Find an equation for a tangent plane z=e-x2-y2 at the point (0,0,1).

equation of the tangent plane is z = 1

Step-by-step explanation:

Given equation

z = e^(-x²-y²) at point (0,0,1)

now let z = f(x,y)

Δf(x,y) = [ fx, fy ]

= (-2xe^(-x²-y²)), (-2ye^(-x²-y²))

now

Δf (0,0) = [ 0, 0 ] = [ a, b ]

equation of the tangent plane therefore will be

z - z₀ = a(x-x₀) + b(y-y₀)

z - 1 = 0(x-0) + 0(y-0)

z - 1 = 0 + 0

z = 1

Therefore equation of the tangent plane is z = 1

each side has a different length.  

step-by-step explanation:

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap