Answer: A
Step-by-step explanation:
First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.
When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.
Step-by-step explanation:
Angles in the same segment. The angles at the circumference subtended by the same arc are equal. More simply, angles in the same segment are equal.
Picture 1 a° = a° they are the same
Picture 2 p= 52° q= 40° Angles in the same segment are equal. if they ask you to calculate you just show both angles p= 52 and q=40.
Picture 3 + 4 Let the obtuse angle MOQ = 2x.
Using the circle theorem, the angle at the centre is twice the angle at the circumference.
Therefore picture 4 tells us and proves;
Angle MNQ = x and angle MPQ = x.
Picture 5
A cyclic quadrilateral is a quadrilateral drawn inside a circle. Every corner of the quadrilateral must touch the circumference of the circle.
The second shape is not a cyclic quadrilateral. One corner does not touch the circumference.
Picture 6
The opposite angles in a cyclic quadrilateral add up to 180°.
a + c = 180°
b + d = 180°
Picture 7
Example
Calculate the angles a and b.
The opposite angles in a cyclic quadrilateral add up to 180°.
This is a picture that couldn't be added but has a quadrilateral occupying half the triangle consisting of 3 arcs the middle one was 140 and proved part of one triangle. The x (a) missing angle showed below a = 180-60 = 120 °
So hopefully this will help you remember the format here for quadrilateral shapes within a circle.
b = 180 - 140 = 40°
a = 180 - 60 = 120°
