6ax2 + 6ax + 6a = 6a(x2 + x + 1)
Find what you can divide both 6ax and 9a by.They can both be divided by 3, leaving 2ax and 3a.These can both be divided by a, leaving 2x and 3.So 6ax + 9a = 3a (2x + 3)The factors of 6ax + 9a are 3a and (2x + a)
Circle : x^2 + y^2 = 16a^2 , therefore, y = +√(16a^2 - x^2) (above X-axis) Parabola : y^2 = 6ax , therefore, y = +√(6ax) (above X-axis) Intersection point of circle with parabola : √(16a^2 - x^2) = √(6ax) 16a^2 - x^2 = 6ax x^2 + 6ax - 16a^2 = 0 (x - 2a)(x + 8a) = 0 x = 2a (as the only positive zero) Total common area = 2 * common area above X-axis Common area above X-axis = ∫ (6ax) dx (from 0 to 2a) + ∫ √(16a^2 - x^2) dx (from 2a to 4) = {2x√(6ax)/3} (from 0 to 2a) + {(x/2)√(16a^2 - x^2) + 8arctan[x/√(16a^2 - x^2)]} (from 2a to 4) = 8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3 Therefore, total common area = 2{8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3} Area of circle = πr^2 = 16π Therefore, larger area of the circle = 16π - 2{8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3} = (4/3){14π - a^2√(3) - 12√(a^2 - 1) - 12arctan[1/√(a^2 - 1)]} (Note: if a = 1, then arctan[1/√(a^2 - 1)] = π/2)
Since you cannot multiply x by a, in the answer you get ax. 2a * 3x = 6ax
6ax2 + 15ax = 3ax(2x + 5)
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My HSBC branch customer service agent told me, that the SWIFT # for the NatWest Bank Berkeley Square Branch is the same as the BIC #: NWBK GB 2L.
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