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Find the x coordinate of the stationary points of the curve with equation y = 2x^3 - 0.5x^2 - 2x + 4

Firstly, to find the stationary points of a curve you must differentiate the equation of the curve. To do this each x component is multiplied by its current power and then the power is decreased by one. A...

Answered by Bartosz S. Maths tutor
4762 Views

Insert a pair of brackets into this question to make it correct 2 + 5 x -6 = -42

(2+5) x -6 = 42

Answered by Inumidun O. Maths tutor
3219 Views

Find two solutions to the quadratic equation x^2 + 2x - 15 = 0

Firstly, we need to find two numbers that multiply together to make -15 and sum together to make +2. We try a few numbers and find that the solution is -3 and 5, as -3 x 5 = -15 and -3 + 5 = 2.Next you in...

Answered by James G. Maths tutor
3671 Views

Use the substitution u = cos 2x to find ∫(cos^2*(2x) *sin3 (2x)) dx

∫(cos2 2x *sin3 2x)dx u = cos2x - u =(du/dx) = -2sin2x - differentiate u dx = du/(-2sin(2x)) - dx = -1/2 ∫cos22x * sin22x du - sub in dx-1/2 ∫u2(1-u<...

Answered by Will B. Maths tutor
7060 Views

Does the equation x^2 + 2x + 5 = 0 have any real roots?

This equation has no real roots, as when we test the equation with the formula b2 - 4ac, we get 22 -415 = 4 - 20 = -16. As this is negative we can be assured that this equat...

Answered by Sami B. Maths tutor
4303 Views

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