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First we notice that this is a product of two functions of x, so we are going to use the product rule. Recall (uv)'(x)=u'(x)v(x)+v'(x)u(x). Let u(x)=x^(1/2) and v(x)=ln(3x). We need to find u'(x) and v'(x). ...

a = 0.72727272 100a = 72.72727272 100a - a =72.72727272-0.72727272 = 72 = 99a a =72/99 so 0.72727272 = 72/99 0.72727272 refers to 0.72 where the 72 is recurring.

f'(x)=9x 2 ​+4x, and f''(x)=18x+4 (derivatives) f'(x)=0 at x=0 or x=-4/9 when x=0 f''(x)>0 therefore a minimum value, when x=-4/9 f''(x)<0 and thus a maximum value.

This is solving a quadratic equation. There are two ways we can solve this. Using the quadratic equation or completing the square. 1. Quadratic equation-The equation is in the form ax^2 + bx + c. Where a=1 b...

Start by rearranging the inequality - make sure the sign next to the x 2 term is positive to make it easier: 2x 2 -5x - 12< 0 Next step is to factorise this quadratic. To do this, remember that the genera...