<h2>
<u>AnswEr </u><u>:</u></h2>
Provided Equation
As we know the Standard Equation of a Circle is
where,
r radius of circle.
(a,b) centre .
This equation can be further written as
Now completing the square ( by adding 4 & 9 on both side ) .
again this equation can be further written as
Now comparing this equation with standard equation of circle ( mentioned above) and we will get
- Centre = ( a,b) = (3, -2 )
- Radius = r = 3
Therefore,
- <u>Centre </u><u>of </u><u>circle </u><u>is </u><u>(</u><u>3</u><u>,</u><u>-</u><u>2</u><u>)</u><u> </u><u>and </u><u>radius </u><u>is </u><u>3</u>
Answer:
The answer is the option C
Step-by-step explanation:
we know that
An isosceles triangle has two equal sides and two equal angles
The two equal angles are called the base angles, and the third angle is called the vertex angle
In this problem we have the isosceles triangle TIC
∠I= -----> vertex angle
∠T=∠C -----> base angles
Remember that the sum of the internal angles of a triangle is equal to
so
∠I+∠T+∠C=
∠I+2∠T=
Substitute
+2∠T=
Solve for angle T
2∠T=
∠T=
∠T=
Given:
The given polynomial is:
To find the roots of the given polynomial.
To find the roots we have to take
So,
Formula
By quadratic formula, the root of the equation is,
and
Now,
Putting, we get,
and
and
Hence,
The values of the roots of the given polynomial are and
Hence, Option A and F are the correct answer.
Is that suppose to be the problem?