V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
It's just asking you to sit down and COUNT the little squares in each sector.
It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it. That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.
(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)
The answer is Trend Line.
Answer:
the new resister is 11 ohms.
Explanation:
Set it up like this.
1/x + 1/1.1 = 1 Subtract 1/1.1 from both sides
1/x = 1 - 1/1.1
1 - 1/1.1 = 1/11
1/x = 1/11 Cross multiply
11 = x
If 1/11 bothers you, you could do it it another way.
1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1 Multiply top and bottom by 10
0.1*10/(1.1 * 10 ) = 1 / 11
