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Why is the gradient of a curve at a point the same as the gradient of the tangent if you can't use gradient formula on a curve?

This may be initially confusing to get your head around but gradient is really just an expression of "rate of change" on a curve the rate of change is always changing at every point, this abou...

BD
Answered by Brendan D. Maths tutor
3927 Views

Solve the following simulateous equations: 3x+y=11 and 2x+y=8

Minus 2x+y=8 from 3x+y=11 so that the y's cancel out (y-y=0).
Then we get the equation 3x-2x=11-8.So we get x=3.We can substitute this value back into one the orginal equations: y=11-3(3)=2 - so y=2....

CG
Answered by Cameron G. Maths tutor
2896 Views

Find the coordinates of the point of intersection of the lines y = 3x - 2 and x + 3y = 1.

(A-level Maths)
Simultaneous equations:
1) y = 3x -2 2) x + 3y = 1
Substitute y = 3x - 2 into equation 2:...

KB
Answered by Kiran B. Maths tutor
6761 Views

The circle C has centre (2,1) and radius 10. The point A(10,7) lies on the circle. Find the equation of the tangent to C at A and give it in the form 0 =ay + bx + c.

Whenever possible it is always a good idea to draw a graph to give a better understanding of what is going on in the question. As we have already been given a point on the tangent, namely A(10,7), it is a...

GG
Answered by George G. Maths tutor
4652 Views

Find partial fractions of : (x+7) / ((x-3)(x+1)^2)

First, separate the algebraic fraction:(x+7) / ((x-3)(x+1)2 = A/(x+3) + B/(x+1) + C/(x+1)2 Multiply both sides by (x-3)(x+1)2: (x+7) = A(x+1)2+ B(x+1)(x+3) + C(...

JV
Answered by Janni V. Maths tutor
2879 Views

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