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The circumference  =  2pi (8)   =  16 pi inches

And 62.8 ft  =  62.8 (12)   = 753.6 inches

So....

The revolutions  =    753.6  /  ( 16 pi )  ≈  15 revs

Here's a quick way......the distance between a point (a,b)  and a line in the form of

Ax + By + C  = 0  =

abs (A(a) + B(b) + C)

_________________       so we have        

sqrt (A^2 + B^2 )

abs ( 4(3) + 3(4) + 7 )                    31                    31     

___________________   =          ______ =         ___   = 6.2 units

sqrt (4^2 + 3^2 )                           sqrt(25)              5

tan(2θ) + 2sinθ = 0

2tanθ

__________ + 2sinθ   = 0

1 - tan^2θ

2tanθ + 2sinθ(1 - tan^2θ)  = 0

2tanθ + 2sinθ - 2sinθtan^2θ = 0

tanθ + sinθ - sinθtan^2θ  = 0

sinθtan^2θ - tanθ - sinθ = 0

sinθsin^2θ        sinθcosθ        sinθcos^2θ

_________  -  __________  - __________  =   0

cos^2θ             cos^2θ             cos^2θ

sin^3θ  - sinθcosθ  - sincos^2θ  = 0

sinθ   [ sin^2θ - cosθ - cos^2θ] = 0               sinθ  =  0     at  0   and  pi

1-cos^2θ - cosθ - cos^2θ  =  0

-2cos^2θ - cosθ + 1  = 0

2cos^2θ + cosθ - 1  = 0

(2cosθ -1)(cosθ + 1)  = 0

2cosθ - 1  = 0                   and          cosθ + 1  = 0

2cosθ = 1                                          cosθ  = - 1

cosθ = 1/2                                        And this happens at pi

And this happens at

pi/3  and 5pi/3

So the solutions are  θ  =  0, pi/3, pi  and 5pi/3

Can you do the other one, AT???

Yes that sounds reasonable.

You say there are 47 trailing zeros

Our number is  2012!

43^2 = 1849

Since 1849 is smaller than 2012 so I would expect there to be more than 44 trailing zeros when 2012! is written in base 43.

I know I have told you many times before but I really like your little lines of code that you figure out and run.

It is a really helpful check mechanism.

I'd really like it if you became a member.

You could call yourself  CodeDude    OR    Cplusplus ( I think C++ is the code you use??)   LOL     

OR you could chose whatever you want.   

AHH Really? Yay!!!

I would have never thought to use Vieta's formula, I am sorry I should have known how to this

We can use Vieta's formula and find that (-1)^3(-1/3)=your answer. 

\(1/3\)

Here is one way:

4x+3y=-7

3y = -4x-7

y = -4/3 x- 7/3     so the slope is -4/3      Perpindicular slope is   3/4

Find line with 3/4 slope and point 3,4

4 = 3/4 (3) + b

b=  7/4

y = 3/4 x + 7/4

Now equate the two lines:

3/4x + 7/4 =   -4/3 x - 7/3

9/12 x + 7/4 = - 16/12 x - 7/3

25/12x = - 28/12 - 21/12

25/12 x = -49/12

x = -49/25

y= 196/75  - 7/3 = 196-7(25) /75 =  7/25

Now use distance formual between   3,4   and   -49/12 , 7/25

d^2 =   (3 -  - 49/25)^2     +    (4 - 7/25)^2

      =     124/25  ^2         +   93/25  ^2                  =    24025/625        take the square root   = 155/25 = 6.2

Here is another way:

     https://brilliant.org/wiki/dot-product-distance-between-point-and-a-line/

Note: To count the number of trailing zeros of N! in any base above 36, simply divide N by the powers of the base that are <= N.

Example: 2012! in base 43, we simply divide:2012 / 43^1 + 2012 / 43^2 =46 + 1= 47 zeros. Just take the INTEGER part.

2012! in base 47: 2012 / 47^1 + 2012 / 47^2 = 42 + 0 = 42 zeros. [47^2 is > than 2012, hence = 0]

  x+4/x where x≠0, 2/3

The last answer is correct

ElectricPavlov, That doesn't mean trial 4.  It means that there were 4 green marbles in 5 of the drawings.  If the numbers represented trials, there should be 12 of them.  The dots represent the trials. 

A total of 120 marbles were drawn.  Owing that 60% of the marbles in the jar were green, we would expect close to 72 green marbles out of the 120.  However, only 41 marbles were green.  That's too far off, even allowing for chance.  Something's wrong here.  Having to choose from the four, I'd choose

The number generator is not fair. In most of the experiments, considerably less than 60% of the selected marbles are green.

As below:

Like so:

As follows:

.

The weight is 'springing' between   +9   and -9     0 is the 'resting position'

at 1.25   it is at 0

  then at 2.5 it goes to -9

    then back to 0 at 3.75

      then to +9 at 5 sec

        then finally back to 0 at 6.25 seconds     This is one complete cycle     6.25 sec - 1.25 sec = 5 seconds

bottom row , just to the right of the % key.....

If ten marbles ae drawn each time , you would expect 60% of them to be green as a whole ...that would be  6 marbles out of ten....only in five trials  does this even come close to 60 % ... where 4 out of 10 are green (40%)...and only one trial gets 50%   and only one trial gets 80%

so:

The number generator is not fair. In most of the experiments, considerably less than 60% of the selected marbles are green.

(edited...thanx  , Guest !)

Hello Melody: I wrote a very short code to convert 2012! to base 43 in its entirety: Here are just the beginnings of the number and its trailing zeros. They count to 47 zeros.
n=2012!;m=43; a=n%m;printa,", ",;b=int(n/m);n=b;if(n>m, goto2, discard=0;printb,
(6, 38, 32, 10, 35............   14, 21, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

P.S. I also converted it to base 47 and it gives 42 zeros.

As Guest notes, the denominator should be Σf, so:

Σ(fx)/Σf = 3.667,  or mean number of pets per person ≈ 3.7

I'm not sure about this, but wouldn't the denominator be    248 + 567 + 1402 + 728 + 419 + 456 + 203   ?

    This would give you the average (mean) number of animals owned by the surveyed population.   

Let y = x².   Then y² - 2y - 7 = 0.

Use the standard quadratic formula to find that y = 2√2 + 1  and  y = 1 - 2√2

Since x = ±√y we must have positive y for real roots.  Only 2√2 + 1 is positive, hence there are just two real roots of the original equation.