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Show how you can rewrite (x+1)(x-2)(x+3) into the form of ax^3 + bx^2 + cx + d

Split the first equation into three parts, i.e. (x+1), (x-2) and (x+3). Multiply the first two parts to get x 2 - x - 2 , then multiply the result with the third part to get x 3 + 2x 2 - 5x - 6 . All that is...
GM
Answered by Gustas M. • Maths tutor
4360 Views

Find the solutions of the equation: sin(x - 15degrees) = 0.5 between 0<= x <= 180

Consider a sin graph and all the points on the graph which are equal to 0.5. You'll find that, between one period of a positive sin graph, invsin(0.5) may equal 30 degrees or 150 degrees. The equation can no...
MT
Answered by Mahir T. • Maths tutor
4905 Views

What is differentiation?

Differentiation is a way of measuring a rate of change. Formally, we can look at two points on a curve and as an example, I would use the x^2 curve. We can plot on a point (x,x^2) and a point (x+h, (x+h)^2)....
IT
Answered by Ivana T. • Maths tutor
3574 Views

Integrate sin7xcos3x

Let A=7x and B=3x( From the formulae we know that sinxcosx= 1/2(sin(A+B)+sin(A-B))So if we replace A=7x and B=3x in this equation: 1/2(sin(7x+3x)+sin(7x-3x))= = 1/2 (sin(10x)+sin(4x))Now we just integrate 1/...
FP
Answered by Frances P. • Maths tutor
6104 Views

Given that y= 1/ (6x-3)^0.5 find the value of dy/dx at (2;1/3)

Let u=6x-3 , then y=u^-0.5hence, du/dx=6 and dy/du= -0.5u^-3/2then, as dy/dx =dy/du * du/dx dy/dx=(-0.5u^-3/2 )*6= -3(6x-3)^-3/2substitute x=2 to give the required value required value : -1/9
PN
Answered by Polina N. • Maths tutor
4368 Views