Alan

Use v² = u² + 2as, where v = final speed (0), u = initial speed (10000/3600 m/s), a = acceleration (-0.8m/s²) and s = distance (metres)

0 = (10000/3600)² -2*0.8*s

s = (100/36)²/1.6 m

$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{100}}}{{\mathtt{36}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{1.6}}}} \Rightarrow {\mathtt{s}} = {\mathtt{4.822\: \!530\: \!864\: \!197\: \!530\: \!9}}$$

s ≈ 4.8m

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For the first part use s = ut + (1/2)at², where s = distance, u = initial speed, a = acceleration, t = time:

$${\mathtt{s}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{3.5}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{{\mathtt{3.5}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{17.15}}$$

So distance = 17.15m

For the second part find the speed from v = u + at:

$${\mathtt{v}} = {\mathtt{0}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{v}} = {\mathtt{9.8}}$$

v = 9.8m/s

At this speed the sprinter will cover the remaining distance in time t2 = (100 - 17.15)/9.8 seconds

$${\mathtt{t2}} = {\frac{\left({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{17.15}}\right)}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t2}} = {\mathtt{8.454\: \!081\: \!632\: \!653\: \!061\: \!2}}$$

t2 ≈ 8.45 s

So, total time for the 100m is 3.5 + 8.45 seconds = 11.95 seconds

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Use the binomial expansion, which, for (a + x)ⁿ = aⁿ + na^n-1x + n(n-1)a^n-2x²/2! + n(n-1)(n-2)a^n-3x³/3! + ...

Here: a = 4, n = 1/2 (because √(4 + x) = (4 + x)^1/2), and you don't need to go further than the terms written above because these go as far as the x³ term.

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There are 3 even numbers, so probability of even or 5 is (3+1)/6 = 4/6 = 2/3

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$${\frac{\left({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{4}}\right){\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{12}}}{{\mathtt{6}}}} = {\mathtt{150}}$$

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To find x take logs of both sides:

ln(0.000061) = ln(0.5^x)

Using the fact that ln(a^b) = b*ln(a) you have:

ln(0.000061)=x*ln(0.5)

or x = ln(0.000061)/ln(0.5)

$${\mathtt{x}} = {\frac{{ln}{\left({\mathtt{0.000\: \!061}}\right)}}{{ln}{\left({\mathtt{0.5}}\right)}}} \Rightarrow {\mathtt{x}} = {\mathtt{14.000\: \!831\: \!231\: \!761\: \!287\: \!9}}$$

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If you have the coordinates of the end points simply average each of them. i.e. if the end points are (x1,y1) and (x2,y2) then the midpoint is ( (x1+x2)/2, (y1+y2)/2 )

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if f(x) = c (a constant) then

c = 2ln(x-3)

Divide both sides by 2

(c/2) = ln(x-3)

Raise both sides to the power e

e^c/2 = e ^ln(x-3)

But, e^ln(a) = a so

e^c/2 = x-3

Add 3 to both sides

x = 3 + e^c/2 

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If you have something like 

$$\frac{\pi}{x}=2$$

then multiply both sides by x

$$\pi=2*x$$

then divide both sides by 2

$$\frac{\pi}{2}=x$$

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Alternatively, if you have y = ax² + bx +c, then the vertex is at x = -b/(2a)

Plug this value of x back into the equation to find the corresponding value of y.

Put two other values of x into the equation to find two other values of y for the other points.

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