1. **Problem 1: Identify the true statement about exponential functions.** - Exponential functions have the general form $$f(x) = a^x$$ where $$a$$ is a positive constant base (not equal to 1) and $$x$$ is the variable exponent. - **Step:** Review each statement: - "It always involves inequalities." — Not necessarily true; exponential functions are defined for all real numbers and do not inherently involve inequalities. - "It has a constant exponent and a variable base." — False; the exponent is variable, the base is constant. - "It has a variable exponent and a constant base." — True; this is the defining feature of exponential functions. - "It represents a linear relationship." — False; exponential functions are nonlinear. - **Answer for Problem 1:** "It has a variable exponent and a constant base." 2. **Problem 2: Calculate interest earned by Lino after 5 years with 5.5% annual rate.** - Given principal $$P = 5000$$, rate $$r = 5.5\% = 0.055$$, time $$t = 5$$ years. - Assuming simple interest unless otherwise stated: $$\text{Interest} = P \times r \times t = 5000 \times 0.055 \times 5 = 1375$$ - None of the options exactly match 1375, so we test compound interest (compounded annually): $$A = P (1 + r)^t = 5000 (1 + 0.055)^5 = 5000 (1.055)^5$$ - Calculate: $$(1.055)^5 \approx 1.305\quad \Rightarrow\quad A \approx 5000 \times 1.305 = 6525$$ - Interest earned: $$6525 - 5000 = 1525$$ (approximately) - Closest given option to interest earned is **₱1,534.80**, which matches compound interest calculation better. - **Answer for Problem 2:** ₱1,534.80 3. **Problem 3: Calculate amount Mark Jared will have after 2 years and 6 months with 4% interest compounded annually.** - Given: - Principal $$P = 10000$$ - Annual interest rate $$r = 4\% = 0.04$$ - Time $$t = 2.5$$ years (2 years and 6 months) - Since compounding is annually, partial years cannot be compounded fully. We calculate for full 2 years and then add simple interest for 0.5 year or apply compound interest approximately. - Step 1: After 2 years: $$A_2 = P (1 + r)^2 = 10000 (1.04)^2 = 10000 (1.0816) = 10816$$ - Step 2: For the remaining 0.5 year with simple interest on $$A_2$$: $$\text{Interest for 0.5 year} = 10816 \times 0.04 \times 0.5 = 216.32$$ - Total amount after 2.5 years: $$A = 10816 + 216.32 = 11032.32$$ - Alternatively, using continuous compounding or interpolation this approximates one of the options. The closest given choices are: - Php 11073.55 - Php 11073.54 - Php 11030.20 - Php 11030.19 - The closest is **Php 11 073.55**, which likely assumes compounding exactly for the half year. - **Answer for Problem 3:** Php 11 073.55