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Cell biology
Biological molecules
Organisation
Infection and response
Energy transfers (a2 only)
Homeostasis and response
Responding to change (a2 only)
The control of gene expression (a-level only)
Substance exchange
Bioenergetics
Genetic information & variation
Inheritance, variation and evolution
Genetics & ecosystems (a2 only)
Ecology
Cells
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1c the tudors: england, 1485-1603
1l the quest for political stability: germany, 1871-1991
Inter-war germany
1f industrialisation and the people: britain, c1783-1885
Britain & the wider world: 1745 -1901
2n revolution and dictatorship: russia, 1917-1953
2j america: a nation divided, c1845-1877
The cold war
World war two & the holocaust
World war one
Medieval period: 1066 -1509
The fight for female suffrage
2m wars and welfare: britain in transition, 1906-1957
2d religious conflict and the church in england, c1529-c1570
Britain: 1509 -1745
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09/11/2022
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maxima OPTIMISATION minima Ex A rectangular area, has to be fenced off with a total length of 280m of fencing. a) Given that x is the length of the rectangle, show that the area is given by A (x) = 140x-x² x x 19 Area = xy Perimeter = 2x +2y Watch for factorising A'(x)=140-2x For st pts (et A'(x) = 0 140-2x = 0 140=2x x=70 y=140-x Hink for a MAX b)Hence find the greatest area that can be enclosed and the dimensions of the rectangle. A(x) = 140x-x² x60 70 80 A') + O Slope V=lxbxh 108=x*x*y 108=x²y y=10€ S(x)=x²+432 S(x)=x²+432x" S'(x)=2x-432x=² S'(x) = 2x - 432 For st pts let s'(x) = 0 * 2x - 432=0 Perimeter = 280m 280=2x+2y 2y=-2x+280 Ex2 Open rectangular bins with square bases and a volume of 108m² are to be made using a minimum of metal. If x metres is the length of a side of the base and height of each bing show that the total surface area of each bin is given by S(x) = x² + 43²282 6) Calculate the dimensions of then bin so the metal used is a minimum. metres is the y Total SA 5(x)=x² +- 2x = 432 2x³=432 x³ =216 x = = 3√216 x=6 Base = xxx = x² Sides = 4* x*y * S = 4x+10 x 5¹(x) Slope S= 432 x 1 610 - 0 :.Max at x = 70 Area =x (140-x) A(x) = 140x-x² +y=108 • min at x = 6 Use in area Find dimensions At x=70 A(x) = 140x-x² A (70) - 140x70-(70) ² = 9800 - 4900 = 4900 • The greatest...
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area is 4900m² and the length is 70m by 70m y value S(x) = x² + 432 5 (6) = 6² +432 6 S(6)=36+72 S(6)=108m 432 Final dimension y=-1083/2 y=108280 y=108808 y=3m :. The minimum SA is 108m² when the length is 6m and breath is 6m and height is 3m