Area of the shaded region
square cm
Perimeter of the shaded region
cm
Solution:
Radius of the quarter of circle = 12 cm
Area of the shaded region = Area of quarter of circle – Area of the triangle



square cm.
Area of the shaded region
square cm
Using Pythagoras theorem,



Taking square root on both sides of the equation, we get
cm
Perimeter of the quadrant of a circle = 

cm
Perimeter of the shaded region =
cm
cm
Hence area of the shaded region
square cm
Perimeter of the shaded region
cm
Answer: 
<u>Step-by-step explanation:</u>

Answer:
sin(A-B) = 24/25
Step-by-step explanation:
The trig identity for the differnce of angles tells you ...
sin(A -B) = sin(A)cos(B) -sin(B)cos(A)
We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.
And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.
Then ...
sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25
The desired sine is 24/25.
Answer:
<h2>3x + 21 = 6x - 60
</h2>
Step-by-step explanation:
parallel angles are equal
3x + 21 = 6x - 60
combine similar terms:
3x - 6x = -60 - 21
-3x = -81
x = 81/3
x = 27
now, plugin the value of x=27 back into the equation to get the angle:
= 3x + 21
= 3(27) + 21
= 102
= 6x - 60
= 6(27) - 60
= 102
as you noticed, both values are equal 102 = 102 because its a parallel angles
so the equation is 3x + 21 = 6x - 60
================================
i am pretty sure its correct.