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Acceleration is change in velocity over time.  In this example it will be a NEGATIVE acceleration as the car slows.

(Vf-Vo ) /  time   =  (45 - 85 m/s) / 4.5 s =  - 8.89 m/s^2

what should i pay you

This is the greatest site I have ever seen!

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25600cm³

25600cm³

If I have a list of 5 numbers each of them following another [1,2,3,4,5] and I want to multiply all of them together. 

How do I write (1*2)+(1*3)+(1*4)+(1*5)+(2*3)+(2*4)+(2*5)+(3*4)+(3*5)+(4*5)

But with 4000 numbers in the list instead of 5 without having to write every instance out?

Is it possible to make a rule and calculate it in a common calculator? 

Table:

$\begin{array}{|r|r|r|r|r|r} \hline \text{multiply} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & \color{red}1 & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline 2 & \color{grey}2 & \color{red}4 & \mathbf{6} & \mathbf{8} & \mathbf{10} \\ \hline 3 & \color{grey}3 & \color{grey}6 & \color{red}9 & \mathbf{12} & \mathbf{15} \\ \hline 4 & \color{grey}4 & \color{grey}8 & \color{grey}12 & \color{red}16 & \mathbf{20} \\ \hline 5 & \color{grey}5 & \color{grey}10 & \color{grey}15 & \color{grey}20 & \color{red}25 \\ \hline sum & \color{blue}15 & \color{blue}30 & \color{blue}45 & \color{blue}60 & \color{blue}75 & \color{blue}\text{arithmetical sequence} \\ \hline \end{array}$

$\text{Let the sum of all numbers in the table $\mathbf{S} = \color{blue}{15+30+45+60+75} = \color{blue}225 $} \\ \begin{array}{rcll} \text{Let the sum we are looking for } \mathbf{s=} \\ && (1*2)+(1*3)+(1*4)+(1*5)\\ && +(2*3)+(2*4)+(2*5) \\ && +(3*4)+(3*5) \\ && +(4*5) \\ &=& \mathbf{2+3+4+5} \\ && \mathbf{+6+8+10} \\ && \mathbf{+12+15} \\ && \mathbf{+20} \\ &=& 85\\ \end{array} \\ \begin{array}{rcll} \text{Let the sum of all square numbers in the diagonal of the table $\mathbf{q} =$} \color{red}{1+4+9+16+25}=\color{red}55 \\ \end{array}$

So we see the formula: $\boxed{S = 2s + q}$  or $\boxed{s = \dfrac{1}{2}\cdot (S-q) }$

$\mathbf{ S=\ ?}$

$\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)d \quad | \quad d=a_1 \\ &=& a_1+(n-1)a_1 \\ &=& a_1+ na_1-a_1 \\ &=& na_1 \quad | \quad a_1=(1+n) \dfrac{n}{2} \\ &=& (1+n) \dfrac{n^2}{2} \\\\ S &=& (a_1 + a_n)\cdot \dfrac{n}{2} \\ &=& \left((1+n) \dfrac{n}{2} +(1+n) \dfrac{n^2}{2} \right)\cdot \dfrac{n}{2} \\ &=& \Big((1+n) +(1+n) n \Big)\cdot \dfrac{n^2}{4} \\ &=& (n^2+2n+1 )\cdot \dfrac{n^2}{4} \\ &=& (n+1)^2\cdot \dfrac{n^2}{4} \\ \mathbf{S} & \mathbf{=} & \mathbf{\left[\dfrac{(n+1)\cdot n}{2}\right]^2} \\ \hline \end{array}$

$\mathbf{ q=\ ?}$

$\begin{array}{|rcll|} \hline \mathbf{q} & \mathbf{=} & \mathbf{\dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3}} \quad | \quad \text{from the formula collection} \\ \hline \end{array}$

$\mathbf{ s=\ ?}$

$\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{ \dfrac{1}{2}\cdot (S-q) } \\ &=& \dfrac{1}{2}\cdot \left(\left[\dfrac{(n+1)\cdot n}{2}\right]^2 - \dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3} \right) \\ &=& \dfrac{(n+1)\cdot n}{4}\cdot \left[ \dfrac{(n+1)\cdot n}{2} - \dfrac{(2n+1)}{3} \right] \\ &=& \dfrac{(n+1)\cdot n}{24}\cdot \Big[ 3n(n+1) - 2(2n+1) \Big] \\ &=& \dfrac{(n+1)\cdot n\cdot ( 3n^2-n-2)}{24} \\ &=& \dfrac{(n+1)\cdot n\cdot (3n+2)\cdot (n-1)}{24} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(n-1)\cdot n \cdot (n+1)\cdot (3n+2)}{24} } \\ \hline \end{array}$

Example: $[1,2,3,4,5]$

So $n = 5$

$\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(5-1)\cdot 5 \cdot (5+1)\cdot (3\cdot 5+2)}{24} } \\ &=& \dfrac{4\cdot 5 \cdot 6\cdot 17}{24} \\ &=& 5 \cdot 17 \\ &\mathbf{=}& \mathbf{85}\ \checkmark\\ \hline \end{array}$

Example:$[1,2,3,4,5, \ldots ,4000]$

So $n = 4000$

$\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(4000-1)\cdot 4000 \cdot (4000+1)\cdot (3\cdot 4000+2)}{24} } \\ &=& \dfrac{3999\cdot 4000 \cdot 4001\cdot 12002 }{24} \\ &=& 1333\cdot 1000 \cdot 4001\cdot 6001 \\ &\mathbf{=}& \mathbf{32\ 005\ 331\ 333\ 000} \\ \hline \end{array}$

I assume you mean "factor" rather than "fracture"!

You can type   factor(N)   (where N is an integer), or  something like factor(x^2+2x+1) for example, into the input region and then press =

7 years - that is a long time ago!

The forum looked completely different back then. 

I didn't even find the forum until Sept that year.  

Ahr  the memories :)

I do not believe the question can be answered. 3.5 is too small there is no solution.

I think that 3.5 was an approximation, not an exact value.  The exact value that they wanted was sqrt(12.8) which is approx 3.5776.  

This is an approximation of the smallest possible perimeter of the octagon.

Ok now with the proper explanaton, But I am replacing 3.5 with  P which will eventually be replaced with sqrt(12.8)

I discovered lots about rotational symmetry while I was answering this question. It is really cool!

For starters why don't I say that the perimeter of the octagon is P units and work from there.  

$a+2b=1\\ a=1-2b\\~\\ 8\sqrt{a^2+b^2}=P\\ \sqrt{1+4b^2-4b+b^2}=\frac{P}{8}\\ \sqrt{5b^2-4b+1}=\frac{P}{8}\\ 5b^2-4b+1=\frac{P^2}{64}\\ 5b^2-4b+\frac{64-P^2}{64}=0\\ \triangle=16-\frac{20(64-P^2)}{64}\\ \triangle=\frac{16*64-20(64-P^2)}{64}\\ \triangle=\frac{4*64-5(64-P^2)}{16}\\ \triangle=\frac{-64+5P^2}{16}\\$

There can be no real solution if the discriminant is negative so 5P^2 must be bigger or equal to 64

$5P^2\ge64\\ P^2\ge12.8\\ P\ge \sqrt{12.8} \quad \text{which is approximately 3.5777}$

So I am going to assume that the question intended the perimeter of the octagon to be  sqrt(12.8)

which makes the discriminant 0

$5b^2-4b+\frac{64-P^2}{64}=0\\ P=\sqrt{12.8}\\ b=\frac{4\pm0}{10}=\frac{2}{5}\\ a=1-2b=\frac{1}{5}\\ P=8\sqrt{\frac{1}{25}+\frac{4}{25}}=\sqrt{\frac{64*5}{25}}=\sqrt{12.8}\;\;\;\text{Excellent}$

Area of the octogon = $1- 4 * 0.5* a*b = 1-2*0.2*0.4 = 1-0.16 = 0.84 units^2$

How many zeroes are in the solution of 65^20-65^16

$\begin{array}{|rcll|} \hline && 65^{20}-65^{16} \\ &=& 65^{16}(65^4-1) \\ &=& 65^{16}(65^2-1)(65^2+1) \\ &=& (5\cdot 13)^{16}(4224)(4226) \quad \text{Decomposition into prime factors} \\ &=& (5\cdot 13)^{16}(2^7\cdot 3 \cdot 11)(2\cdot 2113 ) \quad 10~\text{is the factor of $5$ and $2$ } \\ &=& 2^8\cdot 5^{16}\ldots \quad \text{The minimum of $16$ and $8$ is $8$ is the number of zeroes in the product. } \\ \hline \end{array}$

In the solution of $\mathbf{65^{20}-65^{16}}$ are $\mathbf{8}$ zeroes.

Proof  $65^{20}-65^{16} = 1\ 812\ 454\ 482\ 098\ 982\ 309\ 886\ 289\ 062\ 500\ 000\ 000$

Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2.

Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3.

Find $$PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$$

$\text{Let $A_n=\dbinom{x_a}{y_a}$, $~x_a =2\cos\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big)$, $~y_a =2\sin\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big) $, $~ n=1,2,\ldots 11$ }\\ \text{Let $P=\dbinom{x_p}{y_p}$, $~x_p =3\cos(\beta)$, $~y_p =3\sin(\beta)$ } \\ \text{Let $PA^2 = (x_p-x_a)^2 + (y_p-y_a)^2$}$

$\begin{array}{|rcll|} \hline PA_1^2 &=& \left[ 3\cos(\beta) -2\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_2^2 &=& \left[ 3\cos(\beta) -2\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_3^2 &=& \left[ 3\cos(\beta) -2\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ \ldots \\\\ PA_{11}^2 &=& \left[ 3\cos(\beta) -2\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right)\right]\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && PA_1^2 + PA_2^2 + PA_3^2 + \dots + PA_{11}^2 \\\\ &=& 11\cdot 13 \\ && -12\cdot \Big\{ \cos(\beta) \Big(\underbrace{ \cos(0\cdot \dfrac{360^{\circ}}{11}) + \cos(1\cdot \dfrac{360^{\circ}}{11}) + \cos(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \cos(10\cdot \dfrac{360^{\circ}}{11}) }_{=0} \Big) \\ && +\sin(\beta) \Big(\underbrace{ \sin(0\cdot \dfrac{360^{\circ}}{11}) + \sin(1\cdot \dfrac{360^{\circ}}{11}) + \sin(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \sin(10\cdot \dfrac{360^{\circ}}{11}) \Big) }_{=0} \Big\} \\\\ &=& 11\cdot 13 -12\cdot \Big( \cos(\beta)\cdot 0 +\sin(\beta)\cdot 0 \Big) \\ &=& 11\cdot 13 -12\cdot 0 \\ &=& 11\cdot 13 \\ &\mathbf{=}& \mathbf{143} \\ \hline \end{array}$

$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \mathbf{143}$

I'm not sure if this is what you're asking about but...

cos(0)=1

Looking at the unit circle at 0°, we see that the coordinate is at (1,0). I would recomend memorizing the coordinates of 0° or 360°, 90°, 180°, and 270°. Since cos represents the x-coordinate, this means that at (1,0) or 0°, cos(0)=1.

n = 98

No, the factor that is in all the products is 15, which is why it is highlighted.

3 hours x 65 miles/hour = 195 miles

195 miles/60mph = 3.25 hours

Originally 3 hours  now 3.25 hours       .25 hours longer with stop    .25 hr = 15 minutes.

Well yah......22.5           it's just not a 'factor'          ooops....fixed..

.Miscopied from my scratch pad....  

.......AND...... you lost the bet.  

Your question states the counterfit one weighs 'different'  .....if you do not know in what manner the counterfit one weighs 'different'  (i.e. it could weigh MORE than gold ....maybe depleted uranium or something) , then the minummwould be as follows:

  .

      split the bars up two per side.......One side will be DIFFERENT......

       One side split up will be either:      =.......we know those are both genuine     (second weighing)

                                                             0r not =   (means the other two were genuine)

      third weighing with a genuine one and one from the second pair   if =   both genuine and the other from the second pair is counterfit

                                                                                                                       if not = then this one is cfit

Sooooo.... it will take THREE weighings if you do not know how the counterfit bar differs from the others (heavier or lighter)

11.49^5x+2=(1/7)^11-x

$11.49^5x+2=(1/7)^{11}-x\\ 11.49^5x+x=(1/7)^{11}-2\\ x(1+11.49^5)=(1/7)^{11}-2\\ x=\frac{(1/7)^{11}-2}{(1+11.49^5)}$

x approx equal  (1/7^11-2)/(1+11.49^5) = -0.0000099868306047

There might be some magnetic interference. You know, from all the magnets in the other question ... LOL   

90 is divisible by 4? 

x+y = -15   y = -15-x

xy=-34

x((-15-x) = -34

-x^2 -15x + 34 = 0

x^2 +15x -34 = 0

(x-2)(x+17) = 0          x = 2 and -17

Total circumf = 2 pi r

                     = 40 pi

Fraction of circumference  =   45/40pi        

2pi radians in a full circle.....the fraction of this  is       45/40pi * 2pi = 90/40 R = 2.25R