Aball is dropped off the side of a bridge.
after falling 1.55 s, what is its velocity?

15.19 m/s

Step-by-step explanation:

, where is the final velocity, is the initial velocity, a is the constant gravitational acceleration 9.8 m/s^2, and t is the time (seconds).

Since the ball is dropped off the bridge, its initial velocity is 0:

We can plug in 9.8 m/s^2 into a and 1.55 s into t:

= 15.19 m/s

4

step-by-step explanation:

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: