Find the values of x, y, and λ that satisfy the system of equations. Such systems arise in certain problems of calculus, and λ is called the Lagrange multiplier. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set λ = t and solve for x and y in terms of t.) 3x + λ = 0 3y + λ = 0 x + y − 4 = 0 (x, y, λ) =

Solution (x, y, λ) = (2, 2, - 6)

Step-by-step explanation:

From the above information given:

3x + λ = 0 eqn (1)

3y + λ = 0 eqn (2)

x + y − 4 = 0 eqn (3)

Now, from equation 1 and 2:

x= - λ / 3

y= - λ / 3

Substitute the values of X and y into equation 3

-λ/ 3 - λ/ 3 - 4 =0

-λ - λ - 12= 0

-2λ - 12= 0

-2λ = 12

λ = - 6

x= - λ/ 3

x = - (- 6) / 3

x= 2

y= - λ/3

y= - (-6)/3

y= 2

Solution (x, y, λ) = (2, 2, - 6)

Therefore,

x = 2

y= 2

λ = - 6

6e14 is the answer.

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: