Iwill marked brainliest if you could answer this! a business uses straight-line depreciation to determine the value of an automobile over a 6-year period. suppose the original value (when t=0) is equal to $17,400 and the salvage value (when t=6) is equal to $3000. write the linear equation that models the value, s, of this automobile at the end of year t.

s = -2400t + 17400

Step-by-step explanation:

Let's say t is the x value on a coordinate plane, and s is the y. Then, we have the points (0, 17400) and (6, 3000). The slope of these is 14400/-6 or -2400.

Now we just have the equation y = -2400x + b, and from the point (0, 17400) we can find that b is 17400. So, we have y = -2400x + 17400. Convert these back into t and s and you get your answer, s = -2400t + 17400.

47 - 90 = 43

43 - 180 = x

x = 137

the answer is c: divide the arc's degree measure by 360°, then multiply by the circumference of the circle.