A truck moves 3 km in 30 min and and returns at same distance in 20 min find average speed

Step-by-step explanation:

the truck mover 3 km one way, 3 km back, so he moves 6 km

one way the truck needs 30 min, on the way back 20 min

so we have 6 km in 50 min

v=distance/time

v one way=3km/30 min=0.1 km/min

v on the way back=3 km/20min=0.15 km/min

v average =0.1+0.15=0.125 km/min

or 7.5 km/hour

-1/3(x+1)^5

i’m pretty sure this is correct

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.