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Welcome......!!!

Hi hectictar: Just for my own curiosity, I took up you suggestion and hastily wrote this very short code to sum up the digits of any number. It can easily add up the first 1,000 digits of Pi, for example. In this short code, it sums up:123456789 and counts them as well:

n=123456789;s=0;p=0;cycle: s=s+(n%10);p=p+1;n=int(n/10);if(n!=0, goto cycle,0);print"Total Sum =",s;print"Total Num =",p

Total Sum = 45
Total Num = 9

Thank You CPhill

27t + 38  = 200      subtract 38 from both sides

27t  =  162      divide both sides by 27

t  =  6

[ So.....200 people entered during the 6th hour, killersteve  !!]

Second one

XA  =

[2 -1 ]  *   [1/2    1/4 ]     =   [  2*1/2 + -1*-2      2*1/4 + -1* -1/2 ]  =   [  3   1     ]

[6  -4]       [ -2    -1/2]          [  6*1/2 + -4*-2      6*1/4  + -4*-1/2 ]       [  11  7/2  ]     

Note that if these are inverses we should get the identity matrix  [ 1   0   ]

                                                                                                      [  0   1  ]

So......

IS NOT        ARE NOT

First one  ....for a 2 x 2 matrix  [ a  b ]

                                                 [  c  d ]

The inverse  is given by

  1               [   d     -b ]

______       [   -c    a  ]

ad - bc

So for the matrix   [ -4  3]              the inverse is

                             [ -2  4]

       1                      [   4   -3  ]

____________       [   2   -4 ]

-4*4 -  (3 * -2)

    1             [  4   -3 ]

_______     [  2   -4 ]

-16 + 6

(-1/10)  [  4    -3  ]

            [  2    -4  ]

[ -4/10    3/10  ]  =   [ -2/5   3/10 ]

[ -2/10   4/10  ]        [ -1/5   2/5  ]

That's really cool!! 

If you want to try it, here's another (similar, maybe easier) challenge:

Make a function that will return the sum of the digits. So for examples:

f(123) = 6

f(100) = 1

f(4532)  =  14

f(3) = 3

f(0) = 0

And let's also assume here that  n ≤ 32

\(\text{Your first differential equation!}\\ \text{I don't know how much you've been taught. The usual way to solve this is}\\ f'(x) = 5f(x)\\ \dfrac{f'(x)}{f(x)}=5\\ \text{Integrate both sides}\\ \ln(f(x)) = 5x+c\\ f(x) = e^{5x+c} = Ce^{5x}\)

\(i = e^{i \pi/2+2\pi k},~i \in \mathbb{Z}\\ z^2 = e^{i (\pi/2+2\pi k)}\\ z = \exp\left(i\dfrac{\frac \pi 2 + 2\pi k}{2}\right) =\\ \exp\left(i\left(\frac \pi 4 + \pi k\right)\right),~k = 0,1 = \\ e^{i\pi/4},~e^{i5\pi/4}=\\ \pm \dfrac{1}{\sqrt{2}}(1+i)\)

Thanks, Rom......I learned something new today from this !!!

To see how Rom arrived at his answer, look at this easy-to-understand proof  :

http://mrhonner.com/archives/8957

The second one is correct  !!!!

First one :

AB   =

[7 1 ]  *   [ 1 -2 5  ]  =  [ 7*1  + 1*2     7*-2 + 1*4      7*5 + 1*1 ]   =   [ 9  -10  36 ]

[3 1 ]       [2  4  1 ]       [ 3*1 + 1*2 +   3*-2 + 1*4      3*5 + 1*1 ]        [ 5   -2   16]

(AB)C  =

[ 9  -10  36 ]   *   [ 2 ]   =   [ 9*2 +  -10 * -1  + 36 * 0]  =     [  28   ]      

[ 5  -2   16]          -1           [ 5*2 +   -2*-1  +   16 * 0 ]        [  12   ]    

                          [ 0 ]

And

BC  = [1  -2  5  ] *  [ 2 ]  =   [ 1*2  +  -2*-1   + 5*0 ]   =    [ 4 ]   

          [ 2  4  1  ]      -1         [  2*2 +   4*-1  +  1*0 ]          [ 0 ]

                              [ 0 ]    

A(BC) =  [7 1]  *   [ 4 ]  =   [ 7*4  + 1 * 0 ]  =    [ 28  ]

               [3 1]      [ 0]       [ 3*4 +   1 * 0 ]        [  12 ]

240, 272, 280, 285, 295, 285, 300, 290, 310, 305, 285, 305

Order the data from low to high

240, 272, 280, 285, 285, 285, 290, 295, 300, 305, 305, 310

The median is [285 + 290 ] / 2  = 287.5

The first quartile, Q1, is  [ 280 + 285]/2  = 282.5

The third quartile, Q3, is  [ 300+305] /2  = 302.5

The interquartile range  is   Q3 - Q1  =  302.5 - 282.5  =  20

Inner fence

Lower bound  =  Q1 - 1.5 (Q3 - Q1) =  282.5 - 1.5 (20)  = 252.5

Upper  bound =  Q3 + 1.5 (Q3 - Q1)  = 302.5 + 1.5(20)  =  332.5

Any data point falling outside these bounds (inner fence) is a minor outlier

So.....240  would be a minor outlier

Outer Fence

Lower Bound  =  Q1 - 3 (Q3 - Q1)  = 282.5 - 3(20)  = 222.5

Upper Bound  =  Q3 + 3(Q3 -Q1)  = 302.5 + 3(20)  = 362.5

Any data point falling  outside these bounds (outer fence)  is a major outlier

Note that no value is a major outlier

so:

\(cos(\alpha)= \cos\angle BAC=-0,52\overline 7 \)

\(\angle BAD=\frac{1}{2} \angle BAC=\frac{1}{2}arccos (-0.52\overline 7)\\ \color{blue}\angle BAD= 60.928° \)

\(cos \angle BAD=0.48591\)

  !

\(64x^3+125\)

\(=(4x)^3+5^3\)

\(a^3+b^3\\ =(a+b)(a^2-ab+b^2)\\ a=4x;\ b=5\)

\((4x)^3+5^3\\ =(4x+5)(16x^2-20x+25)\)

  !

Mathematical equivalents

Could you please explain?

Thanks so much

Where is the list of numbers from which to choose?

.7071 + .7071 i

Sorry asinus the question had a typo it should be "Find \(\cos \angle BAD\)"

Thanks for your reply. I want to say that I already know Python and Javascript, and I know how to convert decimal numbers to binary using programming, but I'm interested in doing this mathematically.

Is there an image to show?

LOL thanks!

Problem 3: Suppose you drive a car 392 miles on one tank of gas. The tank holds 14 gallons of gas. (Assume the car travels the same distance for each one gallon of gas)

The number of miles traveled varies directly with the number of gallons of gas you use.

Write an equation that relates miles traveled to gallons of gas used. (Use any variable you like in the equation.

How far can you drive with 3.7 gallons of gas? (Make sure to show the calculations you did to determine this answer)

392 miles /14 gals = 28 m/g      or  28 mpg

x =gallons used       miles travelled = 28x

Miles travelled = 28 (3.7) = ?

Problem 2:  Fill in the boxes below to solve.

Find the slope of  -7x + 2y = 8 using slope intercept form

-7x + 2y = 8

2y = 7x + 8                    Add       7x   to both sides

 y =  (7/   2      )  x + 4      Divide both sides by  2

Is this equation now in slope intercept form?   Yes

The slope of the line is     7/2