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## A(2,8), B(6,4) and C(-6,y) are collinear points find y?

Tony B
Featured 4 months ago

${P}_{C} \to \left({x}_{c} , {y}_{c}\right) = \left(- 6 , + 16\right)$

Full details shown. With practice you will be able to do this calculation type with very few lines.

#### Explanation:

$\textcolor{b l u e}{\text{The meaning of 'collinear'}}$

Lets split it into two parts

#color(brown)("co"->"together".# Think about the word cooperate
$\textcolor{w h i t e}{\text{ddddddddddddd}}$So this is 'together and operate.'
$\textcolor{w h i t e}{\text{ddddddddddddd}}$So you are doing some operation (activity)
$\textcolor{w h i t e}{\text{ddddddddddddd}}$together

$\textcolor{b r o w n}{\text{liniear".->color(white)("d}}$ In a strait line.

$\textcolor{b r o w n}{\text{collinear}} \to$ co =together, linear =on a strait line.

$\textcolor{b r o w n}{\text{So all the points are on a strait line}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{p u r p \le}{\text{Determine the gradient (slope)}}$

The gradient for part is the same as the gradient for all of it

Gradient (slope) $\to \left(\text{change in y")/("change in x}\right)$

Set point ${P}_{A} \to \left({x}_{a} , {y}_{a}\right) = \left(2 , 8\right)$
Set point ${P}_{B} \to \left({x}_{b} , {y}_{b}\right) = \left(6 , 4\right)$
Set point ${P}_{C} \to \left({x}_{c} , {y}_{c}\right) = \left(- 6 , {y}_{c}\right)$

The gradient ALWAYS reads left to right on the x-axis (for standard form)

So we read from ${P}_{A} \text{ to } {P}_{B}$ thus the we have:

Set gradient$\to m = \text{last "-" first}$

$\textcolor{w h i t e}{\text{d")"gradient " -> m=color(white)("d")P_Bcolor(white)("d")-color(white)("d}} {P}_{A}$

$\textcolor{w h i t e}{\text{dddddddddddd")m=color(white)("d,}} \frac{{y}_{b} - {y}_{a}}{{x}_{b} - {x}_{a}}$

#color(white)(dddddddddddddddddddd") (4-8)/(6-2) = -4/4=-1#

Negative 1 means that the slope (gradient) is downward as you read left to right. For 1 across there is 1 down.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{p u r p \le}{\text{Determine the value of } y}$

Determined that $m = - 1$ so by direct comparison

${P}_{C} - {P}_{A} = m = \frac{{y}_{c} - {y}_{a}}{{x}_{c} - {x}_{a}} = - 1$

$\textcolor{w h i t e}{\text{dddddddddddd.d}} \frac{{y}_{c} - 8}{- 6 - 2} = - 1$

$\textcolor{w h i t e}{\text{dddddddddddddd.}} \frac{{y}_{c} - 8}{- 8} = - 1$

Multiply both sides by (-8)

$\textcolor{w h i t e}{\text{ddddddddddddddd.}} {y}_{c} - 8 = + 8$

$\textcolor{w h i t e}{\text{ddddddddddddddddd.")y_c color(white)("d}} = + 16$

## How do I solve this quadratic equation?

Jacobi J.
Featured 4 months ago

$x = - \frac{1}{2} , - \frac{2}{3}$

#### Explanation:

We can solve this quadratic with the strategy factoring by grouping. Here, we will rewrite the $x$ term as the sum of two terms, so we can split them up and factor. Here's what I mean:

$6 {x}^{2} + \textcolor{b l u e}{7 x} + 2 = 0$

This is equivalent to the following:

$6 {x}^{2} + \textcolor{b l u e}{3 x + 4 x} + 2 = 0$

Notice, I only rewrote $7 x$ as the sum of $3 x$ and $4 x$ so we can factor. You'll see why this is useful:

$\textcolor{red}{6 {x}^{2} + 3 x} + \textcolor{\mathmr{and} a n \ge}{4 x + 2} = 0$

We can factor a $3 x$ out of the red expression, and a $2$ out of the orange expression. We get:

$\textcolor{red}{3 x \left(2 x + 1\right)} + \textcolor{\mathmr{and} a n \ge}{2 \left(2 x + 1\right)} = 0$

Since $3 x$ and $2$ are being multiplied by the same term ($2 x + 1$), we can rewrite this equation as:

$\left(3 x + 2\right) \left(2 x + 1\right) = 0$

We now set both factors equal to zero to get:

$3 x + 2 = 0$

$\implies 3 x = - 2$

$\textcolor{b l u e}{\implies x = - \frac{2}{3}}$

$2 x + 1 = 0$

$\implies 2 x = - 1$

$\textcolor{b l u e}{\implies x = - \frac{1}{2}}$

Our factors are in blue. Hope this helps!

## What is the distance between the points (-6,7) and (-1,1)?Round to the nearest whole unit.

CountryGirl
Featured 4 months ago

The distance is $8$

#### Explanation:

The easiest way is to use the distance formula, which is kinda tricky:
#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2#

That looks really complex, but if you take it slowly, I'll try and help you through it.

So let's call $\left(- 6 , 7\right)$ Point 1. Since points are given in the form $\left(x , y\right)$ we can deduct that

$- 6 = {x}_{1}$ and $7 = {y}_{1}$

Let's call $\left(- 1 , 1\right)$ Point 2. So:

$- 1 = {x}_{2}$ and $1 = {y}_{2}$

Let's plug these numbers into the distance formula:
#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2#
#d = sqrt((-1 - -6)^2 + (1 - 7)^2#
#d = sqrt((5)^2 + (-6)^2#
#d = sqrt(25 + 36#
$d = \sqrt{61}$
$d \approx 7.8$ rounded to the nearest whole unit is $8$

This is quite the hard subject, and is best taught by someone who knows how to explain well! This is a really good video about the distance formula:

## If m,n are the roots of the equation ax^2+bx+c then find the roots of the equation cx^2+bx+a?

Sunayan S
Featured 3 months ago

Answer:-$\text{ } \textcolor{red}{\frac{1}{n}}$ and #color(red)(1/m#

#### Explanation:

• If the roots of an equation #color(red)(ax^2+bx+c=0# is #color(blue)(alpha,beta#, then we can write as per rule that

• $\textcolor{red}{\alpha + \beta} = - \frac{b}{a}$
• $\textcolor{red}{\alpha \cdot \beta} = \frac{c}{a}$
• As per given condition, we can write that

• $\textcolor{red}{m + n} = - \frac{b}{a}$
• $\textcolor{red}{m \cdot n} = \frac{c}{a}$
• We will determine some values now for further use.

• $\textcolor{red}{- \frac{b}{c}} = \frac{- \frac{b}{a}}{\frac{c}{a}} = \frac{m + n}{m \cdot n}$
• #color(red)(a/c)=1/(m cdot n#
• If the roots of the equation #color(red)(cx^2+bx+a=0# is #color(blue)(alpha,beta#, then,

• $\textcolor{red}{\alpha + \beta} = - \frac{b}{c} = \frac{m + n}{m \cdot n} \text{ } \ldots \left(1\right)$
• #color(red)(alpha cdot beta)=a/c=1/(m cdot n#
• $\textcolor{red}{\alpha - \beta}$

$= \sqrt{{\left(\alpha + \beta\right)}^{2} - 4 \cdot \alpha \cdot \beta}$

$= \sqrt{{\left(\frac{m + n}{m \cdot n}\right)}^{2} - \frac{4}{m \cdot n}}$

$= \frac{m - n}{m \cdot n} \text{ } \ldots \left(2\right)$

• From $\left(1\right) \text{ & } \left(2\right)$,
• $\textcolor{red}{\alpha} = \frac{\frac{m + n}{m \cdot n} + \frac{m - n}{m \cdot n}}{2} = \frac{1}{n}$
• $\textcolor{red}{\beta} = \frac{\frac{m + n}{m \cdot n} - \frac{m - n}{m \cdot n}}{2} = \frac{1}{m}$

Hope this helps....
Thank you...

## Jimmy has 28% more pens than Kate. If Jimmy gives 7 pens to Kate, they will have the same amount of pens. How many pens does Kate have? ___ pens

Evan
Featured 1 month ago

Kate has 50 pens.

#### Explanation:

If Jimmy gives 7 pens to Kate, that means Kate got 7 pens from Jimmy. Let the number of pens Jimmy and Kate have be ${P}_{j}$ and ${P}_{k}$ respectively.

From the question, we can deduce that

#P_j=P_k*128%rarr#equation 1

${P}_{j} - 7 = {P}_{k} + 7 \rightarrow$equation 2

From equation 1,

${P}_{j} = 1.28 {P}_{k}$

From equation 2,

${P}_{j} = {P}_{k} + 14$

Since ${P}_{j} = {P}_{j}$,

$1.28 {P}_{k} = {P}_{k} + 14$

Hence,

$0.28 {P}_{k} = 14$

Solve,

${P}_{k} = 50$

## How do you simplify #-50-4((-10-4(-3+1)^2)/ (-2)) #?

Shantelle
Featured 1 month ago

$- 102$

#### Explanation:

$- 50 - 4 \left(\frac{- 10 - 4 {\left(- 3 + 1\right)}^{2}}{- 2}\right)$

To simplify this, we will use PEMDAS, shown here:

This is a common method to simplify expressions, and you can remember it using:
Please Excuse My Dear Aunt Sandy

We also start from inside to outside.

First thing we do is simplify the parenthesis:
$- 50 - 4 \left(\frac{- 10 - 4 {\left(- 2\right)}^{2}}{- 2}\right)$

Exponent:
$- 50 - 4 \left(\frac{- 10 - 4 \left(4\right)}{- 2}\right)$

Multiplication:
$- 50 - 4 \left(\frac{- 10 - 16}{- 2}\right)$

Subtract on numerator:
$- 50 - 4 \left(\frac{- 26}{- 2}\right)$

Division:
$- 50 - 4 \left(13\right)$

Multiplication:
$- 50 - 52$

And lastly, subtraction:
$- 102$

Hope this helps!

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