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differentiating: xy'+y-2x+(x/y)y'+log(y)=0rearranging: y'=y(2x-y-log(y))/x(1+y)at (2,1): y'=3/4 so gradient of normal at (2,1) is -4/3so the equation of the normal is y-1=(-4/3)(x-2)which is equivalent to...
Because both of the equations are equal to y, the first thing we do is make them equal to each other:x^2 + 3x= x + 8We then want to rearrange the equation so that everything is on the same side of the equ...
First express as a single logarithm as follows. The number in front of the logarithm remembering log rules can be rewritten as the power of the number in the bracketsSo rewriting the LHS log3TSAnswered by Theranjit S. • Maths tutor7827 Views
first recognise that if you double the first equation the both equations will be equal to 8.so now we have 6x + 10y = 8 ; 7x - 3y = 8now we can set the equations equal to each other to get 6x + 10y = 7...
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