we can see that the center is (-3, 3) and the radius is 9 units.
<h3 /><h3>
How to find the center and radius of the circle?</h3>
The general circle equation, for a circle with a center (a, b) and radius R is given by:
(x - a)^2 + (y - b)^2 = R^2
Here we have the equation:
x^2 + y^2 + 6x = 6y + 63
Let's complete squares:
x^2 + y^2 + 6x - 6y = 63
(x^2 + 6x) + (y^2 - 6y) = 63
(x^2 + 2*3x) + (y^2 - 2*3y) = 63
Now we can add and subtract 9, (two times) so we get:
(x^2 + 2*3x + 9) - 9 + (x^2 - 2*3x + 9) - 9 = 63
(x + 3)^2 + (y - 3)^2 = 63 + 9 + 9 = 81 = 9^2
(x + 3)^2 + (y - 3)^2 = 9^2
Comparing with the general circle equation, we can see that the center is (-3, 3) and the radius is 9 units.
If you want to learn more about circles:
brainly.com/question/1559324
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Answer:
A. One solution
Step-by-step explanation:
A. One solution
The answer is -3
The answer is B
sin=Opposite/Hypotenuse
sin=28/100
Divide numerator and denominator by 4
sin=7/25
Cosine=adjacent/hypotenuse
cos=96/100
Divide numerator and denominator by 4
cos=24/25
Tangent=opposite/adjacent
tan=28/96
Divide numerator and denominator by 4
tan=7/24
Answer:
=
1
1
=
−
1
1
Step-by-step explanation:
Answer:
ΔNAS≅ΔSEN by SSA axiom of congruency.
Step-by-step explanation:
Consider ΔNAS and ΔSEN,
NS=SN(Common ie . Both are the same side)
SA=NE( Given in the question that SA≅ NE)
∠SNA=∠NSE( Due to corresponding angle property where SE ║ NA)
Therefore, ΔNAS ≅ΔSEN by SSA axiom of congruency.
∴ NA≅SA by congruent parts of congruent Δ. Hence, proved.
