Rearranging Harder Formulae
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Either paper 3 marks
A solid cone of radius \(r\), height \(h\) and density \(\rho\) has mass \(M\), where
\(M = \frac{ 1 }{ 3 }\pi \rho r^2 h\)
Make \(r\) the subject of the formula.
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Either paper 6 marks(a)
Make \(x\) the subject of \(8x + y = 2x + 9\)
(b)Make \(x\) the subject of \(y\left(x - 3\right) = 3x + 8\)
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Either paper 4 marks
Make \(x\) the subject of \(y = \frac{ x - 3 }{x - 5 }\)
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Either paper 6 marks(a)
Rearrange \(y = \sqrt{ 5x - 14 }\) to make \(x\) the subject.
(b)Rearrange \(y = 2\sqrt{x} + 3\) to make \(x\) the subject.
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Either paper 6 marks(a)
Rearrange \(y = \frac{ 2 }{ x } + 1\) to make \(x\) the subject.
(b)Rearrange \(y = \frac{ 17 }{x - 2 }\) to make \(x\) the subject.
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Either paper 5 marks
A car moving in a straight line starts at a speed of \(u\) metres per second and finishes at a speed of \(v\) metres per second. Its acceleration is \(a\) metres per second squared and it travels a distance of \(s\) metres.
Those four are connected by the formula
\(v^2 = u^2 + 2as\)
(a)Make \(u\) the subject of the formula.
(b)Work out \(u\) when \(v = 11\), \(a = 8\) and \(s = 7\).
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Either paper 4 marks
The time period \(T\) of a pendulum of length \(l\) is given by
\(T = 11\pi\sqrt{ \frac{ l }{ g } }\)
Make \(g\) the subject of the formula.
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Either paper 3 marks
The volume, \(V\), of a solid hemisphere of radius \(r\) is given by
\(V = \frac{ 2 }{ 3 }\pi r^3\)
Find a formula for \(r\).
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Either paper 6 marks(a)
Rearrange \(6x + y = 4x - 2\) to make \(x\) the subject.
(b)Rearrange \(y\left(x + 1\right) = 4x - 5\) to make \(x\) the subject.
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Either paper 4 marks
Make \(x\) the subject of \(y = \frac{ 3x + 5 }{x - 8 }\)
- \(r = \sqrt{ \frac{ 3M }{ \pi \rho h } }\)
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(a) \(x = \frac{ 9 - y}{ 6 }\)
(b) \(x = \frac{ 8 + 3y }{y - 3 }\) - \(x = \frac{ -3 + 5y }{y - 1 }\)
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(a) \(x = \frac{y^2 + 14 }{ 5 }\)
(b) \(x = \frac{\left(y - 3\right)^2}{ 4 }\) -
(a) \(x = \frac{ 2 }{y - 1 }\)
(b) \(x = \frac{ 17 }{ y } + 2\) -
(a) \(u = \sqrt{v^2 - 2as}\)
(b) \(3\ \text{m/s}\) - \(g = \frac{ 121\pi^2 l }{ T^2 }\)
- \(r = \sqrt[3]{ \frac{ 3V }{ 2\pi } }\)
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(a) \(x = \frac{ -2 - y}{ 2 }\)
(b) \(x = \frac{ -5 -y }{y - 4 }\) - \(x = \frac{ 5 + 8y }{y - 3 }\)
Takes a couple of seconds the first time a sheet is made.