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##x^3-3x^2+4x-2## We know that the complex roots always occur in conjugate pairs. One complex root is ##1+i## so there must be its conjugate i.e. ##1-i## as […]

##3## a geometric series can be defined as: ##a_n = a_1*(r)^(n-1)## where ##a_1## is the first value of the series and ##r## is the common ratio. […]

Look at the sequence of differences finding that it is a geometric sequence with common ratio ##2## and hence derive the recursive formula: ##a_1 = 3## […]

Given ##color(white)(XXX)f(color(blue)(x))=color(blue)(x)^2-1## and ##color(white)(XXX)g(color(red)(x))=color(red)(x)+1## Note that ##(f@g)(x)## can be written ##f(g(x))## and that ##(g@f)(x)## can be written ##g(f(x))## ##(f@g)(x) = f(color(blue)(g(x))) = color(blue)(g(x))^2-1## ##color(white)(XXXXXX)=(color(blue)(x+1))^2-1## ##color(white)(XXXXXX)=x^2+2x## Since […]

##f(x)=x^3-5x^2+x-5## I have assumed that ##{5i-i}## are intended to be the of the function. In this case the function can be factored as: ##color(white)(XXX)(x-5)((x-i)(x+i)## ##color(white)(XXX)=(x-5)(x^2-i^2)## ##color(white)(XXX)=(x-5)(x^2+1)## […]

##log 12 ~~ 1.0792## Since I remember (as is useful to remember): ##log 2 ~~ 0.30103## ##log 3 ~~ 0.47712125## I can calculate: ##log 12 = […]

you need first to try and find the value for both x and y hence initially you set ##y=0## then solve for ##x##. lets try to […]

##x=7/9 y= -1/9 z=-2/3##

For ellipses ##a >= b## (when ##a = b## we have a circle) ##a## represents half the length of the major axis while ##b## represents half […]