A man 6 ft tall walks at a rate of 6 ft/sec away from a lamppost that is 24 ft high. At what rate is the length of his shadow changing when he is 75 ft away from the lamppost

2 ft/s

Step-by-step explanation:

The lamppost is 24 ft. tall, and the man is 6 ft. tall. So, we will use a proportion to find the shadow.

Let s is the length of the base of the lamppost to the shadow while x is the length of the base of the lamppost to the man, so the length of the shadow is s - x.

Using triangular ratio, we have;

24/6 = s/(s - x)

4 = s/(s - x)

We cross multiply and distribute to get;

4s - 4x = s

4s - s = 4x

3s = 4x

s = 4x/3

Taking the derivative of both sides according to time, we have;

ds/dt = (4/3)dx/dt

Now, dx/dt is given as 6 ft/s

So;

ds/dt = (4/3) × 6

ds/dt = 8 ft/s

For us to find the rate of length of the shadow according to time, we recall that the shadow = s - x, so we will just take the derivative of each and subtract. Thus;

d(s - x)/dt = ds/dt - dx/dt

Plugging in the relevant values, we have;

ds/dt - dx/dt = 8 - 6 = 2 ft/s

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2