A car travels along a straight line at a constant speed of 55.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 26.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
(c) What is the average speed for this new trip?

a. 17.03 mi/hr b. 0 mi/hr or 18.99 mi/hr c. 17.03 mi/hr

Step-by-step explanation:

Av Vel =(V1+V2)/2 =26

Av Speed =d/t=55=> d=55t1

D=d+d 2d =2(55t1)=110t1

av Vel =110ti/(t1+t2)

t2=3.23t1 => total time =4.23 t1

a. V2=55t1/3.23t1 =17.03 mi/hr

b. velocity as vector quantity (v2-v1)/2 =18.88 or displacement d-d=0 and av velocity = (d-d)/2= 0/2

c. since all information are standing but opposite direction which doesn't affect speed as scalar av speed is the same

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

black lol

step-by-step explanation:

h

i can tell you jamal now has 3 cookies xd

tommy has 2.

this made me lol.