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What other A Level subjects would be useful towards applying to do a Mathematics degree?

To compliment Mathematics, Further Mathematics would be an ideal subject to take. It is not essential, however the additional practice would benefit your understanding of Mathematics, easing you into even...

VF
Answered by Vene F. Maths tutor
3630 Views

The point P lies on the curve C: y=f(x) where f(x)=x^3-2x^2+6x-12 and has x coordinate 1. Find the equation of the line normal to C which passes through P.

First we must find the y coordinate of the point P: We know the x-coordinate is x1=1 so the y coordinate must satisfy the equation y1=f(1) which gives y1=-7. So we now know P is at (1,-7).

We now n...

KH
Answered by Kieran H. Maths tutor
10949 Views

Find all solutions of x^2-x-6 using the quadratic formula

From the given quadratic, we have a=1, b=-1, and c=-6. We substitute these values into the quadratic formula, x=-b+-sqrt(b^2-4ac)/2a, giving us x=-1+-sqrt(-1^2-41-6)/2*1. This simplifies to x=1+-...

JH
Answered by Jack H. Maths tutor
4191 Views

Express square root of 48 in the form n x square root of 3 , where n is an integer

We know that 3 x 16 = 48 This is equivalent to saying 3 x 4^2 = 48 Therefore the square root of 48 can be written as square root of (4^2 x 3) = 4 x square root 3

HW
Answered by Helena W. Maths tutor
9646 Views

A curve has equation y=x^2 + (3k - 4)x + 13 and a line has equation y = 2x + k, where k is constant. Show that the x-coordinate of any point of intersection of the line and curve satisfies the equation: x^2 + 3(k - 2)x + 13 - k = 0

When we deal with points of interception, this immediately indicates that these two equations have to equal. Therefore, begin by equaling these two equations: x^2 + (3k - 4)x + 13 = 2x + k Bring all figur...

HW
Answered by Helena W. Maths tutor
12388 Views

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