_oe^{10k}\\\\2=e^{10k}\\\\ln(2)=10k\\\\\dfrac{ln(2)}{10}=k\\\\\\\bullet P(90)=50,000\\\bullet P_o=P_o\\\bullet t=9\\\\\underline{\text{Find}\ P_o:}\\\\50,000=P_oe^{\frac{ln(2)}{10}90}\\\\\dfrac{50,000}{e^{9ln(2)}}=P_o\\\\\boxed{8015}=P_o\qquad \longrightarrow \qquad P(t)=8015e^{\frac{ln(2)}{10}t}$

$P(5\cdot 60)=8015e^{\frac{ln(2)}{10}(5\cdot 60)}\\.\qquad \qquad = 8015 e^{30ln(2)}\\.\qquad \qquad =\boxed{8,606,040,719,360}$