2. in a chemistry experiment, the temperature of a gas, in degrees celsius, followed this pattern.\ntemperatu…

2. in a chemistry experiment, the temperature of a gas, in degrees celsius, followed this pattern.\ntemperature of gas\n| number of seconds after reaction (t) | temperature of gas (°c) |\n| ---- | ---- |\n| 0 | 300 |\n| 1 | 285 |\n| 2 | 270 |\n| 3 | 255 |\nwhich of these expressions describes the pattern?\na 300 - 15t\nb 300 + 15t\nc 300t - 15\nd 300t + 15

2. in a chemistry experiment, the temperature of a gas, in degrees celsius, followed this pattern.\ntemperature of gas\n| number of seconds after reaction (t) | temperature of gas (°c) |\n| ---- | ---- |\n| 0 | 300 |\n| 1 | 285 |\n| 2 | 270 |\n| 3 | 255 |\nwhich of these expressions describes the pattern?\na 300 - 15t\nb 300 + 15t\nc 300t - 15\nd 300t + 15

Answer

Explanation:

Step1: Find the initial - temperature

When (t = 0), the temperature of the gas is (300^{\circ}C). All the expressions start with (300) which is consistent with the initial - condition.

Step2: Find the rate of change of temperature

When (t = 1), the temperature is (285^{\circ}C). The change in temperature from (t = 0) to (t = 1) is (300-285 = 15^{\circ}C) decrease. When (t = 2), the temperature is (270^{\circ}C), and the change from (t = 1) to (t = 2) is also (285 - 270=15^{\circ}C) decrease. So the temperature is decreasing by (15^{\circ}C) per second. The rate of change of temperature with respect to time (t) is (- 15). The general form of a linear relationship is (y=mx + b), where (m) is the slope (rate of change) and (b) is the initial value. Here, (b = 300) (initial temperature) and (m=-15) (rate of change of temperature), so the expression for the temperature (T) in terms of (t) is (T = 300-15t).

Answer:

A. (300 - 15t)