Question 11:

The problem describes an infection spreading in a community, with the number of infected people increasing by a factor of 1.5 each day. This indicates exponential growth. If the outbreak is uncontrolled, exponential growth continues indefinitely, meaning the number of infections will never stop increasing. Thus, the correct answer is:

B. The total number of infections will continue to increase indefinitely.

Question 12:

This problem asks you to express a function as partial fractions: $f(x) = \frac{5}{(1 - 2x)(1 + 3x)}$

We'll decompose the fraction into the form: $\frac{A}{1 - 2x} + \frac{B}{1 + 3x}$

Multiply both sides by the denominator $(1 - 2x)(1 + 3x)$: $5 = A(1 + 3x) + B(1 - 2x)$ Expanding both terms: $5 = A + 3Ax + B - 2Bx$ Group the terms by powers of $x$: $5 = (A + B) + (3A - 2B)x$ For this to hold for all values of $x$, the coefficients of $x$ and the constant terms must be equal. So, we get two equations:

1. $A + B = 5$
2. $3A - 2B = 0$

From the second equation, solve for $A$: $3A = 2B \quad \Rightarrow \quad A = \frac{2}{3}B$ Substitute this into the first equation: $\frac{2}{3}B + B = 5 \quad \Rightarrow \quad \frac{5}{3}B = 5 \quad \Rightarrow \quad B = 3$ Now substitute $B = 3$ into $A = \frac{2}{3}B$: $A = \frac{2}{3} \times 3 = 2$ So, the partial fraction decomposition is: $f(x) = \frac{2}{1 - 2x} + \frac{3}{1 + 3x}$

Would you like more details on these solutions or have any other questions?

5 Related Questions:

1. What are the differences between linear growth and exponential growth?
2. How does the growth factor affect the spread of infections over time?
3. Can exponential growth occur indefinitely in real-world scenarios? Why or why not?
4. How do you solve systems of equations arising from partial fraction decomposition?
5. What are some applications of partial fraction decomposition in calculus?

Tip: When dealing with exponential growth, always check if it can be realistically sustained or if external factors (e.g., medical interventions) might reduce the rate of spread.