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## How do you solve #\frac { x - 11} { 6} = - 1#?

Mia
Featured 3 months ago

$x = + 5$

#### Explanation:

$\text{ }$
$I f \text{ " color(red)( a/b = c/d)" " then" } \textcolor{red}{a \times d = b \times c}$
$\text{ }$
$\frac{x - 11}{6} = - 1$
$\text{ }$
$\Rightarrow \frac{x - 11}{6} = \frac{- 1}{1}$
$\text{ }$
$\Rightarrow \textcolor{red}{\left(x - 11\right) \times 1 = - 6 \times 1} \text{ }$Applying above method
$\text{ }$
$\Rightarrow x - 11 = - 6$
$\text{ }$
$\Rightarrow x = - 6 + 11$
$\text{ }$
$\Rightarrow x = + 5$

## What is the value of the x-coordinate of the solution of the system of equations #2x+7=6, y-3=x#?

Nathan
Featured 1 month ago

$\left(- 0.5 , 2.5\right)$

#### Explanation:

$2 x + 7 = 6$
$y - 3 = x$?
..................................................................................................................

Our goal in this problem is to solve for the x and y coordinate values where the two lines intersect. We can do that in a variety of ways.

I personally find substitution is the fastest and easiest method based on the way the equations are arranged.

..................................................................................................................
We are trying to solve for values of x and y that work in BOTH equations.

You may have noticed that in the second equation, we are given the value of x.

$x = y - 3$

What we need to do for substitution, is plug in $\left(y - 3\right)$ for x in the first equation. That looks like this.

$2 x + 7 = 6 \text{ }$ Our original Equation

$2 \left(y - 3\right) + 7 = 6 \text{ }$ Plugging in the value for x

$2 y - 6 + 7 = 6 \text{ }$ Using the distributive property

$2 y + 1 = 6 \text{ }$ Simplifying like terms

$2 y = 5$ Rearranging the equation by subtracting 1 from both sides

$y = 2.5 \text{ }$ Solving for y by dividing both sides by 2

Now, we have a numerical value for y.

We can plug that value into our second equation.

$x = y - 3$

$x = 2.5 - 3$

$x = - 0.5$

That leaves us with an answer of $\left(- 0.5 , 2.5\right)$

Hope I helped!

## How do you find the slope that is perpendicular to the line #2x +3y = 5#?

Geoff K.
Featured 1 month ago

Take the negative reciprocal of the given line's slope. The new slope will be $\frac{3}{2}$.

#### Explanation:

Lines that are perpendicular will have negative reciprocal slopes. Meaning, if one line's slope is $m$, then a perpendicular line will have a slope of $- \frac{1}{m}$.

Why? A line's slope is equal to its rise over its run—also written as $m = \frac{\Delta y}{\Delta x}$. If we rotate that line 90° counterclockwise (making it perpendicular to its old self), the run (to the right) becomes a rise (up), and the rise (up) becomes a backwards run (to the left):

In math terms:

$\Delta {y}_{\text{new"=Delta x" "and" "Delta x_"new}} = - \Delta y$

thus

#m_"new"=(Delta y_"new")/(Delta x_"new")=(Delta x)/(-Delta y)=-(Delta x)/(Delta y)=-1/m#

(Note: if we rotate this new line another 90° (180° total from the beginning), this 3rd line will have a slope of $\frac{- 1}{- \frac{1}{m}}$, which simplifies to $m$—the same slope of the first line, which is what we would expect.)

Okay, great—so what's the slope of $2 x + 3 y = 5$? If we rearrange this into slope-intercept form, we get

$y = - \frac{2}{3} x + \frac{5}{3}$,

meaning that for every step of "2 down", we have a step of "3 right".

The negative reciprocal of the slope $m = - \frac{2}{3}$ is

${m}_{\text{new}} = - \frac{1}{m} = \frac{- 1}{- \frac{2}{3}} = \frac{3}{2}$,

meaning that, for a perpendicular line, a step of "3 up" comes with a step of "2 right".

## Write each repeating decimal as a fraction #0.bar(09)#, #0.bar(8)#, #0.0bar(6)# ?

Arno W.
Featured 1 month ago

$0. \dot{0} \dot{9} = \frac{1}{11}$, $0. \dot{8} = \frac{8}{9}$, $0. \dot{0} \dot{6} = \frac{2}{33}$

#### Explanation:

Let's start with $0. \dot{8}$, where the 8 repeats.
Let $x = 0. \dot{8}$
When only one digit repeats, one multiplies $x$ with 10:
$10 x = 8. \dot{8}$
$10 x - x = 8. \dot{8} - 0. \dot{8}$
$9 x = 8$
$x = \frac{8}{9}$

If two digits repeat, say in $0. \dot{0} \dot{9}$ one multiplies with 100.
Let $x = 0. \dot{0} \dot{9}$
$100 x = 9. \dot{0} \dot{9}$
$100 x - x = 9. \dot{0} \dot{9} - 0. \dot{0} \dot{9}$
$99 x = 9$
$x = \frac{9}{99}$
$x = \frac{1}{11}$

Let $x = 0. \dot{0} \dot{6}$
$100 x = 6. \dot{0} \dot{6}$
$100 x - x = 6. \dot{0} \dot{6} - 0. \dot{0} \dot{6}$
$99 x = 6$
$x = \frac{6}{99}$ (You could have seen this from knowing that $0. \dot{0} \dot{9} = \frac{9}{99}$)
$x = \frac{2}{33}$

If in this case it is only the 6 that repeats, one only multiplies by 10, but you end up with a decimal to get rid of:
Let $x = 0.0 \dot{6}$
$10 x = 0.6 \dot{6}$
$10 x - x = 0.6 \dot{6} - 0.0 \dot{6}$
$9 x = 0.6$
$z = \frac{0.6}{9}$
$= \frac{6}{90}$
$= \frac{1}{15}$

## In which row and column will the number 2008 appear?

Parzival S.
Featured 3 weeks ago

column A (assumes 4 columns), row 251

#### Explanation:

We have a series of numbers that starts with:

$\left(\begin{matrix}A & B & C & D \\ 8 & 6 & 4 & 2 \\ 10 & 12 & 14 & 16 \\ 24 & 22 & 20 & 18 \\ 26 & 28 & 30 & 32 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

(the question has a column label E but no numbers within it, so for this initial answer, I'm ignoring it)

The pattern is that of a "snake" - with the initial number, 2, being in column D, then the pattern increases by 2 for each number moving from D to A, then drops down to the next row and increases by 2 again as it moves from A to D.

We're looking for the column that 2008 will appear in.

We can describe the numbers in the various rows and columns this way:

$\left(\begin{matrix}A & B & C & D \\ 2 + \left(2\right) 3 & 2 + \left(2\right) 2 & 2 + \left(2\right) 1 & 2 + \left(2\right) 0 \\ 2 + \left(2\right) 4 & 2 + \left(2\right) 5 & 2 + \left(2\right) 6 & 2 + \left(2\right) 7 \\ 2 + \left(2\right) 11 & 2 + \left(2\right) 10 & 2 + \left(2\right) 9 & 2 + \left(2\right) 8 \\ 2 + \left(2\right) 12 & 2 + \left(2\right) 13 & 2 + \left(2\right) 14 & 2 + \left(2\right) 15 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

And so with the pattern, we can find where 2008 will sit.

$2008 = 2 + 2 \left(n\right)$

$2 n = 2006$

$n = \frac{2006}{2} = 1003$

We can now divide by 8 - the remainder will tell us what column:

$\frac{1003}{8} = 125 \frac{3}{8} \implies \text{column A}$

With the column settled, let's now find the row. We can divide 2008 by 8:

$\frac{2008}{8} = 251$

We can see that 8 is on the first row, 16 on the second, 24 on the third, and so on. 2008 is evenly divisible 251 times, and so it's on the 251st row.

~~~~~
Another way to approach finding the column is to utilize bases, specifically base 8. Let's look at the chart we have, but in base 8 numbers:

$\left(\begin{matrix}A & B & C & D \\ 10 & 6 & 4 & 2 \\ 12 & 14 & 16 & 20 \\ 30 & 26 & 24 & 22 \\ 32 & 34 & 36 & 40 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

We can convert 2008 from base 10 to base 8 and use this chart to find our number

Changing from base 10, which has places such as $1 \left({10}^{0}\right) , 10 \left({10}^{1}\right) , 100 \left({10}^{2}\right)$, and so on, to base 8, which has #1 (8^0), 8 (8^1), 64 (8^3), and so on, involves finding the adjusted digits.

${8}^{3} = 512 \implies 512 \times \textcolor{red}{3} = 1536 \implies 2008 - 1536 = 472$
${8}^{2} = 64 \implies 64 \times \textcolor{g r e e n}{7} = 448 \implies 472 - 448 = 24$
${8}^{1} = 8 \implies 8 \times \textcolor{b r o w n}{3} = 24 \implies 24 - 24 = 0$

${2008}_{10} = \textcolor{red}{3} \textcolor{w h i t e}{00} \textcolor{g r e e n}{7} \textcolor{w h i t e}{00} \textcolor{b r o w n}{3} \textcolor{w h i t e}{00} 0$
$\textcolor{w h i t e}{00000000} \overline{{8}^{3}} \textcolor{w h i t e}{0} \overline{{8}^{2}} \textcolor{w h i t e}{0} \overline{{8}^{1}} \textcolor{w h i t e}{0} \overline{{8}^{0}}$

The last two digits are 30, and so that puts us in column A.

To find the row, we can take the digits, drop the final digit (because it's a 0, we're not in the middle of a series and not needing to add 1 additional row as a "final partial row", and re-evaluate the number:

$3 \left({8}^{2}\right) + 7 \left({8}^{1}\right) + 3 \left({8}^{0}\right) = 3 \left(64\right) + 7 \left(8\right) + 3 \left(1\right) = 251$

## Prove that #if u# is an odd integer, then the equation #x^2+x-u=0# has no solution that is an integer?

Steve
Featured 1 week ago

Proposition
If $u$ is an odd integer, then the equation ${x}^{2} + x - u = 0$ has no solution that is an integer.

Proof
Suppose that there exists a integer solution $m$ of the equation:

${x}^{2} + x - u = 0$

where $u$ is an odd integer. We must examine the two possible cases:

$m$ is odd; or
$m$ is even.

First, let us consider the case where $m$ is odd, then there exists an integer $k$ such that:

$m = 2 k + 1$

Now, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
# :. (2k + 1)^2 + (2k + 1) − u = 0 #
# :. (4k^2 + 4k + 1) + (2k + 1) − u = 0 #
# :. 4k^2 + 6k + 2 − u = 0 #
$\therefore u = 4 {k}^{2} + 6 k + 2$
$\therefore u = 2 \left(2 {k}^{2} + 3 k + 1\right)$

And we have a contradiction, as $2 \left(2 {k}^{2} + 3 k + 1\right)$ is even, but $u$ is odd.

Next, let us consider the case where $m$ is even, then there exists an integer $k$ such that:

$m = 2 k$

Similarly, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
# :. (2k)^2 + (2k) − u = 0 #
# :. 4k^2 + 2k − u = 0 #
$\therefore u = 4 {k}^{2} + 2 k$
$\therefore u = 2 \left(2 {k}^{2} + k\right)$

And, again, we have a contradiction, as $2 \left(2 {k}^{2} + k\right)$ is even, but $u$ is odd.

So we have proved that there is no integer solution of the equation ${x}^{2} + x - u = 0$ where $u$ is an odd integer.

Hence the proposition is proved. QED

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