Texaschic! : ) 1. sam is observing the velocity of a car at different times. after three hours, the velocity of the car is 51 km/h. after five hours, the velocity of the car is 59 km/h. part a: write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. show your work and define the variables used. (5 points) part b: how can you graph the equation obtained in part a for the first six hours? (5 points) mypart a: y - 51 = 59 - 51 / 5 - 3 (x-3) y - 51 = 4 (x-3) y = 4x -12 + 51 y= 4x + 39 where x is the time and y is the velocity of car at any time x. part b: y =4 * 6 + 39 = 24 + 39 = 63 ~km/h

(3,51)(5,59)
slope = (59 - 51) / (5 - 3) = 8/2 = 4

y = mx + b
slope(m) = 4
(3,51)...x = 3 and y = 51
sub and find b, the y int
51 = 4(3) + b
51 = 12 + b
51 - 12 = b
39 = b

so ur equation is : y = 4x + 39...but u need it in standard form

y = 4x + 39
-4x + y = 39
4x - y = -39 < this is ur equation...x = number of hrs and y = total velocity

ur x axis is the number of hrs
ur y axis is total velocity in km

y = 4x + 39slope = 4y int = (0,39)x int is (- 9.75,0) or -9 3/4 for graphing purposes)so start at (-9.75,0)...and since the slope is 4, go up 4 and to the right 1plot that...then up 4 and to the right 1...plot that...u will cross the y axis at (0,39)and just connect ur plotted points and u have ur lineor what u can do since it is asking for the first 6 hrssub in 1 for x and solve for ythen sub in 2 for x and solve for y...keep doing this up till 6 and u will have ur line up till 6 hrs