1. The problem is to determine the algebraic equation for the given number pattern and decide if it is linear. 2. The table shows term numbers $t = 1, 2, 3, 4$ and term values $v = 1.2, 2.0, 2.8, 3.6$. 3. To check if the pattern is linear, calculate the differences between consecutive term values: $$2.0 - 1.2 = 0.8, \quad 2.8 - 2.0 = 0.8, \quad 3.6 - 2.8 = 0.8$$ All differences are equal, so the pattern is linear. 4. The general form of a linear equation is: $$y = mx + b$$ where $m$ is the slope (rate of change) and $b$ is the y-intercept (starting value). 5. Since the difference between terms is $0.8$, the slope $m = 0.8$. 6. Use the first term to find $b$: $$1.2 = 0.8 \times 1 + b \implies b = 1.2 - 0.8 = 0.4$$ 7. Therefore, the algebraic equation is: $$y = 0.8x + 0.4$$ 8. This confirms the pattern is linear because it fits the form $y = mx + b$ with constant slope $m$. Final answer: $$y = 0.8x + 0.4$$