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Sketch the curve y = x^2 - 6x + 5, identifying roots and minima/maxima.

Remeber the formula: (a - b)² = a² - 2ab + b². Notice that y = x² - 23x + 5, so we want to write this using (x - 3)

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a) Let u=(2,3,-1) and w=(3,-1,p). Given that u is perpendicular to w, find the value of p. b)Let v=(1,q,5). Given that modulus v = sqrt(42), find the possible values of q.

a) u is perpendicular to w, so u • w =0. This means that 23 + 3(-1) + (-1)*p = 0, so 6-3-p=0, hence p=3. b) The magnitude of v ...

Answered by Cristiana R. • Maths tutor

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Solve for x 2x +3 + (4x-1)/2 = 10

2x +3 + (4x-1)/2 = 10 The divisor (2) is the biggest problem right now so we need to get it on its own...

(4x-1)/2 = 10-2x-3 Equation has been rearranged!

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Answered by Shannon H. • Maths tutor

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Use the substitution u=3+(x+4)^1/2 to find the integral of 1/(3+(x+4)^1/2) dx between 0 and 5.

We will call the integral I, so I = integral of 1/(3+(x+4)^1/2) dx between 0 and 5. First substitute u=3+(x+4)^1/2 into the equation to get I = integral of 1/u dx between 0 and 5 Next ...

Answered by Calum B. • Maths tutor

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Find the centre and radius of the circle with equation: x^2 + y^2 -4x +8y = 5, and determine whether the point (7,-4) lies on the circle.

First we complete the square on the equation of the circle to obtain: (x-2)^2 -4 + (y+4)^2 -16 = 5. Re arrange : (x-2)^2 + (y+4)^2 = 25 General equation of a circle: (x-a)^2 + (y-b)^2 = r^2, with centre ...

Answered by Alec S. • Maths tutor

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