You had the right idea using the Pythagorean theorem to solve for b.
Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.
Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.
5² + b² = (2√2)²
25 + b² = 2²(√2)²
25 + b² = 4(2)
25 + b² = 8
b² = 8 - 25
b² = - 17
b = √-17
b= (√17i)
Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.
Answer: 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
Answer:
C. II only
Step-by-step explanation:
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Answer:
I don't know what the question is but...
Step-by-step explanation:
-2t+5-13= -2t-8
-2t+5+13=-2t+18
-2t+5*13= -2t+65
-2t+5/13= -26t-5/13
I have included the answer and work in the picture attached below (the side length for the bottom leg is cut off, but it is 16 sqrt(2)):
