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Find the integral of [ 2x^4 - (4/sqrt(x) ) + 3 ], giving each term in its simplest form

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...

Answered by Prahlad M. • Maths tutor

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y = p x q^(x - 1), When x = 1, y = 10, and when x = 6, y = 0.3125. Find the value of 'y' when x = 3

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...

Answered by Prahlad M. • Maths tutor

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Write 16 × 8^2x as a power of 2 in terms of x

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

Answered by Prahlad M. • Maths tutor

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Express the following in partial fractions: (x^2+4x+10)/(x+3)(x+4)(x+5)

(x^2+4x+10)/(x+3)(x+4)(x+1) = A/(x+3) + B/(x+4) + C/(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)]/(x+3)(x+4)(x+1) = (x^2+4x+10)/(x+3)(x+4)(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)] = (x^2+4x+10) A(...

Answered by Alessandro S. • Maths tutor

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Curve D has equation 3x^2+2xy-2y^2+4=0 Find the equation of the tangent at point (2,4) and give your answer in the form ax+by+c=0, were a,b and c are integers.

Differentiate the equationdy/dx=6x+2x(dy/dx)+2y-4y(dy/dx)Set this equal to zero and solve for dy/dx which gives:dy/dx=(2y+6x)/(4y-2x)For x=2 and y=4 dy/dx=5/3(y-y_o)=m(x-x_o)y-4=(5/3)(...

Answered by Vasileios D. • Maths tutor

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