3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
3x(x + 2) - 1(x + 2) = 0
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
+ 1 + 1 - 2 - 2
3x = 1 or x = -2
3 3 1 1
x = ¹/₃ or x = -2
f(x) = 3x² + 5x - 2
f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
f(¹/₃) = ¹/₃ - ¹/₃
f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)
f(x) = 3x² + 5x - 2
f(-2) = 3(-2)² + 5(-2) - 2
f(-2) = 3(4) - 10 - 2
f(-2) = 12 - 12
f(-2) = 0
(x, f(x)) = (-2, 0)
Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
Oblique Asymptote: No Asymptotes
Answer:
yes
Step-by-step explanation:
Given:
The x and y axis are tangent to a circle with radius 3 units.
To find:
The standard form of the circle.
Solution:
It is given that the radius of the circle is 3 units and x and y axis are tangent to the circle.
We know that the radius of the circle are perpendicular to the tangent at the point of tangency.
It means center of the circle is 3 units from the y-axis and 3 units from the x-axis. So, the center of the circle is (3,3).
The standard form of a circle is:
Where, (h,k) is the center of the circle and r is the radius of the circle.
Putting 
, we get
Therefore, the standard form of the given circle is .
Answer is C.
All real numbers
hope it helps
Answer:
70
Step-by-step explanation:
