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Binomially expand the equation (2+kx)^-3

(2+kx)^-3 = (2^-3)(1+kx/2)^-3= (2^-3)(1+(-3)(kx/2) + [(-3)(-4)]/2! (kx/2)² + [(-3)(-4)(-5)]/3! (kx/2)³+... )
= 1⁄8 [1 -(3kx/...

Answered by Christopher H. • Maths tutor

9096 Views

How do you find the integral of (2+5x)e^3x ?

If we were to expand the brackets for this question we might be able to get to an answer however there is a much simpler way of solving the problem. As this question is made up of two expressions of x mul...

Answered by Olivia C. • Maths tutor

5830 Views

If a particle of mass m is launched vertically upwards from the ground with velocity u m/s, how long will it take to return to the ground in terms of m, u and g?

Taking upwards to be positive, and using the 'suvat' equation s=ut+1/2at² we know that u=u, a=-g and s=0 when the particle returns to the ground. Then we solve for t:
0=ut-1/2gt²

Answered by Jackie V. • Maths tutor

3167 Views

Use the substitution u=x^2-2 to find the integral of (6x^3+4x)/sqrt( x^2-2)

First use the substitution to find du/dx which is 2x. From this we now know that dx= du/2x (just re-arranging.) Substituting that into the integral we now get (6x³+4x)/ (sqrt (u) x 2x) du. Canc...

Answered by Khalil S. • Maths tutor

9723 Views

Let f(x)=x^3 - 2x^2 + 5. For which value(s) of x does f(x)=5?

Although it can be tempting to dive straight into such a question, its always useful in mathematics to take a step back and assess the question before scribbling down panicked answers.

For inst...

Answered by Jed G. • Maths tutor

3483 Views