Answer:
pls I don't understand can u pls expatriate
part A)



part B)
well, we know y = -2x+8.... so.. what's the runner's velocity after 5hours? well, x = 5, thus y = -2(5) +8 ---> y = -2
to graph it, well, is a LINEar equation, meaning the graph is a LINE, and to graph a line, all you need is two points, and by now, you have more than two.. so graph it away.
Answer:
308 would be the number you are looking for.
Step-by-step explanation:
(6,2)
Mark brainliest please
Hope this helps you
y = 0.6x - 3.2
The general form of the desired equation is
y = mx + b
where
m = slope of the line
b = y intercept of the line
If two lines are parallel, their slopes will be the same, Since the slope of the
given line "y = 0.6x +3" is 0.6, that will also be the slope of the desired line.
So our equation becomes:
y = 0.6x + b
Now we can substitute the x and y value of the desired point we want the new line to pass through and find b. So
y = 0.6x + b
-5 = 0.6(-3) + b
-5 = -1.8 + b
-3.2 = b
So the desired equation is now
y = 0.6x - 3.2