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A curve has parametric equations x=t(t-1), y=4t/(1-t). The point S on the curve has parameter t=-1. Show that the tangent to the curve at S has equation x+3y+4=0.

To anwser this question we need to find a linear equation of the form y=mx+c which we can rearrange to give the desired equation. Firstly, we must find the gradient at the point S given by dy/dx. Using th...

Answered by Marcus F. Maths tutor
7310 Views

A "day return" train ticket is £6.45. A "monthly saver" is £98.50. Sue worked for 18 days last month. She bought a day return ticket each day she worked. A monthly saver ticket is cheaper than 18 day return tickets. How much cheaper?

This is a simple GCSE question which might feature at the beginning of the paper. It tests the student's attention to detail by asking to find the DIFFERENCE and not just the price that Sue paid. Also, gi...

Answered by Ecaterina A. Maths tutor
5322 Views

Find the value of the discriminant of x2 + 6x + 11

For a quadratic equation in the form ax^2+bx+c=0 the determinant is given by b^2-4ac and is used to help identify the types of roots of the equation. In this case, the determinant is (6)^2-4(1)(11) which ...

Answered by Emily C. Maths tutor
8170 Views

Find the derivative of the function y=3x^2e^(2x)sin(x).

y is the product of three different function, so we would use the product rule in order to calculate the derivative of the curve. In order to apply the product rule we need to find the derivatives of each...

Answered by Shreya R. Maths tutor
5892 Views

f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.

(a) Let y=f(x). Then y = (2x+1)/(x-1). Rearrange the equation to get x in terms of y to obtain the inverse function. This gives x=(1+y)/(y-2). So the inverse of f is f-1(x)=(1+x)/(x-2)<...

Answered by Lutfha A. Maths tutor
4385 Views

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