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What about say y=x^(1/3). Isnt there some restriction on how the gradient of the tangent vs the gradient of the graph? I.e. at some pts on the graph, near the root, im sure it wont acquire better estimate. Note: its due to tangent reaching a pt on the x axis, that is further away from the...

I dont get this question at all. Prove that if the ratio \frac{z-1}{z-i} \\ is purely imaginary, the point z lies on the circle whose centre is at the point 1/2(1+i) and whose radius is 1/sqrt 2 Shouldnt it it mean the ratio \frac{z-1}{z-i} \\ = Z (upper case) (since z-1/z-i is a complex...

countless moles, and no dimples :(

.... i dont wanna go back!!!! need more time...

Thxs Gurmies, much appreciated. Ive got more tho. \text{If } (x+iy)^{n} = a + ib\\ \text{Find } a^2+b^2

Re: Sydney Easter Show Do people at uni have time to work at Easter Show?

Oh my tireless soldier of Christ. Only if the world was filled with your sprit, your zeal for the Christ and His name, then it would truly be Eden on Earth.

n=5\times 12=60\\ \text{Rate Increase} = (\frac{0.15}{12}+1)^{12}\Rightarrow r \\ \text{Since percentage rate is charged monthly, with repayment only yearly. Hence compound interest of month will apply for a year}\\ \text{Rest ensues}\\ P_{1} = 6000r - R \\ P_{2} = 6000r^2 - Rr -R\\ P_{3} =...

Um, sorry, its 0.18. You can use whatever. I just didnt wanna type it again and again so i juts used a sub. Yea i get your point addikaye03. My fault. Im too lazy and careless, unlike you:D.

Using Newton's Method, under what conditions will a better estimate NOT be reached?

T_{1}=2000 \times (1+\frac{0.18}{12})^3 = 2091.35 \rightarrow A n=4\times 12 - 3 = 45 T_{1}=2000 \times (1+\frac{0.18}{12})^3 = 2091.35 \rightarrow A (1+\frac{0.18}{12}) \Rightarrow B P_{1} = A\times B - R\\ P_{2} = A\times B^2 - R \times B - R\\P_{3} = A\times B^{3} - R\times B^{2} -...

6 Last step where it says, a+b+c>0, and it assumes it so. But wouldnt that mean solution only applies to positive integers?

Graph is concaving down. f''(x)<0 = the slope of f'(x) <0 Stationary Pts of f(x). When f'(x) = 0, there is a turning pt. If it is a decreasing from right to left, then there is a gradient at right and negative gradient at left. A pt of inflexion. When f''(x) = 0 a pt of inflexion exist...

What about Divide x^3-2-2i by x+1-i

Also Find x in the domain 0<x<pi/2 \frac{\sqrt{2}(cosx-isinx)}{{2+i}} = \frac{1-3i}{5}

Question on Complex Nos. (x+iy)^{1/3} = X + iY Show that 4(X^2-Y^2) = \frac{x}{X} + \frac{y}{Y}

Im referring to second question. Prove E(x)>0 for all real x. Un said if all t.p laid above y, then it is true. But thats not as you said. If only their E"(x)>0, then it would be the case.

but turning point can be concave downwards. Needs rigour from Trebla.

how then do we prove that e(x) >0 for all x

I dont get how x1x2x3 = 1. Could someone explain?