V=square root (3/2)/2
Or V=-square root (3/2)/2
X^2 + y^2 = (3x^2 + 2y^2 - x)^2
2x + 2y f'(x) = 2(3x^2 + 2y^2 - x)(6x + 4y f'(x) - 1) = 36x^3 + 24x^2yf'(x) + 24xy^2 + 16y^3f'(x) - 4y^2 - 18x^2 - 8xyf'(x) + x
f'(x)(2y - 24x^2y - 16y^3 + 8xy) = 36x^3 + 24xy^2 - 4y^2 - 18x^2 - x
f'(x) = (36x^3 + 24xy^2 - 4y^2 - 18x^2 - x)/(2y - 24x^2y - 16y^3 + 8xy)
f'(0, 0.5) = -4(0.5)^2/(2(0.5) - 16(0.5)^3) = -1/(1 - 2) = -1/-1 = 1
Let the required equation be y = mx + c; where y = 0.5, m = 1, x = 0
0.5 = 1(0) + c = 0 + c
c = 0.5
Therefore, the tangent line at point (0, 0.5) is
y = x + 0.5
Answer:99.33
Step-by-step explanation:
The answer is 49,152.
You double the numbers as you go, it goes up to 49,152.
Answer:
2/36 (5.556%)
I believe so correct me if I am wrong
