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a cat started at a distance of 5.4 m away from your front door and began walking directly away from it. 2 minutes later, the cat was 9.3 m from your door. if d is the distance of the cat from the door, what is (delta)d
A. 3.9
B. 9.3
C. 5.4
D. 2

Respuesta :

cryssatemp

Answer: A. 3.9 m

Explanation:

The variation in distance [tex]\Delta d[/tex] is given by:

[tex]\Delta d=d_{f}-d_{o}[/tex]  (1)

Where [tex]d_{f}=9.3m[/tex] is the final distance of the cat from the door and [tex]d_{o}=5.4m[/tex] is the initial distance of the cat from the door.

Solving (1) with the known values:

[tex]\Delta d=9.3m-5.4m=3.9m[/tex]  (2)

Hence:

[tex]\Delta d=3.9m[/tex]  

kadiekfernandez

3.9 is right i got it right on the test for A P E X

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