Answer:
1.
Original square side: s
Area = s^2
Modified square:
Length = 1.3s
Width = 0.82s
Area = 1.3s * 0.82s = 1.066s^2
(1.066 - 1)/1 * 100% = 6.6%
The area increases by 6.6%
2.
[tex] \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{15} [/tex]
[tex] \dfrac{y}{xy} + \dfrac{x}{xy} = \dfrac{2}{15} [/tex]
[tex] \dfrac{x + y}{xy} = \dfrac{2}{15} [/tex]
[tex] x + y = \dfrac{2xy}{15} [/tex]
[tex] \dfrac{7y + 5xy + 7x}{y + x + 2xy} = [/tex]
[tex] = \dfrac{7x + 7y + 5xy}{x + y + 2xy} [/tex]
[tex] = \dfrac{7(x + y) + 5xy}{x + y + 2xy} [/tex]
[tex] = \dfrac{7(\frac{2xy}{15}) + 5xy}{\dfrac{2xy}{15} + 2xy} [/tex]
[tex] = \dfrac{15}{15} \times \dfrac{7(\frac{2xy}{15}) + 5xy}{\dfrac{2xy}{15} + 2xy} [/tex]
[tex] = \dfrac{14xy + 75xy}{2xy + 30xy} [/tex]
[tex] = \dfrac{89xy}{32xy} [/tex]
[tex] = \dfrac{89}{32} [/tex]
3.
[tex] \dfrac{x - 3}{x^3 + 3x} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 3} [/tex]
[tex] = \dfrac{A(x^2 + 3)}{x(x^2 + 3)} + \dfrac{(Bx + C)x}{(x^2 + 3)x} [/tex]
[tex] = \dfrac{Ax^2 + 3A}{x(x^2 + 3)} + \dfrac{Bx^2 + Cx}{(x^2 + 3)x} [/tex]
[tex] = \dfrac{Ax^2 + 3A + Bx^2 + Cx}{x(x^2 + 3)} [/tex]
[tex] \dfrac{x - 3}{x^3 + 3x} =\dfrac{(A + B)x^2 + Cx + 3A}{x(x^2 + 3)} [/tex]
[tex] A + B = 0 [/tex]
[tex] C = 1 [/tex]
[tex] 3A = -3 [/tex]
[tex] A = -1 [/tex]
[tex] -1 + B = 0 [/tex]
[tex] B = 1 [/tex]
[tex] A + B + C = -1 + 1 + 1 = 1 [/tex]
