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Maths
A Level

Find the equation of the tangent to the curve y = (2x -3)^3 at the point (1, - 1), giving your answer in the form y = mx + c.

y = (2x -3)^3

y = (2x)^3 + 3.((2x)^2)(-3) + 3.(2x).(-3)^2 + (-3)^2 using Pascal's Triangle.

y = 8x^3 - 36x^2 + 54x - 27 

dy/dx = 24x^2 - 72x + 54

at point (1,-1); dy/dx = 24 -7...

RS
Answered by Robert S. Maths tutor
12203 Views

Find the first derivative of y=2^x

There is an initial subtle difficulty to this question, and it highlights understanding of the relationship between natural logarithms and the exponential function. One of the ways to solve this question,...

AM
Answered by Alex M. Maths tutor
3927 Views

Find the general solution of 2 dy/dx - 5y = 10x

Try y=Aebx diffrentiate this (dy/dx = Abebx) and sub into 2dy/dx -5y = 0 to find complementary function. 2Abebx - 5Aebx = 0 2b - 5 = 0 b = 2.5 Find the particul...

AH
Answered by Amy H. Maths tutor
4762 Views

Given that y > 0, find ∫((3y - 4)/y(3y + 2)) dy (taken from the Edexcel C4 2016 paper)

This can't be integrated directly as y appears in the numerator and denominator. This is an indication that you must integrate by parts. A/y + B/(3y+2) = (3y - 4)/y(3y + 2) A and B must be found. Multiply...

SS
Answered by Saskia S. Maths tutor
11019 Views

The complex conjugate of 2-3i is also a root of z^3+pz^2+qz-13p=0. Find a quadratic factor of z^3+pz^2+qz-13p=0 with real coefficients and thus find the real root of the equation.

z-2+3i times z-2-3i = z2-4z+13. z3-2z2+5z+26 divided by z2-4z+13 = z+2. Therefore the real root is z=-2.

WN
Answered by William N. Maths tutor
5453 Views

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