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Maths
GCSE

A is the point (4,1). B is the point (10,15). Find the perpendicular bisector or of AB.

First, find the gradient, m, of the line AB. Gradient is given by: change in y / change in xChange in y = 15-1 = 14Change in x = 10-4 = 6Gradient = 14/6 = 7/3The ...

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Answered by Ollie W. Maths tutor
7510 Views

Find the coordinates of the turning point of the curve y=x^2+3x+7

We know that turning points occur when the gradient is equal to zero. Hence, we differentiate this curve. dy/dx = 2x+3 and we set this equal to zero. This gives 2x+3=0, we then rearrange to get 2x=-3 and ...

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Answered by Rosie H. Maths tutor
4860 Views

Find the roots of the following equation x^2 + 6x + 5 = 0

There are a few ways to do this, firstly you notice that there is no coefficient in front of the x^2. This means you can factorise it with relative ease, by finding numbers that add to give 5 but multiply...

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Answered by Callum H. Maths tutor
3064 Views

The equation of a curve is y=(x+3)^2 +5, what are the co-ordinates of the curve's turning point?

Differentiate y with respect to x:dy/dx = 2(x+3) = 2x+6When the above equation is equal to 0, this is where the turning point of the curve is.2x+6 = 02x = -6x = -3Therefore, at x = -3, the curve has a tur...

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Answered by Harry H. Maths tutor
3599 Views

How do I rearrange and make y the subject in equations such as "(y/4) - X = 1"?

The aim is to finish with an equation that looks like "y = .....". We want Y and X on separate sides.The main thing to remember is that whatever you do to one side of the equation, you also do t...

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Answered by Dulcie S. Maths tutor
3318 Views

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