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Student uses the definition of area [A = 1/2 integral r(theta)^2 d theta], and proceeds using standard integration techniques to give a quadratic solvable for alpha. [alpha^2 = 25] Thus, alpha = 5.

2x²y + 2x + 4y - cos(πy) = 45Applying implicit differentiation:4xy + 2x²(dy/dx) + 2 + 4(dy/dx) + πsin(πy)(dy/dx) = 0Moving all (dy/dx) terms to one side:2x² (dy/dx) + 4(dy...

We begin by rewriting it in a more workable form: 2x⁴ - 4x^-1/2 + 3. Indices are easier to integrate than fractions.Now, we integrate each term separately. The first term is 2x^4...

When x = 1, y = 10Therefore, p x q^{(1 - 1)} = 10 p x q⁰ = 10 p x 1 = 10, p = 10When x = 6, y = 0.3125 p x q^{(6 - 1)}= 0.3125 10 x q⁵ = 0.3125 q⁵ = 0.0312...

16 = 4² = (2²)² = 2⁴8 = 2³ , therefore 8^2x = 2^3(2x) = 2^6x2⁴ x 2^6x = 2^{4 + 6x}