The slope is already given in the question so we can find the y-intercept.
Slope-intercept form: y = mx + b
3 = 4(-2) + b
3 = -8 + b
3 + 8 = -8 + b + 8
11 = b
Now, we can write the equation.
y = 4x + 11
Best of Luck!
Its D because vertical angles are right across from each other!
Step-by-step explanation:
function f was reflected over the x-axis and translated 2 units left
The area of the shaded region is
.
Solution:
Given radius = 4 cm
Diameter = 2 × 4 = 8 cm
Let us first find the area of the semi-circle.
Area of the semi-circle = 


Area of the semi-circle =
cm²
Angle in a semi-circle is always 90º.
∠C = 90°
So, ABC is a right angled triangle.
Using Pythagoras theorem, we can find base of the triangle.




cm
Base of the triangle ABC =
cm
Height of the triangle = 4 cm
Area of the triangle ABC = 

Area of the triangle ABC =
cm²
Area of the shaded region
= Area of the semi-circle – Area of the triangle ABC
= 
= 
Hence the area of the shaded region is
.