(-x-14)^2 = x^2 +14x+14x+196
which turns into x^2+28x+196.
hope this helps
and give thanks
Answer:
690
Step-by-step explanation:
Round from the tens place.
Answer:
aₙ = 1/2 x aₙ₋₁ n≥2
Step-by-step explanation:
aₙ = 1/2 x aₙ₋₁ n≥2
a₁ = 64
a₂ = 1/2 x 64 = 32
a₃ = 1/2 x 32 = 16
Answer:
Midpoint of AB = (0 + 2a / 2 , 0 + 0 / 2) = (2a / 2 , 0 / 2) = (a,0)
x coordinate of point c = a
N = (0 + a / 2 , 0 + b / 2) = (a / 2 , b / 2)
M = ( 2a + a / 2 , 0 + b / 2) = (3a / 2 , b / 2)
MA = √(3a / 2 - 0)² + b / 2 - 0)²
= √(3a / 2 )² + (b / 2) = 9a² / 4 + b² / 4
NB = √(a / 2 - 2a)² + (b / 2 - 0 )²
= √( a / 2 - 4a / 2)² + (b / 2 - 0)²
= √(-3a / 2)² + (b / 2)² = √9a² / 4 + b² / 4
Step-by-step explanation:
I tried my best hope its correct :0
The given equation of the ellipse is x^2
+ y^2 = 2 x + 2 y
At tangent line, the point is horizontal with the x-axis
therefore slope = dy / dx = 0
<span>So we have to take the 1st derivative of the equation
then equate dy / dx to zero.</span>
x^2 + y^2 = 2 x + 2 y
x^2 – 2 x = 2 y – y^2
(2x – 2) dx = (2 – 2y) dy
(2x – 2) / (2 – 2y) = 0
2x – 2 = 0
x = 1
To find for y, we go back to the original equation then substitute
the value of x.
x^2 + y^2 = 2 x + 2 y
1^2 + y^2 = 2 * 1 + 2 y
y^2 – 2y + 1 – 2 = 0
y^2 – 2y – 1 = 0
Finding the roots using the quadratic formula:
y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1
y = 1 ± 2.828
y = -1.828 , 3.828
<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828)
and (1, 3.828).</span>
