I NEED HELP!! What is the interquartile range (IQR) of the data?
1 answer:
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Answer:
6.78ft/sec
Step-by-step explanation:
From the question, dx/dt= 3.9 ft/sec
We are looking for Dy/dt
From geometry,vof this case the relationship between x and y is needed here, there is two similar triangle that exhibited by the man and the lamb
12/y= 5.1/(y-x)
Then ,cross multiply, we have
12(y-x)=5.1y
12y-12x=5.1y
12y-5.1y=12x
6.9y=12x
y=( 12/6.9)x
Differentiating implicitly the bother sides with respect to t, we have
Dy/dt= ( 12/6.9)dx/dt
But dx/dt= 3.9 ft/sec
Then Dy/dt= ( 12/6.9)× 3.9
Dy/dt=6.78ft/sec
Hence, the rate that the tip of the person's shadow moves away from the lamppost when the person is 9 ft away from the lampost is 6.78ft/sec
CHECK THE ATTACHMENT FOR THE FIQURE
