All Answers 358516 Answers

this may not be right but here goes.

I am going to sit a boy down - anywhere.

there are 3 places for the other 3 boys  that is 3!

and 4 places for the girls that is 4!

but you probably have to cut that in half to cover the clockwise anticlockwise condition.

so maybe it is

$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$

$${\mathtt{130}}$$ degrees

$${\frac{{\mathtt{130}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}$$

$${\frac{{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{18}}}}$$

$${\mathtt{130}}$$ degrees is equal to $${\frac{{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{18}}}}$$ radians

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{9}}}}$$ radians

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{9}}}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right)$$

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{180}}}{{\mathtt{9}}}}$$

$${\frac{{\mathtt{360}}}{{\mathtt{9}}}}$$

$${\mathtt{40}}$$

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{9}}}}$$ radians is equal to $${\mathtt{40}}$$ degrees

ok thanks!

From radian to degree, you multiply by 180º/π and reduce the fraction.

From degree to radian, you multiply by π/180º nad reduce fraction.

$${{\mathtt{16}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}$$

$${\sqrt[{{\mathtt{{\mathtt{4}}}}}]{{\mathtt{16}}}}$$

$${\mathtt{2}}$$

By definition, a 1π radian is equivalent to 180º. Being X the number of radians, by using the rule of three, one has X = (1 rad)(130º)/180º = 13/18π radians.

130º = 13/18 π radians.

Excellent answer anon :)

$${\frac{{{log}}_{{\mathtt{x}}}{\left({\frac{{\mathtt{37}}}{{\mathtt{999}}}}\right)}}{{{log}}_{{\mathtt{2}}}{\left({\mathtt{4\,096}}\right)}}} = -{\mathtt{4}}$$

$${{log}}_{{\mathtt{2}}}{\left({\mathtt{4\,096}}\right)} = {\mathtt{12}}$$

$${\frac{{{log}}_{{\mathtt{x}}}{\left({\frac{{\mathtt{37}}}{{\mathtt{999}}}}\right)}}{{\mathtt{12}}}} = -{\mathtt{4}}$$

$${{log}}_{{\mathtt{x}}}{\left({\frac{{\mathtt{37}}}{{\mathtt{999}}}}\right)} = -{\mathtt{48}}$$

$${{\mathtt{x}}}^{-{\mathtt{48}}} = {\frac{{\mathtt{37}}}{{\mathtt{999}}}}$$

$${\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{48}}}}} = {\frac{{\mathtt{37}}}{{\mathtt{999}}}}$$

$${\mathtt{999}} = {\mathtt{37}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{48}}}$$

$${\mathtt{27}} = {{\mathtt{x}}}^{{\mathtt{48}}}$$

$${\sqrt[{{\mathtt{{\mathtt{48}}}}}]{{\mathtt{27}}}} = {\mathtt{x}}$$

$${\sqrt[{{\mathtt{{\mathtt{48}}}}}]{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}}} = {\mathtt{x}}$$

$${\mathtt{x}} = {\sqrt[{{\mathtt{{\mathtt{16}}}}}]{{\mathtt{3}}}}$$

You mean you posted the same question twice?   :))  I whited the other one.   Don't sweat it. :)

hi Alan,

I often repeat the question first but I don't know why my calculation wasn't there as well.

Just another of life's mysteries I guess.  :))

180° = pi radians  Try to remember this.

1° = pi/180 radians

130° = 130*pi/180 radians

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I wondered why you'd just repeated the question Melody!

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There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton.

what happened to my answer?   It has disappeared!  I must look under the bed!!!

I did have  5!/3! = 20

how annoying!

There are five positions where the 6 could appear.  For each of those there are four positions where the 4 could appear.  The 2's fill in the other positions.  So there are 5*4 = 20 sequences with one 6, one 4 and three 2's.

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Let John define position 1, so Sam defines position 6.

1.  Put Stacey in position 2, then Roger is in position 7.  There are 6! ways of arranging the others.

2.  Switch Stacey and Roger.  There are another 6! ways of arranging the others.

3.  Put Stacey in position 3, then Roger is in position 8.  There are 6! ways of arranging the others.

4. Switch Stacey and Roger.  There are 6! ways of arranging the others.

5. Put Stacey in position 4, Roger in position 9.  There are 6! ways of arranging the others.

6. Switch Stacey and Roger.  There are 6! ways of arranging the others.

7. Put Stacey in position 5, Roger in position 10.  There are 6! ways of arranging the others.

8. Switch Stacey and Roger.  There are 6! ways of arranging the others.

Hence, in total there are 8*6! arrangements.

$${\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{6}}{!} = {\mathtt{5\,760}}$$

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I roll a fair 6-sided die five times. How many sequences of rolls will give me three 2's, one 4, and one 6? (A "sequence of rolls" is an ordered list of rolls, like 1, 1, 2, 1, 2.)

999/7 is just over 142

100/7 is just over 14

so there are (142 - 14) = 128 numbers that are multiples of 7, hence there are 999-99-128 = 772 numbers that are not multiples of 7.

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deleted deleted

2*4! =

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{4}}{!} = {\mathtt{48}}$$

Thanks Chris and Nauseated.

I shall have to have a really proper look when it is not after 3:30am   :)

Yes, from the left.  But the result is infinite.

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I am splitting my regular octagon into 16 sectors so each one will radiate from a 360/16= 22.5 degree angle at the centre.

$$\\tan22.5=\frac{0.5L_1}{a}\\\\
0.4142\approx \frac{0.5L_1}{a}\\\\
a\approx\frac{0.5L_1}{0.4142}\\\\
a\approx1.2071L_1\\\\
also\\\\
tan22.5\approx\frac{0.5L_2}{a+2}\\\\
0.4142\approx \frac{0.5L_2}{a+2}\\\\
a+2\approx \frac{0.5L_2}{0.4142}\\\\
a\approx \frac{0.5L_2}{0.4142}-2\\\\
a\approx 1.2071L2-2\\\\$$

$$\\so\\\\
1.2071L2-2\approx 1.2071L_1\\\\
1.2071L2\approx 1.2071L_1+2\\\\
L_2\approx L_1+\frac{2}{1.2071}\\\\
L_2\approx L_1+1.6569\\\\$$

so YES for a regular octagon, if you put the 2 unit border around it like you said, the side length will increase by a set amount which is approx   1.6569 units

I would think so.