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## How do you solve this system of equations: y < \frac { 1} { 3} x + 4 and y \geq - x + 1?

Sridhar V.
Featured 3 months ago

color(blue)((1/3)x+4 > y >= -x+1

#### Explanation:

Given:

We are given the System of Inequalities:

color(brown)(y < (1/3)x+4  and

color(brown)(y >= (-x)+1

color(green)(Step.1

color(brown)(y < (1/3)x+4  - Image of the graph created using GeoGebra

color(green)(Step.2

color(brown)(y >= (-x)+1 - Image of the graph created using GeoGebra

color(green)(Step.3

color(brown)(y < (1/3)x+4  and color(brown)(y >= (-x)+1 - Image of the combined graphs created using GeoGebra

If you observe closely, you will find the solution in a visual form.

The solution to the system of inequalities is the darker shaded region, which is the overlap of the two individual regions.

color(green)(Step.4

If you want to view just the solutions, please refer to the image below:

Note:

If the inequality is < or >, graph of the equation has a dotted line.

If the inequality is ≤ or ≥, graph of the equation has a solid line.

This line divides the xy- plane into two regions: a region that satisfies the inequality, and a region that does not.

## Beth had planned to run an average of 6 miles per hour in a race. She had a very good race and ran at an average speed of 7 mile per hour, finishing 10 minutes sooner than she would if she had averaged 6 miles per hour. How long was the race?

LM
Featured 2 months ago

$7$ miles

#### Explanation:

$s = \frac{d}{t}$

speed = distance / time

$\frac{d}{t} = 6 m i / h$

here, $d$ is the total distance of the race

and $t$ is the time, in hours, that Beth would have taken, if she ran at $6 m i / h$.

$1 h = 60 m$
she ran $10$ minutes sooner than planned, so the time taken for her to run was $t - \frac{1}{6}$. (where $t$ is in hours)

when she ran for $10$ minutes less than planned, her speed was $7 m i / h$.

$\frac{d}{t - \frac{1}{6}} = 7 m i / h$

this gives the two equations, $\frac{d}{t} = 6$ and $\frac{d}{t - \frac{1}{6}} = 7$

you can solve both by isolating distance $d :$

$\frac{d}{t} = 6$

$d = 6 t$

$\frac{d}{t - \frac{1}{6}} = 7$

$d = 7 \left(t - \frac{1}{6}\right)$

$d = 6 t , d = 7 \left(t - \frac{7}{6}\right)$

this means that $6 t = 7 \left(t - \frac{1}{6}\right)$.

using this, you can solve for time $t :$

$6 t = 7 \left(t - \frac{1}{6}\right)$

$6 t = 7 t - \frac{7}{6}$

$t - \frac{7}{6} = 6 t - 6 t = 0$

$t = \frac{7}{6} h$ (or $7 h$ $10 m$)

then $t$ can be substituted into the speed equation:

$\frac{d}{t} = 6$

$d = 6 t$

$d = 6 \cdot \frac{7}{6}$

$= \frac{7}{1}$

$= 7$

the distance of the race is $7$ miles.

## Point A(-4,1) is in standard (x,y) coordinate plane. What must be the coordinates of point B so that the line x=2 is the perpendicular bisector of ab?

Jane
Featured 2 months ago

Let,the coordinate of $B$ is $\left(a , b\right)$

So,if $A B$ is perpendicular to $x = 2$ then,its equation will be $Y = b$ where $b$ is a constant as slope for the line $x = 2$ is ${90}^{\circ}$, hence the perpendicular line will have a slope of ${0}^{\circ}$

Now,midpoint of $A B$ will be $\left(\frac{- 4 + a}{2}\right) , \left(\frac{1 + b}{2}\right)$

clearly,this point will lie on $x = 2$

So, $\frac{- 4 + a}{2} = 2$

or, $a = 8$

And this will lie as well on $y = b$

so, $\frac{1 + b}{2} = b$

or, $b = 1$

So,the coordinate is $\left(8 , 1\right)$

## If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

Jim G.
Featured 2 months ago

$y = \frac{21}{125}$

#### Explanation:

$\text{the initial statement is } y \propto \frac{x}{z} ^ 2$

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

$\Rightarrow y = k \times \frac{x}{z} ^ 2 = \frac{k x}{z} ^ 2$

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

$y = \frac{1}{6} \text{ when "x=20" and } z = 6$

$y = \frac{k x}{z} ^ 2 \Rightarrow k = \frac{y {z}^{2}}{x} = \frac{\frac{1}{6} \times 36}{20} = \frac{3}{10}$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{3 x}{10 {z}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{when "x=14" and "z=5" then}$

$y = \frac{3 \times 14}{10 \times 25} = \frac{21}{125}$

## How do you evaluate (3+ 5) ^ { 2} - 2\cdot 7?

Joseph C.
Featured 2 months ago

$50$

#### Explanation:

To simplify this expression, all we need to do, is use $P E M D A S$.

First thing we need to know is the definition of $P E M D A S$:

$P$ - Parenthesis

$E$ - Exponents

$M D$ - Multiply/Divide (Left to Right)

$A S$ - Add/Subtract (Left to Right)

This is the order we must do this, from top to bottom. This process is more commonly known as the Order of Operations .

Now that we know the order of the steps we must take to solve this, this is the solution:

${\left(3 + 5\right)}^{2} - 2 \cdot 7$

$= {\left(8\right)}^{2} - 2 \cdot 7$

$= 64 - 2 \cdot 7$

$= 64 - 14$

$= 50$

## How do you factor completely x^3 + y^3?

George C.
Featured 1 month ago

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

#### Explanation:

Given:

${x}^{3} + {y}^{3}$

Note that if $y = - x$ then this is zero. So we can deduce that $\left(x + y\right)$ is a factor and separate it out:

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

We can calculate the discriminant for the remaining homogeneous quadratic in $x$ and $y$ just like we would for a quadratic in a single variable:

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

is in standard form:

$a {x}^{2} + b x y + c {y}^{2}$

with $a = 1$, $b = - 1$ and $c = 1$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{1}\right) = - 3$

Since $\Delta < 0$, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using ${i}^{2} = - 1$ as follows:

${x}^{2} - x y + {y}^{2} = {\left(x - \frac{1}{2} y\right)}^{2} + \frac{3}{4} {y}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} + {\left(\frac{\sqrt{3}}{2} y\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} - {\left(\frac{\sqrt{3}}{2} y i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(\left(x - \frac{1}{2} y\right) - \frac{\sqrt{3}}{2} i y\right) \left(\left(x - \frac{1}{2} y\right) + \frac{\sqrt{3}}{2} i y\right)$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

So:

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

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