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The force is inversely proportional to the square of the distance between them, so:

F2/F1 = (r1/r2)².  

Here you have r1 = 5cm, r2 = 1cm, F1 = 1N

Can you take it from here?

First, Outer, Inner, Last:  when multiplying two binomials together.

Given that $$f(x) = \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - x}}},$$ compute $(f(f( - 2)))^{ - 2}$.

Express your answer as a common fraction.

$\begin{array}{|rcll|} \hline f(-2) &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - (-2) }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 +2 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{3 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{ \dfrac {2}{3 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {3}{ 2 }} \\\\ &=& \dfrac {1}{\dfrac {2}{ 2 } - \dfrac {3}{ 2 }} \\\\ &=& \dfrac {1}{-\dfrac {1}{ 2 } } \\\\ &=& \dfrac {2}{-1} \\\\ \mathbf{f(-2)} &\mathbf{=}& \mathbf{-2} \\ \hline \end{array}$

I assume:

$\begin{array}{|rcll|} \hline && \mathbf{\left(f(f( - 2)) \right)^{ - 2}} \\\\ &=&\mathbf{ \dfrac{1}{ \left(f(f( - 2)) \right)^2} } \quad | \quad f(-2) = -2 \\\\ &=& \dfrac{1}{ \left(f(-2) \right)^2} \quad | \quad f(-2) = -2 \\\\ &=& \dfrac{1}{ \left(-2 \right)^2} \\ \\ &\mathbf{=}& \mathbf{\dfrac{1}{4} } \\ \hline \end{array}$

The Fibonacci sequence is the sequence 0, 1, 1, 2, 3, 5, ...

where the first and second terms are 1 and each term after that is the sum of the previous two terms.

What is the remainder when the 100th term of the sequence is divided by 8?

Fibonacci cycle length mod 8 is 12:

$\small{ \begin{array}{|r|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Cycle} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{Remainder } \pmod{8} & 0 & 1 & 1 & 2 & \color{red}3 & 5 & 0 & 5 & 5 & 2 & 7 & 1 \\ \hline & f_0=0 & f_1=1 & f_2=1 & f_3=2 & f_4=3 & f_5=5 & f_6=8 & f_7=13 & f_8=21 & f_9=34 & f_{10}=55 & f_{11}=89 \\ & f_{12} & f_{13} & f_{14} & f_{15} & f_{16} & f_{17} & f_{18} & f_{19} & f_{20} & f_{21} & f_{22} & f_{23} \\ & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ & f_{96} & f_{97} & f_{98} & f_{99} & \color{red}{f_{100}} \\ \hline \end{array} }$

$\begin{array}{|rcll|} \hline && f_{100} \pmod{8} \\ &\equiv& f_{100\pmod{12}} \pmod{8} \\ &\equiv& f_{4} \pmod{8} \\ &\equiv& 3 \pmod{8} \\ \hline \end{array}$

The remainder when the 100th term of the sequence is divided by 8 is 3

it has to do w/ multiplying to binomials together.  front.  outiside.  inside . last

I got f(0)=4+sqrt(5)

Meant 4).

#3    I count  24 ways if the domino halves are indistiguishable

I can't believe I didn't do this. officially my brain is frozen

I will assume you can have 1 letter words like   'M'  'A'  'T'  and 'H'  

4 p 1  +   4 p 2 +  4 p 3  +  4 p 4=

4                12        24           24        = 64 'words'

1.815 x 10^11 / 8.67 x 10^22 = .20934x 10^-11 = e^-20000/1.987T

                                             -26.892 =- 20000/1.987T

                                              T = -20000/(1.987(-26.892)     T =374.29

please dudes help me :'( i cant do this on calculator. i try every road but....

See my second answer, here : https://web2.0calc.com/questions/please-help-asap_86

[ It is based off my first answer ]

6.Dmitri has a pair of standard dice; one die is blue, and the other die is yellow. He rolls both of his dice. How many ways could the number on the blue die be larger than the number on the yellow die?

We have a total of  6 * 6  =  36 outcomes

6 outcomes will be the same   ....i.e., (1, 1) (2, 2)  , etc.

15 will have the number on the yellow die > the number on the blue die

And the same number of outcomes will have the number on the blue die > the number on the yellow die

3) You can have (Even, Even), (Even, Odd), (Odd, Even). The last (Odd, Odd) will result in odd no matter what...

So, we have 50*50*3=7500 ways?
 

Yes i have an equation but i cant understand :S

b)  Similar to the first

[ x + x + 11] * 12  / 2  =

[2x + 11] * 12 / 2   =

[ 2x + 11] * 6  =

12x + 66  =

[12x + 12(5) ] + 6     

12 [ x + 5 ] + 6     which, when divided by 12,  will always leave a remainder of 6

Thanks, tertre......here's another approach

Let the first integer  = x

Let the 11th integer = x + 10

Sum  =  [ first integer + 11th integer] * number of terms   / 2     =

[ x + x + 10 ] *  11  /  2  =

[2x + 10 ] * 11 / 2   =

2 [ x + 5 ] * 11 /  2    =

[ x + 5 ] * 11       which is divisible by 11

All you have given is an expression.....  T can be any value....

if you have an EQUATION    like     4500 = e^(-20000/(1.987T)      take the ln of both sides of the equation and you have

                                                     ln 4500 = -20000/(1.987T)      then re-arrange

                                                          1.987T = -2000/4500       then divide both sides by 1.987

                                                                   T = -20000/(4500(1.987)) 

Nice, tertre....!!!!

No problem ! 

a): Label the integers: $x-5, x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4, x+5=11x$ . And, this number is a multiple of $11$ , so it is definitely divisible by $11.$

Try the second one on your own!