Aspotlight on the ground shines on a wall 12 m away. if a man 2 m tall walks along the x-axis from the spotlight toward the building at a speed of 1.8 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man=

Therefore,

Differentiate w.r.t t

When the man is 4 m from  the building

Then, we have x=12-4=8 m

Substitute the values in above equation then, we get

Substitute the values then we get

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3