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Maya and Earl live at opposite ends of the hallway in their apartment building. Their doors are 50 ft. apart. Each person starts at his or her own door and walks at a steady pace toward the other. They stop walking when they meet.
Suppose:
???? Maya walks at a constant rate of 3 ft. every second.
???? Earl walks at a constant rate of 4 ft. every second.
1. Graph both people’s distance from Maya’s door versus time in seconds.

Respuesta :

cecylya

Answer:

Maya walks 21,42 feats  and Earl walks 28,58

Step-by-step explanation:

Graph is attached.

Speed is distance divides by time.

S= D/t

Where

S=  Speed

D= distance

t= time.

t =D/S

In this case both person uses the same time to meet each other.  

Maya 3 ft. every second.

t =D/3

Earl 4 ft. every second.

t =(50-D)/4

D/3=(50-D)/4

D= (50-D)*3/4

D= 3/4*50-3/4D

D+3/4D= 3/4*50

7/4D= 3/4*50

D= 3/4*50* 4/7

D=21,42

50-D= 28,58

So,  Maya walks 21,42 feats  and Earl walks 28,58

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