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
The answer is: width = t - 9
A = l * w
A = t² - 4t - 45
l = t + 5
t² - 4t - 45 = (t + 5)w
7 : 21
which can be simplified to 1 : 3 as both sides are divisible by 7
Given:
The two functions are:


To find:
The type of transformation from f(x) to g(x) in the problem above and including its distance moved.
Solution:
The transformation is defined as
.... (i)
Where, a is horizontal shift and b is vertical shift.
- If a>0, then the graph shifts a units left.
- If a<0, then the graph shifts a units right.
- If b>0, then the graph shifts b units up.
- If b<0, then the graph shifts b units down.
We have,


The function g(x) can be written as
...(ii)
On comparing (i) and (ii), we get

Therefore, the type of transformation is translation and the graph of f(x) shifts 2 units up to get the graph of g(x).