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
Next , solve as you would solve any two -step inequality. Since 2 is added to , you can get by itself on one side of the inequality by subtracting 2 from both sides. Then, to get by itself on one side of the inequality, you need to multiply both sides by 2.D
Answer:
A and D
Step-by-step explanation:
Total ice cream bars sold = sum of chocolate sold , vanilla and strawberry ice-creams sold.
=(1/2)x2 + (6/11)x + 8 + (5/9)x2 + (2/3) +(1/3)x2 + 4x +(4/3) (Given in the question)
=(25/18)x2 + (50/11)x + 10 (Adding terms corresponding to x2,x ,constant respectively)
Difference in chocolate and strawberry bars =[ (1/2)x2 + (6/11)x + 8] - [(1/3)x2 + 4x +(4/3)]
= (1/6)x2 - (38/11)x +(20/3)
So, the correct options are A and D
The roots of this equation would be -1/3 and -3.