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Maths
A Level

What are the solutions of (x^3)+6 = 2(x^2)+5x given x = 3 is a solution?

Firstly we will put this into a form equal to zero, by rearranging to get x3-2x2-5x+6 = 0. This is because in order to solve a polynomial we first need to set it equal to zero. We no...

WB
Answered by William B. Maths tutor
6341 Views

When using the trapezium rule to approximate area underneath a curve between 2 limits, what is the effect of increasing the number of strips used?

Ideally to find the exact area under the curve, we would integrate the function and substitute in the bounds given. However, using the trapezium rule gives an approximation whereby using more trapezia inc...

MM
Answered by Manan M. Maths tutor
9769 Views

Prove that sin(x)+sin(y)=2sin((x+y)/2)cos((x-y)/2)

We know that 1. sin(a+b) = sin(a)cos(b)+sin(b)cos(a) and 2. sin(a-b) = sin(a)cos(b)-sin(b)cos(a) Add equations 1. and 2. sin(a+b)+sin(a-b) = 2sin(a)cos(b)+sin(b)cos(a)-sin(b)cos(a) = 2sin(a)cos(b) Let x=a...

AV
Answered by Anna V. Maths tutor
35787 Views

The points A and B have coordinates (3, 4) and (7, 6) respectively. The straight line l passes through A and is perpendicular to AB. Find an equation for l, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

For the line passing through A and B: m = (y2-y1)/(x2-x1) = (-6-4)/(7-3) = -5/2

For the perpendicular line: m = -1/(-5/2) = 2/5 

y - y1 = m*(x - x1)  >>  y - 4 = (2/5)*(x - 3)  >>...

DA
Answered by Deji A. Maths tutor
12214 Views

Find the gradient of the tangent to the curve with the equation y = (3x^4 - 18)/x at the point where x = 3

y = (3x- 18)/x

The gradient of a tangent to a curve is equal to dy/dx 

However, we must simplify this equation before we can differentiate it;

y = 3x3 - 18/x =...

RO
Answered by Rachel O. Maths tutor
4563 Views

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