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I don't know Mellie but I have a theory  lol

If the youngest 2 children get 6 each then there is 1 peice left that can got to any one of the other 3 children.

so there is 3 ways this can happen  3 ways

If the youngest 2 children get 5 peices each then the other 3 peices can be allocated as

(1,1,1 ) [1way]     (1,2,0 ) [3!=6ways]   or    (3,0,0) [3ways]   = 10 ways

If the youngest 2 children get 4 peices each then this is how the other 5 peices can be allocated

(5,0,0) [3!/2!=3 ways] (4,1,0) [3!=6 ways]  (3,2,0) [6ways]  (3,1,1)  [3ways]  (2,2,1) [3ways]  =  21 ways

If the youngest 2 children get 3 peices each then this is how the other 7 peices can be allocated

(7,0,0) 3ways (6,1,0) 6 ways(5,2,0) 6 ways (5,1,1) 3 ways (4,3,0)6 ways (4,2,1) 6 ways (3,3,1) 3 ways (3,2,2) 3 ways   =   3+6+6+3+6+6+3+3 = 36 ways

Now I am going out on a ledge and say that I can see a pattern forming here

3 = 3C3        10= 5C3        21= 7C3       36= 9C3        .............

So maybe if the 2 youngest get 2 peices each then there will be     11C3 = 165 ways to distribute the other 9 peies

and maybe if the 2 youngest get 1 peice each there will be   13C3 = 286 ways to distribute the other 11 peices between the other 3 children.

and if the yougest 2 get none each then there will be 15C3 = 455 ways to distribute the other 13 peices between the other 3 children.

Total = $${\mathtt{455}}{\mathtt{\,\small\textbf+\,}}{\mathtt{286}}{\mathtt{\,\small\textbf+\,}}{\mathtt{165}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{976}}$$    ways altogether.       

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Mmm

If this is correct, I have found that n identical peices of candy can be distribued

between 3 children $$^{(n+2)}C_3$$     It would be good if someone could walk through the logic of why this is so :/

I shall think upon it  

14% = 14/100

P + P*14/100 = 437.64

P*(1+14/100) = 437.64

1.14P = 437.64

P = 437.64/1.14

P = 383.89 kr

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You need to memorise the expressions for sin, cos and tan in a right-angled triangle:

sin = opposite/hypotenuse

cos = adjacent/hypotenuse

tan = opposite/adjacent

and also memorise 

cot = 1/tan

csc = 1/sin

sec = 1/cos

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The quickest way is to use the calculator here:

If you have a head for numbers you might have spotted that the equation could be written as 

(x+1)(x+2)(x+9)=0

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the problem is, i dont understand how you guys even got your answers. this stuff makes absolutely no sense to me

6/√37 can also be written as 6√37/37

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$$x=\frac{-3\pm \sqrt{(3^2-4\times1\times25))}}{2\times1}$$

Because the discriminant (the part under the square root sign) is negative, the values of x will be complex numbers.

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You can write this as 

$$\frac{18}{t}+\frac{16}{t^2}\rightarrow \frac{2}{t}(9+\frac{8}{t})$$

or 

$$\frac{18}{t}+\frac{16}{t^2}\rightarrow \frac{2(9t+8)}{t^2}$$

but are these really any simpler?

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Simply divide 400 by 3.75

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This picture might help:

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Cot a =6

So  adj=6.   Opp=1.     Hyp=sqrt(37)

You can take it from there :)

any way you could put that in a fraction?

This is asking for  tan^-1(11/6)  which can be done on the calculator (atan(11/6))

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{11}}}{{\mathtt{6}}}}\right)} = {\mathtt{61.389\: \!540\: \!334\: \!035^{\circ}}}$$

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Distributive.   You distributed -1 over the terms in parentheses.

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This factors into f(x) = (x + 4)(2x - 1)  so the two roots are x = -4 and x = 1/2.    1/2 is larger than -4, unless by "larger" you mean the one with the larger absolute value.

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First find the angle a by using cot^-1(6)  (cot^-1 is also called acot or arccot).

$${acot}{\left({\mathtt{6}}\right)} = {\mathtt{9.462\: \!322\: \!208\: \!026^{\circ}}}$$  This is within the specified range 

Now 

$${csc}{\left({\mathtt{9.462\: \!322\: \!208\: \!026}}\right)} = {\mathtt{6.082\: \!762\: \!530\: \!311\: \!439\: \!4}}$$

You can try cos a, using any calculator with cos on it.

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The arc length of a segment is proportional to the angle of the arc, so:

length of arc AB/70 = length of circumference/360     (because there are 360° all the way round)

length of arc AB/70 = 25/360

Multiply both sides by 70

length of arc AB = 70*25/360 cm

I'll leave you to crunch the numbers

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3-(4+(4)*(-2))+1

3-(4-8)+1

3-(-4)+1

3+4+1 = 8

Thanks Brittany, that was a wonderful peice of work :)

Thanks Alan, I forgot that travel bags have wheels these days.

To me a suitcase is is a much smaller thing with no wheels

I was really confused     LOL      :)

hi Civonamzuk

If it had of gotten down to something like this, then this is what I would do

$$\\97+\frac{1}{4}-\frac{2}{3}\\\\
97+\frac{3}{12}-\frac{4}{12}\\\\
97+\frac{-1}{12}\\\\
96\frac{11}{12}\\\\$$

We all see something new every day here :)  

Well most of us that are interested do anyway :)

Yes i did :)