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To answer these questions, students must know the values of cos30 and tan60. If you don't know them off by heart, you can work them out using an equilateral triangle.

cos 30° = adj / hyp =...

x₁ = (-b + (b² - 4ac)^1/2)/2a and x₂ = (-b - (b² - 4ac)^1/2)/2a. So with the values a = 1, b = 3 and c = 2: x₁ = (-3 + (3^{2 <...}

x²+2x+1=0, (x+1)(x+1)=0, (x+1)²=0. So x=-1

f'(x)=( sqrt (x^2+1) - x * ( x / sqrt (x^2 +1) ) ) / (x^2+1) = (x^2 + 1 + x^2) / ( (x^2 + 1) * sqrt ( x^2 + 1) ) =  1 / ( (x^2 + 1) * sqrt (x^2 + 1) ). 

f'(x) > 0 for any x => f is increasing...

The point of intersection is where the lines would cross if we drew them on the same graph, in your exam you may be asked to find the coordinates of this point.

To do this we would first draw a ske...