(1 point) The intensity of light at a depth of x meters below the surface of a lake satisfies the differential equation dIdx=(−1.21)I. (a) At what depth, in meters, is the intensity of light half that of L, where L equals the intensity of light at the surface (where x=0)?

0.573 m

Step-by-step explanation:

a. To find the depth, x, we first solve the differential equation to find the expression for I  

dI/dx = (-1.21)I

dI = (-1.21)Idx

dI/I = -1.21dx

Integrating both sides, we have

∫dI/I = ∫-1.21dx

㏑I = -1.21x + C

I = exp(-1.21x + C)

I = exp(-1.21x)exp(C) Let exp(C) = A

I =Aexp(-1.21x)

when x = 0, I = L. Substituting these into the equation, we have

L = Aexp(-1.21 × 0)

L = Aexp(0)

L = A

So, I = Lexp(-1.21x)

we want to find x when I = L/2.

So, L/2 = Lexp(-1.21x)

1/2 = exp(-1.21x)

-1.21x= ㏑(1/2)

-1.21x= -㏑2

x = -㏑2/-1.21

x =  0.693/1.21

x = 0.573 m

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18