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I thought that was the answer too, but it turned out that wasn't correct.

I need a compus to find east, then get some measuring tape and find out, or just drive with a satnav

The six letters, b through to g, can be listed in all possible orders, each followed by an a. There are 6!=720 different ways of doing that.

Just keep going. Put what you have in terms of cos and sin.

You might then simply spot the integral (think of the function of a function rule for differentiation), but if you can't use the (second) substitution $$u=\sin\theta$$.

can u please list the solution how to get it please Thanks

6! Permutations

Thank you Alan to make it clear .. 

thank you Melody and seven up 

oh yes yes .. hhhhhhhh I put sinx insted of cosx  

Thank you Alan =) 

u can use pythagoras theorem. If the long side (north east length) is 200 m, then a squared (north) + b squared (east)=200. But a=b because the guy has to walk as far he goes north as he goes east if its exactly north east.

2a squared= 200 squared

2a squared = 40000

a squared = 20000

a= 141.421356

:) 

-SungEun Choi Regents School Pattaya Grade 7 student

subtract 5 from both sides

then divide both sides by 6

@@ End of Day Wrap    Wed 25/2/15     Sydney, Australia   Time 11:15 am     ♪ ♫

Hi all,

Great answers today from Heureka, Rosala, CPhill, Alan, Yoshi-san, ThisGuy and Geno3141.   Thank you. 

Interest posts:

26/2/15

if sin=1/3 and tan is negative; what is cot and sec?

2nd quadrant

$$\\sin\theta=\frac{opp}{hyp}=\frac{1}{3}\\\\
adj=\sqrt{9-1}=\sqrt8=2\sqrt2\\\\
Cot\theta=\frac{-adj}{opp}=\frac{-2\sqrt2}{1}=-2\sqrt2\\\\
Sec\theta=\frac{-hyp}{adj}=\frac{-3}{2\sqrt2}=\frac{-3\sqrt2}{4}\\\\$$

$$t_n=n^3$$

Derivative of (x^4)÷(y^8) with respect to x, given dy÷dx=v

$$\\\frac{d}{dx}\;\frac{x^4}{y^8}\\\\
=\frac{d}{dx}\;x^4*y^{-8}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}\;\;\frac{dy}{dx}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}*v\\\\
=\frac{4x^3}{y^{8}}+\frac{-8x^4v}{y^{9}}\\\\
=\frac{4x^3y-8x^4v}{y^{9}}\\\\
=\frac{4x^3(y-2xv)}{y^{9}}\\\\$$

Could a mathematician check  this please ?

It is undefined