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But in this case, 1/f(x) is cosx which is defined for for all x. If however you where to do f(x)=lnx, then there would be an open circle at x=0. Atleast this is my understanding of what your ment to graph?

From the diagram, the rope is 15 m long. i.e 8-y+hypotenuse=15 therefore, hypotenuse=y+7 and by pythagorus, x^2 +6^2 =(y+7)^2 When y=3, x=8. Also it is given that dx/dt=2.5m/s x^2+6^2=(y+7)^2 \\ \textup{implicitly differentiating},2x=2(y+7)\frac{\mathrm{d} y}{\mathrm{d} x} \\ \therefore...

For the cone one we know; dh/dt= 2cm/min at h=16cm, and tanA=3/4 now, V=1/3pir^2h, if you draw a pretty picture you should relise that r/h=3/4 i,e. r=3h/4 and V=1/3pi(3h/4)^2 h therefore, dV/dh=9/16pih^2 and at h=16, dV/dh=144cm^3/min now, dV/dt=dV/dh*dh/dt therefore...

Could you be more specific or give us an example? A quadratic will have two roots, not necesarrily unique or real.

h(n+2)-h(n)=(n+2)^4+6(n+2)^2+9-n^4-6n^2-9\\=[(n+2)^2-n^2][(n+2)^2+n^2]+6[(n+2)-n][(n+2)+n] \\=(4n+4)(2n^2+4n+4)+12(2n+2)\\=8(n+1)[(n^2+2n+2)+3]\\=8(n+1)(n^2+2n+5) Now that took alot of time. Too lazy to do the induction part, but its fairly simple.

Black wattle represent.

x^3+y^3=3xy \\ (x+y)(x^2-xy+y^2)=3xy \\ x+y=\frac{3xy}{x^2-xy+y^2}\\ \therefore x+y=\frac{3}{x/y -1 +y/x} \\From \;previous, as\; x\to \infty, y/x\to-1 \\it\; follows\; then\; that \;x+y\to-1, as \;x\to\infty

x^3+y^3=3xy \\ 1+(\frac{y}{x})^3=3 \frac{y}{x^2}\\ \frac{1+(\frac{y}{x})^3}{\frac{y}{x}}=\frac{3}{x} as\; x\to\infty, RHS\to0 \\ \therefore LHS\to0, \;i.e \;y/x\to-1

(-1)^2 =1 ... so yes, the square root of 1 is 1. I;m pretty sure there is a difference between \sqrt{-1^2} \; and \; (\sqrt{-1})^2

jm01, check your solution i think you made a mistake (after second use of integration by parts). using the sums to product identity; \int (sin5xcos3x)dx = \int \frac{(sin8x+sin2x)}{2}dx \\ =\frac{1}{2}\left [ -\frac{1}{8}cos8x -\frac{1}{2}cos2x \right ] +C \\ =-\frac{1}{16}\left [...

do you mean how do you graph arcsin(x^2) arccos(x^2), etc. It's the same process for any other function, its mirrored around the y axis, and its similar to f(x) except its sort of squished around x=1,-1, with the f(x) and f(x^2) intersecting at x=0, +-1. go to wolframalpha.com and just...

Wow, he seems worse then my engineering teacher... I'm pretty sure he cant change the weighting after the assignments have been submitted, so i would complain to the principle and if not the board of studies, or atleast acquire about it. Goodluck with your dodgy engineering teacher, it...

I think the question is wrong. LHS=(1+cos^2(\frac{x}{2})-sin^2(\frac{x}{2})+2icos(\frac{x}{2})sin(\frac{x}{2}))^n \\=(2cos^2(\frac{x}{2})+2icos(\frac{x}{2})sin(\frac{x}{2}))^n \\ =2^ncos^n(\frac{x}{2})(cos(\frac{nx}{2})+isin(\frac{nx}{2})), by \;De \;Moivre's \;Theorem

for ii) the question refers to real parts, not imaginary parts. continueing from shaon0, using the sum of roots we get x+iy+x-iy+A+iB+A-iB+R=0, where R is the real root. 2x+2A=-R from i) we know p(0)=-c<0, therefore, R>0 and -R<0 It follows then that either x, A or both are <0.

11a) x^3+y^3=8 \\ 3x^2+3y^2y'=0 \\ \therefore y'=\frac{-x^2}{y^2} \\When \;x=0, \;y=2, \;y'=0 \\ When \;y=0,\;x=2,\; y'= \infty \\When\; y=x,\;y'=-1 (i didn;t find the equations of the tangents, just the gradient at those points. I might do the rest of the questions later when i have time...

This is question 17. in chapter 10 of the cambridge year 12 3 unit book. Question: Bob is about to hang his eight shirts in the wardrobe. He has four different styles of shirt, two identical of one of each particular style. How many different arrangements are possible if no two identical...

perpendicular distance?

For the last part, notice that y=ln(2x+1)\Leftrightarrow x=\frac{e^y-1}{2} also, when y=ln5, x=2. when y=0, x=0. Since the integral is the area underneath the function, (when its with respect to y, its the area the function makes with the y-axis) \Rightarrow...

Really?, damn. D: i guess im abit tired. probably shouldn;t write an english essay at the same time as doing math then. >< edit: wait nvm, i saw it.

y=\frac{(x+1)\sqrt{x-1}}{x+2} \\ ln(y)= ln(\frac{(x+1)\sqrt{x-1}}{x+2})\\ ln(y) =ln(x+1)+\frac{1}{2}ln(x-1)-ln(x+2)\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x+1}+\frac{1}{2(x-1)}-\frac{1}{x+2} \\ \text{multiplying both sides by y, canceling common factors} \\ \frac{dy}{dx}=...