A. According to Figure 2-1, what is the displacement of the mass during interval A-D?
12 meters
18 meters
20 meters
24 meters

Well, the figure seems to report that velocity is measured in m/s²... That label should say m/s. (Unless of course this is the graph of acceleration over time, but then the answer would probably be more complicated than the given choices.)

If the graph indeed shows velocity, and the unit is just a typo, then the displacement from A to D is equal to the area under the curve.

From A to B, the area is of a triangle with height 4 m/s and base 1 s, hence the area is 1/2 • (4 m/s) • (1 s) = 2 m.

From B to C, it's a rectangle with length 3 s and height 4 m/s, hence with area (3 s) • (4 m/s) = 12 m.

From C to D, it's a trapezoid with "height" 2 s and bases 4 m/s and 2 m/s, hence with area 1/2 • (4 m/s + 2 m/s) • (2 s) = 6 m.

The total displacement is then 2 m + 12 m + 6m = 20 m.

atomic = no of protons.

mass number-atomic number = neutron number

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .