Ed is planning daily walking workouts on the same distance of 12 km in the morning and in the evening. usually he is walking at constant rate, however he planned his rate to be 1 km/h more in the morning than in the evening. given he is willing to spend 5 hours and 24 minutes daily on these workouts, what should his morning rate be, in km/h?

Given is :

Let the rate of walking in the evening be = x km/h

As Ed wants to walk at a rate if 1 km/h more in the morning, then rate in morning becomes = x+1 km/h

So, distance is 12 km, speed is x+1 km/h

Total time is = 5 h 24 m or convert it into hours, it becomes = 5.4 hours

Time in morning becomes =

Time in evening becomes =

So equation becomes=

Solving this quadratic equation, we get, x= -5/9 or x=4

As X cannot be negative so neglect -5/9.

Solving by using x=4, we get rate as x= 4+1 =5 km/h

Hence, rate in the morning is 5 km/h.

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

c. 1,233 miles