What was the content of the lesson? It is hard to answer a question we need context on.
let's say they catch-up to each other after t hour.
so
in t hour distance travelled by Renee
d = speed × time = 50t
in t hour distance travelled by Kim
d = speed × time = 60(t-1) + 0×1 = 60(t-1)
Note: here kim didn't covered any distance in first hour and in rest t-1 hour it travelled all distance
now.as distance travelled by both is same so
60(t-1) = 50t
60t -60 = 50t
adding 60 both sides
60t = 50t+60
subtracting 50t both sides
60t -50t = 50t+60 - 50t
10t = 60
dividing by 10 both sides
t = 60/10 = 6
so in 6 hour both will catch-up to each other
Answer:
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Explanation:
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Answer:
<h3>
<em><u>simple </u></em><em><u>ans</u></em></h3>
Explanation:
<h2>
<em><u>it </u></em><em><u>is </u></em><em><u>because </u></em><em><u>as </u></em><em><u>it </u></em><em><u>has </u></em><em><u>server </u></em><em><u>computer </u></em><em><u>and </u></em><em><u>no </u></em><em><u>one </u></em><em><u>are </u></em><em><u>free </u></em><em><u>as </u></em><em><u>other </u></em><em><u>topology </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>
<h2>
<em><u>client </u></em><em><u>computer </u></em><em><u>should </u></em><em><u>ask </u></em><em><u>server </u></em><em><u>to </u></em><em><u>share </u></em><em><u>any </u></em><em><u>information </u></em><em><u>ideas </u></em><em><u>etc.</u></em></h2>
<em><u>this </u></em><em><u>much.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
<h2>
<em><u>follow</u></em></h2>