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Find the area under the curve of y=1/(3x-2)^0.5 between the limits x=1 and x=2 and the line y=0

This question requires integration since the area under the curve is equal to the integral between these bounds. Initially let u=3x-2 and differentiate with respect to x so then du/dx = 3. Rearrange to dx...

Answered by Callum T. • Maths tutor

3154 Views

At time t = 0 a particle leaves the origin and moves along the x-axis. At time t seconds, the velocity of P is v m/s in the positive x direction, where v=4t^2–13t+2. How far does it travel between the times t1 and t2 at which it is at rest?

First, you have to work out the values of t1 and t2 at which the particle is at rest. This is done by solving the quadratic equation for v, producing values for t of 13/8 +- sqrt(137): 0.1619s and 3.088s....

Answered by Jack M. • Maths tutor

5928 Views

Given that, dy/dx = 6x^2 - 3x + 4, and y = 14 when x = 2, express y in terms of x.

dy/dx = 6x² - 3x + 4
To retrieve the original function y from dy/dx you have to integrate the derivative with respect to x.
y = ∫(dy/dx)dxy = ∫(6x² - 3x + 4)dx
To inte...

Answered by Ciaran B. • Maths tutor

5756 Views

Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)

We must first complete the square for both equations and get the equations in the (x-a)²+(y-b)²=r² with the primary objective of determining the centres of both circles (a...

Answered by Shanti S. • Maths tutor

3394 Views

Factorise x^2 - 6x + 9 = 0

x²- 6x + 9The method for factorising quadratics is to find 2 numbers that add to make the 'x' term (-6x in this case), and multiply together to make the final term (+9 in this case). Factors o...

Answered by Kate B. • Maths tutor

7677 Views