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°))
6z - 12 + 4 = 10
6z -8 = 10
6z = 18
z = 3
Answer:
we know the plane line has 180°
so here it's 111°-180° = 69°
so co- interior angles are equal so 111°=the olace of 13x-6
Perimeter=4*side so
side=p/4 and we are told p=16x+32y so
s=(16x+32y)/4
s=4x+8y
s=4(x+2y)
The equation which models the distance (d) of the weight from its equilibrium after time (t) is equal to d = -9cos(2π/3)t.
<h3>What is the period of a cosine function?</h3>
The period of a cosine function simply means the total length (distance) of the interval of values on the x-axis over which a graph lies and it's repeated.
Since the weight attached is at its lowest point at time (t = 0), therefore, the amplitude of equation will be negative nine (-9)
For the angular velocity at time period (t = 3s), we have:
ω = 2π/T
ω = 2π/3
Mathematically, the standard equation of a cosine function is given by:
y = Acos(ω)t
Substituting the given parameters into the formula, we have;
d = -9cos(2π/3)t.
Read more on cosine function here: brainly.com/question/4599903
