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Hi everyone, I've got a few First Year Commerce textbooks up for sale. The books are listed below: ECON1101 - Principles of Microeconomics 2E (very good condition with some useful highlighting) $65 (RRP: $112.46) ACCT1501 - Financial Accounting An Integrated Approach 4E (very good...

(x+y)^2=1\\x^2+y^2=1-2xy\geq 1-2(\frac{1}{4})=\frac{1}{2}\\Also,\frac{1}{xy}\geq \frac{1}{4}\Rightarrow \frac{1}{x^2y^2}\geq \frac{1}{16}\\(x+\frac{1}{x})^2+(y+\frac{1}{y})^2\\=x^2+y^2+\frac{x^2+y^2}{x^2y^2}+4\\\geq \frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{16}}+4\\=\frac{25}{2}

lolol

If you do any combined degree, you still have to complete the UNSW Part II courses ACTL4001 and ACTL4002 along with your combined degree if you want the part II exemptions. Also, in order to enrol for ACTL4001 and ACTL4002 you need to first: 1) complete your part I 2) average mark of at least...

Alternatively, Consider y=-kx^2 for -2h\leq y\leq 0 when y=-h,x=\pm \sqrt{\frac{h}{k}} when y=-2h,x=\pm \sqrt{\frac{2h}{k}} D1 (distance travelled by stone) =\sqrt{\frac{2h}{k}}+\sqrt{\frac{h}{k}} D2 (distance travelled by bird) =2\sqrt{\frac{h}{k}}...

thanks

Use the fact that f(x)=(ax-b)^2+(cx-d)^2\geq 0 to prove |ab+cd|\leq \sqrt{a^2+c^2}\sqrt{b^2+d^2} (a,b,c,d real) It's obvious if you just use 2abcd\leq a^2d^2+b^2c^2 to prove \sqrt{(ab+cd)^2}\leq \sqrt{(a^2+c^2)(b^2+d^2)} but how would you do the question using f(x) ?

You find y in terms of x and then do the usual differentiation thing. so x/(12-y) = 16/12 (by letting AB=16 and BC=12, works the other way around as well) x/(12-y) = 4/3 3x=48-4y y=(48-3x)/4 A = xy = (48x-3x^2)/4 then differentiate

(i) z^9+1=(z^3+1)(z^6-z^3+1) (ii) roots of (z^3+1) are cis(pi/3), cis(pi), cis(-pi/3) roots of (z^9+1) are cis(pi/9), cis(pi/3), cis(5pi/9), cis(7pi/9), cis(pi), cis(-7pi/9), cis(-5pi/9), cis(-pi/3), cis(-pi/9) so roots of (z^6-z^3+1) are cis(pi/9),cis(5pi/9), cis(7pi/9), cis(-7pi/9)...

yep, gonna have very similar preferences but i cant see the pin number on the letter :confused:

It's just a matter of finding a and b. Sub in x=0 and x=1 and you should have two equations which allow you to find a and b.

holy crap, you've gone 3d. niceee

hm...i actually like jetblacks method, much simpler and less subbing in

f(x)=x^5-ax^2+b\\f'(x)=5x^4-2ax\\let\,root\,be\,x=\alpha \\f'(\alpha)=5\alpha^4-2a\alpha=0\\\alpha^3=\frac{2a}{5}\,since\,\alpha\neq...

quadratic formula would be useful.

(n+1)^4-n^4=4n^3+6n^2+4n+1\\\sum_{n=1}^{N}\left [ (n+1)^4-n^4 \right ]=\sum_{n=1}^{N}\left [ 4n^3+6n^2+4n+1 \right ]\\\sum_{n=2}^{N+1}n^4-\sum_{n=1}^{N}n^4=4\sum_{n=1}^{N}n^3+6\sum_{n=1}^{N}n^2+4\sum_{n=1}^{N}n+N\\(N+1)^4-1=4\sum_{n=1}^{N}n^3+6\left [ \frac{N}{6}(n+1)(2N+1) \right ]+4\left [...

for the first one: (n+1)^3-n^3=3n^2+3n+1\\\sum_{n=1}^{N}\left [(n+1)^3-n^3 \right ]=\sum_{n=1}^{N}\left [3n^2+3n+1 \right...

last question: let A be arg(z1) Using sine rule: |z2 + 1|/sin(180-2A) = |z2|/sinA |z2 + 1| = sin2A/sinA since |z2| = 1 and sin(180-2A)=sin2A =2sinAcosA/sinaA =2cosA =2cos(arg(z1))

if it has 3 real roots, then the y-values of the 2 turning points must be opposite in sign. So: P(1).P(2)<0 (5-k)(4-k)<0 4 < k < 5

Re: New MX2 Game think you gotta consider two cases: when n is even and when n is odd. so when n is even, you take the sin's in pairs (1st and last, 2nd and 2nd last...) and use sums to products.