Leas house is 350 meters from her friends house. Lea walks to her friends house at a constant rate of 50 meters per minute . What equation expresses the relationship between time and the distances Lea has left to wall to her friend's house .
A. Y=350 -50x
B. Y = 350x - 50
C. Y = 50x + 350
D. Y = 50x - 350
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Answer:
AT first divide 350 by 50 = 350 ÷ 50 = 7
On this way we are calculated how much minute she has to walk to the friend's house with constant rate of 50 metres per minute.
As we have five fields in the table for values of time and distance, divide 7 by 5.
= 7.0 ÷ 5 = 1.4
The value of distance for each value of time we calculate as Distance = 50 time
Time (min) Distance
1.4 70
2.8 140
4.2 210
5.6 280
7 350
Answer: She has to walk 7 minutes.
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