Answer:
Step-by-step explanation:
Theorm-The Fundamental Theorem of Algebra: If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots.
Let's verify that the Fundamental Theorem of Algebra holds for quadratic polynomials.
A quadratic polynomial is a second degree polynomial. According to the Fundamental Theorem of Algebra, the quadratic set = 0 has exactly two roots.
As we have seen, factoring a quadratic equation will result in one of three possible situations.
graph 1
The quadratic may have 2 distinct real roots. This graph crosses the
x-axis in two locations. These graphs may open upward or downward.
graph 2
It may appear that the quadratic has only one real root. But, it actually has one repeated root. This graph is tangent to the x-axis in one location (touching once).
graph 3
The quadratic may have two non-real complex roots called a conjugate pair. This graph will not cross or touch the x-axis, but it will have two roots.
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
Answer:
x = 127°
Step-by-step explanation:
in each triangle, the value of each exterior angle is equal to the sum of its two non-adjacent interior angles..
47° + 80° = 127°
or
the third angle in the triangle:
180° - (47° + 80°) = 53°
exterior angle:
180° - 53° = 127°
2 sqrt3 + 3 sqrt3
= 5 sqrt3
Since this contains the irrational sqrt3 it is irrational
