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/
Just plug the inequality in desmos and you get 7<x<13
The second one because for every x value there is one and only one y value. If you plotted the points and graphed it, you would know it is not a function if it doesn't pass the vertical line test. Notice the same x values show up repeatedly in the other ordered pairs with different y values. Only one y value for every x value
Answer:
A
Step-by-step explanation:
6 is a power. So both C and D are wrong because 6 is being treated as an ordinary integer.
So the answer must be either A or B.
B is not correct because to move an expression from the denominator to the numerator changes the sign in the base (which is m in this question).
A is correct. the power is changed from 6 in the denominator to -6 in the numerator. m is shifted to the numerator as well.
<span>Both variables are categorical. We analyze an association through a comparison of conditional probabilities and graphically represent the data using contingency tables. Examples of categorical variables are gender and class standing.</span>
