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## How do you find a standard form equation for the line with (6, -1) and is perpendicular to the line whose equation is 5x + 9y - 61 = 0?

Nam D.
Featured 1 month ago

$y = \frac{9}{5} x - \frac{59}{5}$

#### Explanation:

First, convert $5 x + 9 y - 61 = 0$ into $y = m x + b$

$5 x + 9 y - 61 = 0$

Therefore, $y = - \frac{5}{9} x + \frac{61}{9}$

The perpendicular line to $y = - \frac{5}{9} x + \frac{61}{9}$ will have a slope $m = \frac{9}{5}$, because the product of the perpendicular line's slope and the original line's slope (${m}_{1} {m}_{2}$) will equal $- 1$.

${y}_{2} = \frac{9}{5} x + c$ where $c$ is the y-intercept

We know that it passes through $\left(6 , - 1\right)$. Plug that in to ${y}_{2}$.

$- 1 = \frac{54}{5} + c$

$c = - \frac{59}{5}$

Therefore, the equation of the line that passes through $\left(6 , - 1\right)$ and is perpendicular to $5 x + 9 y - 61 = 0$ is

$y = \frac{9}{5} x - \frac{59}{5}$

## How do you write an equation of slope for #m = -2#?

Alan P.
Featured 1 month ago

$y = \left(- 2\right) x + b$ for any constant $b$

#### Explanation:

The general slope-intercept form (for example) is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} x + \textcolor{b l u e}{b}$
(where $\textcolor{b l u e}{b}$ is the value of the y-intercept).

Sample equations that would meet your requirement:
$\textcolor{w h i t e}{\text{XXX}} y = - 2 x + 7$

$\textcolor{w h i t e}{\text{XXX}} y = - 2 x - 84$

$\textcolor{w h i t e}{\text{XXX}} y = - 2 x$
$\textcolor{w h i t e}{\text{XXXXXXXXX}}$this would be the same as $y = - 2 x + 0$

## Given that y varies directly with x, and x = 6, y = 102, how do you write a direct variation equation that relates x and y?

Jim G.
Featured 1 month ago

$y = 17 x$

#### Explanation:

$\text{the initial statement is } y \propto x$

$\text{to convert to an equation multiply by k the constant}$
$\text{of variation}$

$\Rightarrow y = k x$

$\text{to find k use the given condition}$

$x = 6 \text{ when } y = 102$

$y = k x \Rightarrow k = \frac{y}{x} = \frac{102}{6} = 17$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = 17 x} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

## How do you graph #f(x)=(2x^2+1)/(x^3-x)# using holes, vertical and horizontal asymptotes, x and y intercepts?

Alexander L.
Featured 1 month ago

We can graph this equation using a sign chart.

#### Explanation:

First, let's identify the holes, vertical and horizontal asymptotes and x and y intercepts.

Horizontal Asymptotes
Observing the formula, we can see the degree of the denominator is higher than the numerator (The degree is the largest exponent in a polynomial). This means there will be an asymptote at $y = 0$.

Vertical Asymptotes
In order to find vertical asymptotes, we muct factor the denominator. Factoring out an $x$ from ${x}^{3} - x$, we will get $x \left({x}^{2} - 1\right)$. Equating both answers to 0 will allow us to find our asymptotes.

$x = 0$

${x}^{2} - 1 = 0$
${x}^{2} = 1$
$\sqrt{{x}^{2}} = \sqrt{1}$
$x = \pm 1$
$x \ne - 1 , 0 , 1$

$x$ can not equal any of these values as they all result in a 0 in the denominator. This will make the function undefined aka asymptotes.

Holes
Holes occur when there is a zero in both the numerator and denominator that will cancel. There are none in this formula.

X-Intercepts
To find x-intercepts, substitute 0 for $f \left(x\right)$ and solve.

$0 = \frac{2 {x}^{2} + 1}{{x}^{3} - x}$

There are no x-intercepts as the equation above would result in imaginary numbers as answers ($\frac{i \sqrt{2}}{\sqrt{2}} , - \frac{i \sqrt{2}}{\sqrt{2}}$).

Y-Intercepts
To find y-intercepts substitute 0 for $x$ and solve.

$y = \frac{2 {\left(0\right)}^{2} + 1}{{\left(0\right)}^{3} - \left(0\right)}$

As the denominator equals zero the equation is undefined. Therefore there are no y-intercepts.

Sign Chart
Using the values above, also known as critical values, we will create a sign chart. We do this by sorting the critical values from least to greatest, and surrounding them with $- \propto$ and $\propto$.

We will then select a value for $x$ in between the critical numbers.

We will test the $x$ values for positivity by substituting them for $x$ in the formula and write either the positive or negative symbol in the spaces we created. I'll do $x = 10$ as an example below.
$y = \frac{2 {\left(10\right)}^{2} + 1}{{\left(10\right)}^{3} - \left(10\right)}$
$y = \frac{67}{330}$

We don't necessarily care about the value, just whether it is positive or negative. Since it is positive, we will mark a $+$ between $1$ and $\propto$. We will do this for all the values we have chosen.

We are now able to graph now that we know where the function is positive and negative and we have the asymptotes.
graph{(2x^2+1)/(x^3-x) [-10, 10, -5, 5]}
Notice how the graph is positive where we noted positive, and negative where we noted negative.

## A rectangle has an area of 120cm^2 and a diagonal that is 17cm long. Let the length and breath of the rectangle be x and y respectively. What is the length and breath of the rectangle? Find it using Simultaneous Equation.

Jim G.
Featured 1 month ago

$15 \text{ and "8" cm}$

#### Explanation:

$\text{the diagonal 'splits' the rectangle into 2 right triangles}$

$\text{using "color(blue)"Pythagoras' theorem }$

${x}^{2} + {y}^{2} = {17}^{2}$

$\text{area of rectangle } = x y = 120$

$\text{thus we have 2 equations}$

${x}^{2} + {y}^{2} = 289 \to \left(1\right)$

$x y = 120 \to \left(2\right)$

$\text{from equation } \left(2\right) \to y = \frac{120}{x} \to \left(3\right)$

$\text{substitute "y=120/x" into equation } \left(1\right)$

$\Rightarrow {x}^{2} + {\left(\frac{120}{x}\right)}^{2} = 289$

$\Rightarrow {x}^{2} + \frac{14400}{x} ^ 2 = 289$

$\text{multiply through by } {x}^{2}$

$\Rightarrow {x}^{4} + 14400 = 289 {x}^{2}$

$\Rightarrow {x}^{4} - 289 {x}^{2} + 14400 = 0$

$\text{let } u = {x}^{2}$

$\Rightarrow {u}^{2} - 289 u + 14400 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{with "a=1,b=-289" and } c = 14400$

$\text{solve for u using the "color(blue)"quadratic formula}$

$u = \frac{289 \pm \sqrt{83521 - 57600}}{2} = \frac{289 \pm 161}{2}$

$\Rightarrow u = \frac{289 - 161}{2} \text{ or } u = \frac{289 + 161}{2}$

$\Rightarrow u = 64 \text{ or } u = 225$

$u = {x}^{2} \Rightarrow x = \sqrt{u}$

$\Rightarrow x = \sqrt{64} = 8 \text{ or } x = \sqrt{225} = 15$

$\text{substitute these values into equation } \left(3\right)$

$x = 8 \to y = \frac{120}{8} = 15$

$x = 15 \to y = \frac{120}{15} = 8$

$\text{length can be "15" or } 8$

$\text{breadth can be " 8" or } 15$

$\Rightarrow 15 \times 8 \text{ or "8xx15" are the dimensions}$

## X=sqrt(7+sqrt(7+sqrt(7+sqrt(7+...) y=sqrt(7-sqrt(7-sqrt(7-sqrt(7-...) x-y=?

Cesareo R.
Featured 4 weeks ago

$x - y = 1$

#### Explanation:

From

$x = \sqrt{7 + \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}}$

and

$y = \sqrt{7 - \sqrt{7 - \sqrt{7 - \sqrt{7 - \cdots}}}}$

we get

$x = \sqrt{7 + x}$
$y = \sqrt{7 - y}$

now squaring both sides

${x}^{2} = 7 + x$
${y}^{2} = 7 - y$

now subtracting side by side

${x}^{2} - {y}^{2} = x + y \Rightarrow \left(x + y\right) \left(x - y\right) = x + y \Rightarrow x - y = 1$

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