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$Distance travelled by $x = 2\sint-t$ from 0 - $\pi$ seconds.\\1. Firstly, find the velocity equation.$\\x = 2 \sin t - t\\\dot{x}=2 \cos t - 1\\\\$Because the velocity graph crosses the x-axis between 0 and $\pi$, and we want $\textit{distance travelled}$ not $\textit{displacement}$, we have...

We have 9 consonants and 6 vowels. So we can arrange the consonants in 9! ways and vowels in 6! ways. and remember there are duplicated letters !

The area under a velocity-time curve is the displacement, which in many cases, is the distance travelled.

when you take out the 16 you forgot to square root it :) sqrt 512 = sqrt (16 * 32) = 4 sqrt 32 = 16 sqrt 2

If u bring the x^1/2 up the top part becomes x * x^-1/2 using index laws: x* x^-1/2 = x^(1+ (-1/2)) = x^1/2 index law with division states: x^a / x^b = x^(a-b)

no because it is division divide means subtract the powers but y should be x^1/2 / sqrt 2

Subbing x = 4: 1/ sqrt(2*4) - 4 / sqrt (2*4)^3 = 1/ 2sqrt2 - 4 / 16sqrt2 = 1/ 2sqrt2 - 1/4sqrt2 = 2/ 4sqrt2 - 1/4sqrt2 = 1/ 4sqrt2 complicated :/

you will still get 1/4sqrt2 if u sub x = 4

I assume its what the diagram looks like Note that the angle between the radius and the tangent is a right angle you can then use Pythagoras to find the angle at the centre (between the two radii) and then the length of the arc

23. m = \frac{y-2}{x+1}\\mx+m=y-2\\mx-y+m+2=0\\\\x-$intercept: When y = 0:$\\mx+m+2=0\\x=\frac{2-m}{m}\\\\(\frac{-2-m}{m}, 0)\\y-$intercept: When x = 0:$\\-y+m+2=0\\y = m+2\\(0,m+2)\\\\ $a) If (-1,2) is the midpoint of the two intercepts, then the intercepts must be (-2,0) and (0,4).\\ Using...

Re: 2012 Year 9 &10 Mathematics Marathon Pythagoras :) First we know BE = ED (diagonals of parallelogram bisect each other) Then using Pythagoras: {AB}^2+{BE}^2={AE}^2\\{AB}^2+{ED}^2={AE}^2\\{AB}^2={AE}^2-{ED}^2...(1)\\\\{AB}^2+{BD}^2={AD}^2\\{AB}^2+{(2ED)}^2={AD}^2...(2)\\\\$Substitute (1)...

Re: 2012 Year 9 &10 Mathematics Marathon \frac{(1+100)\times 100}{2}\\=101 \times 50\\=5050 learnt that long time ago (i hv seen this question in a yr 7 paper btw)

Re: 2012 Year 9 &10 Mathematics Marathon \left\{\begin{matrix}x^2-xy+y^2=19 ... (1)\\x^4+x^2y^2+y^4=741 ...(2) \end{matrix}\right\\\\$Rearrange $ (1)\\x^2+y^2=19+xy \\(x^2+y^2)^2=(19+xy)^2... (3)\\\\$Rearrange...

Re: 2012 Year 9 &10 Mathematics Marathon i used cos rule we know that $AB = BC = AC = BX+CX\\CX=2BX\\\\${AX}^2=AC^2+CX^2-2 \times AC \times CX \times cos 60\\AX^2=AC^2+CX^2-AC \times...

Re: 2012 Year 9 &10 Mathematics Marathon actually she forgot to minus the small cone

Re: 2012 Year 9 &10 Mathematics Marathon am i supposed to know it ? cuz i dont...

Re: 2012 Year 9 &10 Mathematics Marathon Instead of "plus", it should be a "times" P = \frac{10000}{\left ( \frac{403}{400} \right )^{42} \times \left ( \frac{79}{80} \right )^{18}} \\ = 9163.05$ (2dp)$

Re: 2012 Year 9 &10 Mathematics Marathon for ii. i got 9163.05 instead

Re: 2012 Year 9 &10 Mathematics Marathon hard to explain. i'll try my best If n = number of vertices (or sides) \frac{n(n-3)}{2} n - 3 = a vertex can form diagonals with all other vertices except for the two next to it and itself then times n (for each of the vertices) divide by 2 because...

Re: 2012 Year 9 &10 Mathematics Marathon thanks i was looking at the wrong angle