Search results

I sorta got lost here thinking AED was a straight line Anyway you could assume that 3 of the points, say A, B and C were concyclic, because any three points are. Then prove that D lies on this circle

Don't know if this can be classed as 3 unit but: Question: Given quadrilateral ABCD such that Angle ABC + Angle ADC = 180 degrees Using contradiction or otherwise, prove that A, B, C and D are concyclic points (Given you know nothing about cyclic quadrilaterals)

ahh I see where I went wrong, near the end Angle bisector: y = 1/3(x + 1) y = 1/3x + 1/3 sub x = 2 y = 1 Centre of circle is (2 , 1) Radius is perpendicular distance to x-axis = 1 Equation (x - 2)² + (y - 1)² = 1 Hope I didn't do something hopelessly wrong

3x - 4y + 3 = 0 => y = 3/4x + 3/4 ............. m(1) = 3/4 3x + 4y - 15 = 0 => y = -3/4x + 15/4............m(2) = -3/4 tan x = gradient When solving for x with these gradients, we work out that the solution is the same angle, but in different quadrants Since the triangle has one side as y =...

Question: \lim_{x\rightarrow \infty }\frac{sinx}{x}

Let 2 m above ground level to be "zero" and up to be positive. So the ground has a displacement of -2m Muzzle velocity : 800m/s At 30 degrees , vertical velocity = 800sin30 = 400m/s You want time of flight, so the displacement you want is -2 Looking at just the vertical component: s(y) = -2...

\angle ABP = \angle ABQ=90^{0}$(Angle at the circumference in a semicircle is 90^{0})$\\\angle ABP+\angle ABQ=\angle PBQ\\90+90=\angle PBQ\\\angle PBQ=180 \therefore $P,B and Q are collinear \\\\d/dx a^x

Shooter McGavin: You're in big trouble though, pal. I eat pieces of shit like you for breakfast! Happy Gilmore: [laughing] You eat pieces of shit for breakfast? Shooter McGavin: [long pause] No!

Wasn't absolute value something like |x| = sqrt(x²) Or something about scalar quantity like in complex numbers, modulus

So the 70m distance that the ball lands away from the moment the ball was thrown is not the straight line distance, but from what I can gather, the horizontal distance? Also does the ball land on the ground ie. 150m below the moment the ball was thrown.

One issue u=1-x x=1-u du = -dx \int x\sqrt{1-x}dx = =\int (1-u)\sqrt{u}.-du =\int (u-1)\sqrt{u}du =\int u^{3/2}-u^{1/2}du =2/5u^{5/2}-2/3u^{3/2}+c =2/5(1-x)^{5/2}-2/3(1-x)^{3/2}+c

Gurmies forgot to sub dx = 2udu entirely; forgot to multiply by "u" when subbing in causing the denominator to be carried through.

The use of proper time and proper length sounds very awkward. With reference to the Lorentz transformation equations, rest time refers to the frame of reference in which the event occurs. @adomad Seems correct to me, though the bit about the muon: The muon does not live longer per se, in its...

Apparently not, the Earth is actually a squished sphere with a bulge at the equator. But you are correct in your argument about rotational velocity.

Yes as r increases, F decrerases But when considerring escape velocity, we are looking at Gravitational potential energy. GPE = -GMm/r \Delta GPE = GMm/r By equating change in GPE and Kinetic Energy, we are able to determine the escape velocity \Delta GPE = Kinetic Energy GMm/r = 1/2 mv^2...

It would be advisable to expand and simplify the top bit of the fraction as it looks much nicer as 4 as opposed to (x + 2) - (x - 2). The bottom however needn't be expanded.

Well if it's intuitive for example, graphing ohm's law on a graph, zero voltage would intuitively result in zero current, I would definitely make the graph go through the origin and alter the line of best fit.

Mm there's often 1 day tournaments you could enter a school team into. Maybe check out this website NSW Junior Chess League I've been to one of their school tournaments and it seems alright, but that was in primary school. The thing with chess is that you would need the necessary resources ie...

Oh Awesome, thanks. Long Division does look easier after realizing the denominator. Didn't quite see that before.

Hi, I found this question in a textbook and was wondering how you would divide x³-2-2i by x+1-i The answer is x³-2-2i = (x+1-i)[x²+(-1+i)x-2i] I've tried turning it into a fraction and realizing the denominator, but end up with something horrible. I suppose you could manually alter the...