hectictar

$\phantom{=\quad}\dfrac{x^2+2x^4+3x^6+\cdots+1005x^{2010}}{2x+4x^3+6x^5+\cdots+2010x^{2009}}$

Factor  x  out of all the terms in the numerator and factor  2  out of all the terms in the denominator to get:

${=\quad}\dfrac{x(x+2x^3+3x^5+\cdots+1005x^{2009})}{2(x+2x^3+3x^5+\cdots+1005x^{2009})}$

Now we can cancel out the common factor from the numerator and denominator to get:

${=\quad}\dfrac{x}{2}$

gdfngkdfngkj

The difference between  √N  and  10  is less than 2. So the following two inequalities must both be true:

√N - 10  <  2     and      10 - √N  <  2

Looking at the first inequality:

√N - 10  <  2

√N  <  12

N  <  144          ( and   N > 0 )

Looking at the second inequlaity:

10 - √N  <  2

-√N  <  -8

√N  >  8

N  >  64          ( and   N > 0 )

So  N  must be a positive integer in the interval  (64, 144)

That is  79  integers.

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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60 hours   =   48 hours  +  12 hours

Total pay for the week   =   $17.25 * 48   +   1.5 * $17.25 * 12   =   $1138.50

It looks like just making a box with the \hspace command makes a narrower box like this:

$\boxed{\hspace{10cm}}$

And I tried to use the \vspace command to make it taller but it seems like that doesn't work in math mode :\

Another code I like to use to make things invisible is  \phantom{}

It will makes whatever you put inside invisible! I sometimes use it for spacing

Also there is the  \hspace  command which can be used to make a horizontal space of a specified length.

For example, the code:  a\hspace{3cm}b

makes this:

$a\hspace{3cm}b$

So there are 3 cm of space between the a and the b.

Couldn't have said it any better! :)

a*b  =  2a - b²

a*6  =  2a - 6²  =  8

2a - 6²  =  8

2a - 36  =  8

2a  =  44

a   =   22

1  =  3x - 8

9  =  3x

3  =  x

s(t(3))  =  3² + 8(3) - 2

s(  1  )  =  9 + 24 - 2

s(  1  )  =  31

Since △ABC ~ △ACD,

AC / AD   =   AB / AC

AC / AD  =  AB / AC

x / 12  =  27 / x

x * x / 12   =   27

x * x   =   27 * 12

x²   =   324

x   =   sqrt(324)

x   =   18

a/c   =   a/b * b/c

$\frac{a}{c}\ =\ \frac{a}{b}\cdot\frac{b}{c}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt{10}}{\sqrt{21}}\cdot\frac{\sqrt{14}}{\sqrt{15}}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt{10\ \cdot\ 14}}{\sqrt{21\ \cdot\ 15}}\\~\\ \frac{a}{c}\ =\ \sqrt{\frac{10\ \cdot\ 14}{21\ \cdot\ 15}}\\~\\ \frac{a}{c}\ =\ \sqrt{\frac49}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt4}{\sqrt9}\\~\\ \frac{a}{c}\ =\ \frac23$

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