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Alan and heureka: Do you guys have any new insights to this question??

I want to caculate 11power 850 mod 1643 

\(\begin{array}{|rcll|} \hline && 11^{850} \pmod{1643} \\ & \equiv & 11^{4\cdot 212+2} \pmod{1643} \\ & \equiv & (11^4)^{212}\cdot 121 \pmod{1643} \quad & | \quad 11^4\pmod{1643} &\equiv& -146 \pmod{1643} \\ & \equiv & (-146)^{212}\cdot 121 \pmod{1643} \\ & \equiv & (-146)^{2\cdot 106}\cdot 121 \pmod{1643} \\ & \equiv & [(-146)^2]^{106}\cdot 121 \pmod{1643} \quad & | \quad (-146)^2\pmod{1643} &\equiv& -43 \pmod{1643} \\ & \equiv & (-43)^{106}\cdot 121 \pmod{1643} \\ & \equiv & (-43)^{4\cdot 26+2}\cdot 121 \pmod{1643} \\ & \equiv & [(-43)^4]^{26}\cdot (-43)^2 \cdot 121 \pmod{1643} \quad & | \quad (-43)^4 \pmod{1643} &\equiv& -282 \pmod{1643} \\ & \equiv & (-282)^{26}\cdot (-43)^2 \cdot 121 \pmod{1643} \quad & | \quad (-43)^2 \pmod{1643} &\equiv& 206 \pmod{1643} \\ & \equiv & (-282)^{26}\cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & (-282)^{2\cdot 13}\cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & [(-282)^2]^{13} \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad (-282)^2 \pmod{1643} &\equiv& 660 \pmod{1643} \\ & \equiv & 660^{13} \cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & 660^{2\cdot6 + 1} \cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & (660^2)^6\cdot 660 \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad 660^2\pmod{1643} &\equiv& 205 \pmod{1643} \\ & \equiv & 205^6\cdot 660 \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad 660 \cdot 206 \cdot 121\pmod{1643} &\equiv& -199 \pmod{1643} \\ & \equiv & 205^6\cdot (-199) \pmod{1643} \\ & \equiv & 205^{2\cdot 3}\cdot (-199) \pmod{1643} \\ & \equiv & (205^2)^3\cdot (-199) \pmod{1643} \quad & | \quad 205^2\pmod{1643} &\equiv& -693 \pmod{1643} \\ & \equiv & (-693)^3\cdot (-199) \pmod{1643} \\ & \equiv & (-693)^{2+1}\cdot (-199) \pmod{1643} \\ & \equiv & (-693)^2\cdot(-693)\cdot (-199) \pmod{1643} \quad & | \quad (-693)\cdot (-199)\pmod{1643} &\equiv& -105 \pmod{1643} \\ & \equiv & (-693)^2\cdot(-105) \pmod{1643} \quad & | \quad (-693)^2\pmod{1643} &\equiv& 493 \pmod{1643} \\ & \equiv & 493\cdot(-105) \pmod{1643} \\ & \equiv & 493\cdot(-105) \pmod{1643} \quad & | \quad 493\cdot(-105) \pmod{1643} &\equiv& -832 \pmod{1643} &\equiv& 811 \pmod{1643} \\ & \equiv & 811 \pmod{1643} \\ \hline \end{array}\)

I assume the 10 in the denominator is in seconds. The numerator is the distance travelled in 10 seconds, so if we divide this by 10 seconds we get the average speed (in ft/sec) over the first 10 seconds.    

Note that for h(10) you have: -16*10^2 + 294.4*10 + 15  and for h(0), just 15

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We know that cos(A)=sin(90-A)

We also know that if sin(A)=sin(B) then B=A+360K or B=180-A+360K (think about it)

sin(x+37)=cos(2x+8)=sin(90-8-2x)=sin(82-2x)

FIRST OPTION:

x+37=82-2x+360K   / add 2x, and subtract 37

3x=45+360K    / divide by 3

x=15+120K

SECOND OPTION:

x+37=180-(82-2x)+360K=98+2x+360K   / subtract (37+2x)

-x=61+360K   / divide by -1

x=-360K-61

Nicely done Melody!  Here's a similar, but step by step version.  Again, X and Y values are plotted as coordinates within a square box:

. 

If the method used for the number of 3's between 1 and 1,000 and between 1 and 10,000 is exactly the same as between 1 and 10^12, which everybody agrees that it is 300, and 4,000 threes respectively, without dividing by 2, then what is so special about the number of 3's between 1 and 10^12????. Some rigorous explanation is needed!!!. 

I have just leaned a new technique!!    Someone very kind taught me :))

A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?

Let the lengths be  x,   5-y,    and    y-x   

Obviously x and y are between 0 and 5

and  y>x

If you add these together you will see that they add to 5.

So the entire region where x and y can be is a triangle, (the big green one in the contour map)

Now

If x is less then 1, the condition is met.        x<1

If x is more than 4 the condition is met.       x>4

If    5-y>4    or if    5-y<1    the condition is met.

i.e.     if     y<1   or if      y>4     the condition is met

If           0< y-x<1     or if      4

i.e.        x< y<1+x     or if      4+x

i.e.       y>x and  y<1+x     or if      y>4+x  and y<5+x    the condition is met

Ok lets try mapping this on a boolean contour map.

I've done all that and the only place that the conditions are not met is the little red triangle in middle.

So all possible outcomes are within the green area which is  0.5*5*5=12.5units squared

The area of red triangle represents all triads where the smallest is bigger than 1

Area =  0.5*2*2 = 2 units squared

So the probability that the shortest side is more than 1 is    2/12.5 =    4/25 = 16%

Half zeroes? could you prove that? And How did you infer from that that there are 6*10¹¹ threes? it doesnt make sense. I know alot of the digits are zeroes, but it doesnt change anything.

Please post your proof.

cot(-1)(1.6)=?

Unfortunately, all of you noobs would be wrong.

The reason that we have geometry in math class is because of the same guy that said e^(pi*i)=-1, so e^(pi*i) + 1 should equal 0.

Plus, whenever you put a positive number to a power, the output is always negative.

2 conflicting pieces of logic just THREW down your solutions.

Question: What are the ranges of the adult tickets and student tickets the theater can sell?

Answer:

{s + a <= 800} and {4.5s + 7.5a >= 3000}

400-a <= s <= 800-a

ceil((2000-3s)/3) <= a <= 2400-3s

400 <= a <= 800

667 <= s <= 800

I just want to be nice and doing math is fun

Be careful with this stuff! Guess the answer to the following equation:

ln(x)^e

Highlight the black text below to see if you simplified correctly!

It is not always x! Note where the e is: outside of the ln.

The number must be odd

If it is a prime number modulo 4 = 1, it works

I am in 8th grade, so this is most likely an incorrect conjecture. However, here is what I have started with:

Assuming that the above must be true, then the value of the first point, x, is {1 < x < 5}. Now, we know that the second point cannot be within 1 unit of the edges of the line, or within 1 unit of the point x. So, the second division, y, must agree with the following statements: {1 < y < 5}, and: {{y < (x - 1)} or {y > (x + 1)}}.

Go ahead and calculate the probability for yourselves, I don't know how to calculate it.

YOU'RE WELCOME!!!

what is 0.0992125657 as a fraction

\(\begin{array}{|rcll|} \hline 0.0992125657 & \approx & \frac{878473}{8854453} \\ & \approx & {\color{red}0.0992125656}99992986579746936372015301227529244324861174371\ldots \\ \hline \end{array} \)

e^ln z

\(e ^{\ln (z)} = z\)

\(\large{\frac{992, 125, 657}{10, 000 ,000 ,000}}\)

I don't think that this can be reduced any further.

_{IK so we need help} 

thank you

[ 3 / [(x + h) + 2]   -  3/ [ x + 2]  ]   /  h          get a common denominator

[ (3 [x + 2]  - 3 [ (x + h) + 2 ] ] /  [ h (x + h + 2) (x + 2)]       simplify

[ 3x + 6  - 3x - 3h - 6 ] /  [ h (x + h + 2) (x + 2) ]

[ -3h] /  [ h (x + h + 2) (x + 2) ]      the "h's" cancel

[-3] /  [  (x + h + 2) (x + 2) ]

Take  the limit  as h → 0

[-3] / [ (x + 2) ( x + 2) ]  =

-3 / (x + 2)² 

Using

A = P (1 + r/n)^nt

Where P  is the invested amt, r is the yearly interest rate, n is the number of compoundings per year, t is the number of years  and A is the acumulated amt at the end

1700 (1 + .06/360)^360*10  ≈ $3097.45 

What other information are you given?

Area under normal curve between x = 28.5 and x = 60.0 is given by

P(28.5 < x < 60.0) = P( 0.50 < z < 5) =

This stats calc site is great  :)

The address of this calculator as well as others are stored in our 'reference materials' thread that is kept in with the sticky notes!

Hmmm....that's funny how if you try the cost function with years between 0 and 5 you get a negative cost...??

That's a pretty good deal they got then!!