If I'm correct she hiked 1.47 miles of the trail in percentages
Answer:
B
Step-by-step explanation:
Height of the woman = 5 ft
Rate at which the woman is walking = 7.5 ft/sec
Let us assume the length of the shadow = s
Le us assume the <span>distance of the woman's feet from the base of the streetlight = x
</span>Then
s/5 = (s + x)/12
12s = 5s + 5x
7s = 5x
s = (5/7)x
Now let us differentiate with respect to t
ds/dt = (5/7)(dx/dt)
We already know that dx/dt = 7/2 ft/sec
Then
ds/dt = (5/7) * (7/2)
= (5/2)
= 2.5 ft/sec
From the above deduction, it can be easily concluded that the rate at which the tip of her shadow is moving is 2.5 ft/sec.
Answer:
ur answer
and thank u for following me
(i just changed my answer)
sqrt is the square root of
v=sqrt(2(ke)/m)
