1. **Problem 1: Pattern of squares** We are given a pattern where $S$ represents the number of small squares as a function of the step number $n$. The chart shows some values of $S$ for different $n$. 2. **Identify the pattern and formula** From the diagrams and values, it looks like the number of squares forms a sequence related to square numbers or a quadratic function. 3. **Check given values:** - For $n=1$, $S=3$ - For $n=2$, $S=27$ - For $n=3$, $S=102$ - For $n=4$, $S=10002$ (likely a typo, probably $S=1002$ or similar, but we will proceed with given data) 4. **Find a formula for $S$ in terms of $n$** Since the pattern is not linear, assume a quadratic form: $$S = an^2 + bn + c$$ Use the first three points to solve for $a$, $b$, and $c$: For $n=1$: $$a(1)^2 + b(1) + c = 3$$ For $n=2$: $$a(2)^2 + b(2) + c = 27$$ For $n=3$: $$a(3)^2 + b(3) + c = 102$$ This gives the system: $$a + b + c = 3$$ $$4a + 2b + c = 27$$ $$9a + 3b + c = 102$$ 5. **Solve the system:** Subtract first from second: $$(4a + 2b + c) - (a + b + c) = 27 - 3$$ $$3a + b = 24$$ Subtract second from third: $$(9a + 3b + c) - (4a + 2b + c) = 102 - 27$$ $$5a + b = 75$$ Subtract the two new equations: $$(5a + b) - (3a + b) = 75 - 24$$ $$2a = 51 \\ a = 25.5$$ Plug $a=25.5$ into $3a + b = 24$: $$3(25.5) + b = 24 \\ 76.5 + b = 24 \\ b = 24 - 76.5 = -52.5$$ Plug $a$ and $b$ into $a + b + c = 3$: $$25.5 - 52.5 + c = 3 \\ -27 + c = 3 \\ c = 30$$ 6. **Final formula:** $$S = 25.5 n^2 - 52.5 n + 30$$ 7. **Check for $n=4$:** $$S = 25.5(16) - 52.5(4) + 30 = 408 - 210 + 30 = 228$$ This does not match $10002$, so the given $10002$ is likely a typo. 8. **Problem 2: Orange launched by catapult** Given height function: $$h(t) = -16 t^2 + 60 t + 15$$ where $h(t)$ is height in feet and $t$ is time in seconds. 9. **Analyze statements:** a. Domain of $h(t)$ only contains $t \geq 0$ because time cannot be negative. **True** b. Check if $t=10$ is in domain: Calculate $h(10)$: $$h(10) = -16(100) + 60(10) + 15 = -1600 + 600 + 15 = -985$$ Height is negative, which is not physically meaningful, so $t=10$ is outside the realistic domain. **True** c. Height at launch $t=0$: $$h(0) = 15$$ So the orange is 15 feet above ground at launch. **True** d. Find when orange hits ground $h(t) = 0$: $$-16 t^2 + 60 t + 15 = 0$$ Divide by -1: $$16 t^2 - 60 t - 15 = 0$$ Use quadratic formula: $$t = \frac{60 \pm \sqrt{(-60)^2 - 4(16)(-15)}}{2(16)} = \frac{60 \pm \sqrt{3600 + 960}}{32} = \frac{60 \pm \sqrt{4560}}{32}$$ $$\sqrt{4560} \approx 67.53$$ So, $$t = \frac{60 + 67.53}{32} \approx 3.98, \quad t = \frac{60 - 67.53}{32} \approx -0.23$$ Negative time is discarded, so orange hits ground at about 3.98 seconds, not 3 seconds. **False** e. Check if height at $t=1$ equals height at $t=2$: $$h(1) = -16(1) + 60(1) + 15 = -16 + 60 + 15 = 59$$ $$h(2) = -16(4) + 60(2) + 15 = -64 + 120 + 15 = 71$$ Heights are not equal. **False** f. Same as (a), domain only $t \geq 0$. **True** **Final answers:** - Problem 1 formula: $$S = 25.5 n^2 - 52.5 n + 30$$ - Problem 2 true statements: a, b, c, f