Question

sanjay turns on his oven before putting in a batch of peppermint brownies. the temperature of the oven rises over time. this situation can be modeled as a linear relationship. chart: x-axis: elapsed time (minutes) from 0 to 10, y-axis: temperature (°f) from 35 to 350. line passes through (0,70), (3,140), (6,210), etc.

Explanation:

Step1: Identify the linear relationship

The graph shows a linear relationship between time (x - axis) and temperature (y - axis). The general form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. From the graph, when $x = 0$, $y=70$, so $b = 70$.

Step2: Calculate the slope

To find the slope $m$, we can use two points. Let's take $(x_1,y_1)=(0,70)$ and $(x_2,y_2)=(3,140)$. The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Substituting the values, we get $m=\frac{140 - 70}{3 - 0}=\frac{70}{3}\approx23.33$ (or we can see from the graph that for every 3 minutes, the temperature rises by 70°F, so the rate of change (slope) is $\frac{70}{3}$ °F per minute). The equation of the line is $y = \frac{70}{3}x+70$. If we want to find the temperature at a certain time or the time to reach a certain temperature, we can use this equation. For example, if we want to find the temperature at $x = 6$ minutes, substitute $x = 6$ into the equation: $y=\frac{70}{3}\times6 + 70=140 + 70 = 210$, which matches the graph.

Answer:

The linear model for the oven temperature over time is $y=\frac{70}{3}x + 70$ (where $x$ is time in minutes and $y$ is temperature in °F). If we want to find specific values, we can use this equation. For example, at $x = 10$ minutes, $y=\frac{70}{3}\times10+70=\frac{700 + 210}{3}=\frac{910}{3}\approx303.33$ °F. (The answer can be adjusted based on the specific question, but the linear equation is the key model here.)