DB/dt = k*B
dB/(kB) = dt
Integrating both sides:
1/k*ln|B| = t + C
B = C*e^(kt)
At time t = 0, B = 4000 and at time t = 3 (3 hours past 9 AM), B = 4500
4000 = C
4500 = 4000*e^(3k) => k = 1/3*ln(9/8)
At time t = 15 (Midnight), B = 4000*e^(1/3*ln(9/8)*15) = 4000*(9/8)^5 = 7208 bacteria
Answer:
4c
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
5c - c
<u>Step 2: Simplify</u>
- [Subtraction] Combine like terms: 4c
Answer: 1: 45
Hope this helps!
B. because it has three different terms
(if that makes sense)
<em>t would be 8 if you are multiplying .</em>
