Algebraic Biology P1
- Solution: Use the exponential growth formula \(P(t) = P_0e^{rt}\).
- Here, \(P_0 = 100\), doubling time is 3 hours so \(r = \frac{\ln 2}{3}\).
- \(P(9) = 100e^{(\frac{\ln 2}{3}) \cdot 9} = 100e^{3\ln 2} = 100 \cdot 2^3 = 100 \cdot 8 = 800\).
Algebraic Biology P2
- Solution: Use the stoichiometry of the photosynthesis equation: \[
6CO_2 + 6H_2O + \text{light} \rightarrow C_6H_{12}O_6 + 6O_2
\]
- Molar mass of \(CO_2 = 44 \, \text{g/mol}\), \(C_6H_{12}O_6 = 180 \, \text{g/mol}\).
- Moles of \(CO_2 = \frac{48}{44} = 1.09\).
- From the equation, 6 moles of \(CO_2\) produce 1 mole of \(C_6H_{12}O_6\).
- Glucose produced = $ = 32.7 , $.
Algebraic Biology P3
- Solution: Solve the quadratic equation \(0.25x^2 + 0.5x + 0.25 = 1\).
- Simplify: \(0.25x^2 + 0.5x - 0.75 = 0\).
- Multiply by 4: \(x^2 + 2x - 3 = 0\).
- Factor: \((x + 3)(x - 1) = 0\).
- Solutions: \(x = -3\) or \(x = 1\).
Algebraic Biology P4
- Solution: Find the vertex of the parabola \(A(t) = -2t^2 + 4t + 6\).
- Vertex formula: \(t = -\frac{b}{2a}\).
- Here, \(a = -2\), \(b = 4\).
- \(t = -\frac{4}{2 \cdot -2} = 1\).
- Maximum activity occurs at \(t = 1\) hour.