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The line AB has equation 5x+3y+3=0. The line AB is parallel to the line with equation y=mx+7 . Find the value of m.

Rearrange first equation to show y=-5/3x-1 lines are parallel therefore gradient is the same m=-5/3
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Answered by Isabel C. • Maths tutor
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Find the curve whose gradient is given by dy/dx=xy and which passes through the point (0,3)

First "Separate the Variables" by rearranging the equation to get the ys on the LHS and the xs on the RHS: (1/y) dy=x dx Now Integrate: Integral(1/y) dy = Integral(x) dx ln(y)=x 2 /2 + constant of ...
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Answered by Christian C. • Maths tutor
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Express 3/2x+3 – 1/2x-3 + 6/4x^2-9 as a single fraction in its simplest form.

First it is necessary to notice that 4x^2-9 can be written as (2x-3)(2x+3). To solve this question, you first have to write all the fractions in terms of their lowest common denominator. In this case that is...
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Answered by Dhian S. • Maths tutor
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By using partial fractions, integrate the function: f(x) = (4-2x)/(2x+1)(x+1)(x+3)

(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1) + B/(x+1) + C/(x+3) 4-2x = A(x+1)(x+3) + B(2x+1)(x+3) + C(2x+1)(x+1) let x = -1: 4-2(-1) = B(2(-1)+1)((-1)+3) 6 = B(-1)(2) B = -3 let x = -3: 4-2(-3)= C(2(-3)+1)((-3)+1) 10...
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Answered by Oliver F. • Maths tutor
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The velocity of a moving body is given by an equation v = 30 - 6t, where v - velocity in m/s, t - time in s. A) What is the acceleration a in m/s^2? B) Find the expression for the displacement s in terms of t given the initial displacement s(0)=10 m.

A) Acceleration is the rate of change of velocity with respect to time; therefore, in order to calculate it we need to differentiate the given equation of velocity v with respect to time t: a = dv / dt = d( ...
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Answered by Krisjanis P. • Maths tutor
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