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\({u \over v} + {v \over u} = {u^2 \over uv} + {v^2 \over uv} = {u^2 + v^2 \over uv}\)

\({u^2 + v^2 \over uv} = {(u + v)^2 - 2uv \over uv}\)

\(uv = {c \over a} = {7 \over 3}\)

\(u + v = -{b \over a} = -{5 \over 3}\)

\({(-{5 \over 3})^2 - 2({7 \over 3}) \over {7 \over 3}} = {{25 \over 9} - {14 \over 3} \over {7 \over 3}} = {-{17 \over 9}\over {21 \over 9}} = \color{brown}\boxed{-{17 \over 21}}\)

thanks

nvm guys please disregard this question. dyslexica is beating me once again. i put 1,000 in the calculator rather than 10,000. lol.

Note: Where are " The pay rate and hours worked are given below" ??!

It's a fine answer, but it's only numbers.  

The Original Poster asked how to solve. 

2048888  2228888  2408888  2486666  2666666  2846666  Total =  6 such integers.

Note: Your question isn't clear when you say " all the digit in my thousand period are the same all the digit in my units period are the same" !!  Hence, the 6 choices above. Take your pick!.

The cosine rule says that

 \(\displaystyle \cos C= \frac{a^{2}+b^{2}-c^{2}}{2ab}.\)

Make that substitution and, in the same way, substitute for cos B and cos A.

Tidy up and you find that

\(\displaystyle a^{2}+b^{2}=c^{2},\)

so ABC is a right-angled triangle with hypotenuse c.

Like so:

number of people who have heard the rumor after cycle 1   =   3

number of people who have heard the rumor after cycle 2   =   3 + (3 + 3 + 3)

number of people who have heard the rumor after cycle 2   =   3 + 3 * 3

number of people who have heard the rumor after cycle 2   =   3 + 3²

number of people who have heard the rumor after cycle 3   =   3 + 3²  +  3² * 3

number of people who have heard the rumor after cycle 3   =   3 + 3²  +  3³

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number of people who have heard the rumor after cycle 5   =   3 + 3² + 3³ + 3⁴ + 3⁵     =   363

We are given that:

\(2\sqrt{x}+\frac1y=13\)

We can multiply both sides of this equation by  4  to get the following equation:

\(\color{purple}8\sqrt{x}+\frac4y=52\)

We are also given that:

\(8\sqrt{x}-\frac3y=3\)

From this equation, we can subtract  \(\color{purple}8\sqrt{x}+\frac4y\)  from the left side and  \(\color{purple}52\)  from the right side. We can do this because the amounts are equal, and so we are taking the exact same amount from each side, and the resulting equation is still true.

\(8\sqrt{x}-\frac3y-{\color{purple}(8\sqrt{x}+\frac4y)}\ =\ 3-{\color{purple}52}\\~\\ 8\sqrt{x}-\frac3y-8\sqrt{x}-\frac4y\ =\ -49\\~\\ -\frac3y-\frac4y\ =\ -49\\~\\ -\frac{7}{y}\ =\ -49\\~\\ -7\ =\ -49y\qquad\text{and}\quad y\neq0\\~\\ \frac{1}{7}\ =\ y\)

Now that we know what  y  is, we can find  x  by plugging in  1/7  for  y  into either equation.

\(2\sqrt{x}+\dfrac{1}{\frac17}\ =\ 13\\~\\ 2\sqrt{x}+7\ =\ 13\\~\\ 2\sqrt{x}\ =\ 6\\~\\ \sqrt{x}\ =\ 3\\~\\ x\ =\ 3^2\\~\\ x\ =\ 9\)

Now that we know  y  is  1/7  and  x  is  9,  we can find  x/y:

\(\dfrac{x}{y}\ =\ \dfrac{9}{\frac17}\ =\ 9\div\frac17\ =\ 9\cdot\frac71\ =\ 63\)

\(f(x)=\dfrac{2x-3}{x+1}\\~\\ f(3a-2)=\dfrac{2(3a-2)-3}{(3a-2)+1}\\~\\ \ \\~\\ f(3a-2)\ =\ 1\\~\\ \dfrac{2(3a-2)-3}{(3a-2)+1}\ =\ 1\\~\\ \ \\~\\ \dfrac{6a-4-3}{3a-2+1}\ =\ 1\\~\\ \dfrac{6a-7}{3a-1}\ =\ 1\\~\\ \begin{array}{} 6a-7\ =\ 1(3a-1)&\qquad\text{and}\quad3a-1\neq0\\~\\ 6a-7\ =\ 3a-1&\qquad\text{and}\quad3a-1\neq0\\~\\ 6a-3a\ =\ 7-1&\qquad\text{and}\quad3a-1\neq0\\~\\ 3a\ =\ 6&\qquad\text{and}\quad3a-1\neq0\\~\\ a\ =\ 2&\qquad\text{and}\quad a\neq\frac13 \end{array}\\~\\ a\ =\ 2 \)

\(f(x)\ =\ g(x+1)-g(x-1)\\~\\ \color{purple}{f(\frac12)}\ =\ g(\frac12+1)-g(\frac12-1)\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ g(\frac32)-g(-\frac12)\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ 1--1\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ 2\\~\\ \color{Green}{f(-\frac12)}\ =\ g(-\frac12+1)-g(-\frac12-1)\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ g(\frac12)-g(-\frac32)\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ 1--1\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ 2\\~\\ f(\frac12)+f(-\frac12)\ =\ 2+2\ =\ 4\)

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Sequences of consecutive integers:
2  * [127  to 128] = 510
2  * [ 84  to 86]    = 510
2  * [ 49  to 53]    = 510
2  * [ 40  to 45]    = 510
2  * [ 21  to 30]    = 510
2  * [ 10  to 24]    = 510
2  * [ 7  to 23]      = 510

? What are we supposed to do with this?

-5^2-5 = -25 - 5 = -30

Twice the sum of consecutive numbers n and is 510. 

One of the numbers is n  

The next consecutive number is (n + 1)  

so                                            2 • [ (n) + (n + 1) ]  =  510  

next just work it on out             2 • (2n + 1)  =  510  

                                                4n + 2  =  510  

                                                4n  =  508  

                                                  n  = 127   and   (n + 1) = 128 

the consecutive numbers are    127 and 128 

_.

(-45)^(1/3) == - 3.5568933......

(-101)^(1/3)== - 4.6570095.....

Therefore, the only integer between  - 3.5568933......   and     - 4.6570095.....== - 4

oh okay. thank you.