Solve (−7) ⋅ (−4). a −28 b −11 c 28 d 11
2 answers:
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If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region
T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}
Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.
The surface element for the "positive" half is
dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy
dS = √(1 + 0 + 4y²/(16 - y²)) dx dy
dS = √((16 + 3y²)/(16 - y²)) dx dy
The the surface integral along this half is
You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.
Answer:
The square roots of 49·i in ascending order are;
1) -7·(cos(45°) + i·sin(45°))
2) 7·(cos(45°) + i·sin(45°))
Step-by-step explanation:
The square root of complex numbers 49·i is found as follows;
x + y·i = r·(cosθ + i·sinθ)
Where;
r = √(x² + y²)
θ = arctan(y/x)
Therefore;
49·i = 0 + 49·i
Therefore, we have;
r = √(0² + 49²) = 49
θ = arctan(49/0) → 90°
Therefore, we have;
49·i = 49·(cos(90°) + i·sin(90°)
By De Moivre's formula, we have;
Therefore;
√(49·i) = √(49·(cos(90°) + i·sin(90°)) = ± √49·(cos(90°/2) + i·sin(90°/2))
∴ √(49·i) = ± √49·(cos(90°/2) + i·sin(90°/2)) = ± 7·(cos(45°) + i·sin(45°))
√(49·i) = ± 7·(cos(45°) + i·sin(45°))
The square roots of 49·i in ascending order are;
√(49·i) = - 7·(cos(45°) + i·sin(45°)) and 7·(cos(45°) + i·sin(45°))