This diagram is a straightedge and compass construction. A is the center of one circle,
1 answer:
ABC is a equilateral triangle .
Proof :-
Let's assume both circles as C1 and C2 [ as shown in the figure ]
- AB is the radius of circle C1
- AB is the radius of Circle C2
AC is the radius of circle C1.
BC is the radius of circle C2 .
AB and AC both are radius of circle C1 so both are equal ie AB = AC .
AB and BC both are radius of circle C 2 so both are equal ie AB = BC .
Hence we conclude that .
AB = BC = AC.
So the triangle is equilateral triangle.
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Answer:
Option (2).
Step-by-step explanation:
Let f(x) and g(x) are the two functions with domain A and B.
Then (fg)(x) will be the function with domain A∩B
(fg)(x) = f(x)g(x) Domain : A∩B
In this question two functions are,
c(x) = and d(x) = (x + 3)
Domain : (x - 2) > 0 ⇒ x > 2
and (x - 2) < 0 ⇒ x < 2
Or domain of (cd)(x) will be C = (-∞, 2)∪(2, ∞).
Similarly, d(x) = x + 3
Domain : All real numbers
: D = (-∞, ∞)
(cd)(x) = c(x)d(x)
=
Domain of (cd)(x) will be,
Domain : C∩D = (-∞, 2)∪(2, ∞) ∩ (-∞, ∞)
: (-∞, 2) ∩ (2, ∞)
Therefore, domain of (cd)(x) will be all real values except 2.
Option (2) will be the answer.