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i) Make y the subject of the expression x = ((a-y)/b))^1/2 ii) Simplify fully (2x^2 − 8)/(4x^2 − 8x)

i) Square both sides of the expression to make x^2=(a-y)/b

Multiply by b to remove the fraction from the equation and then rearrange to make y the subject.

bx^2=a-y --> y=a-bx^2

i...

Answered by Oscar S. • Maths tutor

4179 Views

Differentiate x^3 − 3x^2 − 9x. Hence find the x-coordinates of the stationary points on the curve y = x^3 − 3x^2 − 9x

To differentiate, we bring the power down and decrease the power by 1. So x³ becomes 3x², -3x² becomes -6x, and -9x (which can be written as -9x¹ ) becomes -9. ...

Answered by Tutor105800 D. • Maths tutor

9746 Views

A semicircle has a diameter of 8cm, what it the area?

Area of a circle = pi × r²

Area of semi-circle = 1/2(Area of circle)

= 1/2(pi × r²)

Diameter = 2r

r = 8÷2 = 4

Area of semi-circle = 1/2(pi × 4

Answered by Annabelle P. • Maths tutor

8084 Views

Evaluate the following integral: (x^4 - x^2 +2)/(x^2(x-1)) dx

This type of question appears over-complicated with limited options, but one must not fear! At first, the numerator seems to be of a higher degree than the denominator (x^4 compared to x^2), but from the ...

Answered by Nicholas H. • Maths tutor

5352 Views

Differentiate the function f(x) = 3x^2/sin(2x)

Using the product rule, f=uv, df = (vu'-uv')/v^2. we first set u = 3x^2 and v = sin(2x). u' = 6x, v'=2cos(2x) Therefore, vu' = 6xsin(2x). uv' = 6x^2cos(2x), v^2 = 4cos^2(2x) Therefore the differe...

Answered by Kilian S. • Maths tutor

5669 Views