Answer: If we define 2:00pm as our 0 in time; then:
at t= 0. the velocity is 30 mi/h.
then at t = 10m (or 1/6 hours) the velocity is 50mi/h
Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:
a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/
Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)
So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/
Answer:
Width is 120ft but may be different if the "times the width" information is given. Follow the same process.
Step-by-step explanation:
To solve the problem, write an expression that relates length and width using the information given to you. "The length of a rectangular park is 3 feet shorter than times its width" means that l = 3 + w. Since no number is given for "times its width" we disregard this portion. If you have the information then you would include it here l = 3 + _w.
So using the expression l = 3 + w, substitute l = 123. Then the width will be:
l = 3 + w
123 = 3 + w
120 = w
-3 / 4 + p = 1/2
LCD 4 and 2 = 4
Multiply by LCD = 4
p. (4)- 3/4 .( 4 ) = 1/2.(4)
4 p - 3 = 2
Add 3 to both sides:
4 p - 3 + 3 = 2 + 3
4 p = 5
Divide both sides by 4 :
4p / 4 = 5 /4
p = 5/4
Um i just started school about three weeks ago and im in advanced but i have not learned that
Answer:
3
1
Step-by-step explanation: