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More generally, if you want the equation of the line in the form y = mx + c, then m = -2/3 and you find c from: 1 = -(2/3)*2 + c   (so c = 7/3).

If the two ponts are (1, 3) and (2, 3) then the line between them is horizontal (the two y-values are the same), so it has zero slope.

In general with points (x1, y1) and (x2, y2) the slope is given by (y2 - y1)/(x2 - x1)

What is the average rate of change?

\(\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)\)

\(n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}\)

\(81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9\)

\(F= \dfrac{260N}{4} = 65N\\ d = 4(0.5m)= 2m\\ \text{Note that the work done }F\cdot d \text{ is constant}\)

\(\text{What are we supposed to vary here in order to get the max and min values?}\\ \text{The order of the numbers? The parens, or lack of them?}\\ \text{This is a very unclear question}\)

ok here is my logic.

\((x^2+\frac{1}{x})^n\)

let a be an integer such that    \(0\le a \le n\)

the a'th term will be

\(\begin{pmatrix}n\\a\end{pmatrix}[x^2]^a\;\left[ \frac{1}{x} \right]^{n-a}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{2a}\;\times x ^{a-n}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{3a-n}\\\ \)

this will be a constant term whenever

  \(3a-n=0\\n=3a\)

Since a is an integer, the expression will have a constant term if n is a multiple of 3

The value of the constant term is already included in my answer.

 2016 NS 28

\(\begin{array}{|ll|} \hline 0^2+1^2+2^2+\ldots +7^2+8^2 &+9^2 +10^2+\ldots +15^2+16^2 \\ +17^2+18^2+19^2+\ldots +24^2+25^2 &+26^2+27^2\ldots +32^2+33^2 \\ \dots \\ +2006^2+2007^2+2008^2+\ldots +2013^2+2014^2 &+2015^2+2016^2 \pmod{17}\\ =\\ \color{red}\underbrace{0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\dots \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}+9^2+10^2 \pmod{17} \\\\ =9^2+10^2 \pmod{17}\\ =11 \pmod{17}\\ \hline \end{array}\)

\(\text{I have a simpler answer.}\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac{1}{(2k)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\)

\(\sum \limits_{k=1}^\infty \dfrac{1}{(2k)^3} = \sum \limits_{k=1}^\infty \dfrac{1}{8k^3} = \dfrac 1 8 \sum \limits_{k=1}^\infty \dfrac{1}{k^3}\)

\(\dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac 1 8 \sum \limits_{k=1}^\infty\dfrac{1}{k^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} + \dfrac 1 8 = 1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} = \dfrac 7 8\)

Plot the point (2,1)

Plot the point 3 left and 2 up from it, i.e. (-1,3)

Draw a line through those two points.

The point 3 right and 2 down works as well, i.e (5,-1)

incorrect answer deleted

That came out of Mathematica.

The squiggle stands for the Riemann Zeta function.

It's really more advanced stuff than I know.

Yeah.  That's a pretty standard result.

I made a frame with two pieces of slat 5" long and two pieces 4" long

The inside perimeter is 14" around  (4+4+3+3) 

As we move from left to right.....the function rises on the  x interval of (0, 1)    and again on the x interval of (2, infinity)

So....these are intervals of increase

(f/g)(x)   =     f(x)           2x^2 - 5x - 3       (2x + 1)(x - 3)             x - 3    

                  ____  =      __________  =   ____________=      _______ 

                    g(x)             2x^2 + x            x(2x + 1)                      x

Note that  x  cannot = 0   or  -1/2    (these make the original denominator 0.....)

Think of it this way:

It costs 12 dollars to get in.

It costs 1.25 dollars for each ride.

You have 27 dollars in your pocket and you're going to spend it all.

How many rides can you go on? 

.

 y=-265x+2800

The company will have no money when y =  0.....so

0 = -265x + 2800      subtract 2800 from both sides

-2800 = - 265x      divide both sides by -265

≈10.56  = x

So....they can hire a max of 10 workers and still remain solvent....(the 11th worker will sink them )

F(x) = 1.25x + 12  

Find x when F(x) = 27 

This isn't too complicated, just substitute 27 in place of F(x) and solve for x

Like this:                                                   27 = 1.25x + 12 

Swap sides of the equation                      1.25x + 12 = 27                 this isn't necessary, I just prefer the x on the left

Subtract 12 from both sides                     1.25x = 27 – 12 = 15

Divide both sides by 1.25                          x = 15/1.25 = 12

F(x) =  1.25x + 12

When F(x) = 27, we have

27 = 1.25x + 12    subtract 12 from both sides

15 = 1.25 x      divide both sides by 1.25

12 = x

This means that the number of rides = 12

(And the admission fee is  $12 )

To see Melody's  answer

Note that......when n = 1

(x^2 + 1/x)^(3*1)  = (x^2 + 1/x)^3  will have the term C(3,2)(x^2)^1 * (1/x)^2  = 3x^2(1/x^2)

Constant term = 3

When n = 2.....

(x^2 + 1/x)^(3*2) = (x^2 + 1/x)^6  will have the term C(6, 4)(x^2)^2 * (1/x)^4 = 15 (x^4)(1/x^4) 

Constant term = 15

When n = 3....

(x^2 + 1/x)(3 * 3)   will have the term  C(9, 6) (x^2)^3 *(1/x)^6  = 84 (x^6)(1/x^6) 

Constant term = 84

So.....it appears the the pattern for the constant term will be  

C(3n, 2n) * (x^2)^n * (1/x)^(2n)      where n is a positive integer

How did you get that? And what is that notation?