6 i’m think i’m not a 100% sure tho
Let ABC the vertices of the EQUILATERAL triangle inscribed in the circle centered in O & with R as radius (sketch it to better understand)
Join OA & OB so that to calculate the area of sector BOA
Mesure Angle C = 1/2 mes Arc AB, but C =60° , then Arc AB = 120°
So the central angle AOB = mes ARC OAB = 120°
The area of a sector = mes angle BOA x R² (mind you angle should be in radian) Angle BOA = 120°or 2π/3 (in radian) the Area Sect = (2π/3).R²
Answer:
y=2x+4
Step-by-step explanation:
Answer:
x = -9, -3
Step-by-step explanation:
x = -3, -9
Simplify the following:
(k^3 k^7)/3 - 5
Combine powers. (k^3 k^7)/3 = k^(7 + 3)/3:
k^(7 + 3)/3 - 5
7 + 3 = 10:
k^10/3 - 5
Put each term in k^10/3 - 5 over the common denominator 3: k^10/3 - 5 = k^10/3 - 15/3:
k^10/3 - 15/3
k^10/3 - 15/3 = (k^10 - 15)/3:
Answer: (k^10 - 15)/3
