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## Question #2e945

Cesareo R.
Featured 2 months 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

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

## What is #2x+3y=12# in point slope form?

Jim G.
Featured 3 weeks ago

$y - \frac{10}{3} = - \frac{2}{3} \left(x - 1\right)$

#### Explanation:

$\text{the equation of a line in "color(blue)"point-slope form}$ is.

#•color(white)(x)y-y_1=m(x-x_1)#

$\text{where m is the slope and "(x_1,y_1)" a point on the line}$

$\text{we require to find "m" and } \left({x}_{1} , {y}_{1}\right)$

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

#•color(white)(x)y=mx+b#

$\text{rearrange "2x+3y=12" into this form}$

$\Rightarrow 3 y = - 2 x + 12 \Rightarrow y = - \frac{2}{3} x + 4$

$\Rightarrow m = - \frac{2}{3}$

$\text{to find a point on the line choose any value for x }$
$\text{substitute into the equation and evaluate for y}$

$x = 1 \to y = - \frac{2}{3} + 4 = \frac{10}{3}$

$\text{using "m=-2/3" and } \left({x}_{1} , {y}_{1}\right) = \left(1 , \frac{10}{3}\right)$

$y - \frac{10}{3} = - \frac{2}{3} \left(x - 1\right) \leftarrow \textcolor{b l u e}{\text{in point-slope form}}$

## What is √18+√12÷√8-√48 in form of A+B√6?

Jim G.
Featured 1 week ago

$- \frac{9}{10} - \frac{2}{5} \sqrt{6}$

#### Explanation:

$\text{using the "color(blue)"rules of radicals}$

#•color(white)(x)sqrt(ab)hArrsqrtaxxsqrtb#

#•color(white)(x)(sqrta+sqrtb)(sqrta-sqrtb)=a^2-b^2#

$\text{let's begin by simplifying the given radicals}$

$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3 \sqrt{2}$

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}$

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}$

$\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4 \sqrt{3}$

$\Rightarrow \frac{\sqrt{18} + \sqrt{12}}{\sqrt{8} - \sqrt{48}} = \frac{3 \sqrt{2} + 2 \sqrt{3}}{2 \sqrt{2} - 4 \sqrt{3}}$

$\text{we now require to "color(blue)"rationalise the denominator}$

$\text{that is, eliminate the radicals from the denominator}$

$\text{multiply numerator/denominator by the "color(blue)"conjugate}$
$\text{of the denominator}$

$\text{the conjugate of "2sqrt2-4sqrt3 " is } 2 \sqrt{2} \textcolor{red}{+} 4 \sqrt{3}$

$= \frac{\left(3 \sqrt{2} + 2 \sqrt{3}\right) \left(2 \sqrt{2} + 4 \sqrt{3}\right)}{\left(2 \sqrt{2} - 4 \sqrt{3}\right) \left(2 \sqrt{2} + 4 \sqrt{3}\right)}$

$\text{expand the factors using FOIL gives}$

$= \frac{12 + 12 \sqrt{6} + 4 \sqrt{6} + 24}{8 - 48}$

$= \frac{36 + 16 \sqrt{6}}{- 40}$

$= \frac{36}{- 40} + \frac{16 \sqrt{6}}{- 40} = - \frac{9}{10} - \frac{2}{5} \sqrt{6}$

## How to prove this? 6) For sets A,B,C prove A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) by showing Left side ⊆ Right side and Right side ⊆ Left side.

Sunayan S
Featured 1 week ago

Proof:-$\text{ } A \cup \left(B \cap C\right) = \left(A \cup B\right) \cap \left(A \cup C\right)$

Let,
$\text{ } x \in A \cup \left(B \cap C\right)$

$\implies x \in A \vee x \in \left(B \cap C\right)$

$\implies x \in A \vee \left(x \in B \wedge x \in C\right)$

$\implies \left(x \in A \vee x \in B\right) \wedge \left(x \in A \vee x \in C\right)$

$\implies x \in \left(A \cup B\right) \wedge x \in \left(A \cup C\right)$

$\implies x \in \left(A \cup B\right) \cap \left(A \cup C\right)$

• $x \in A \cup \left(B \cap C\right) \implies x \in \left(A \cup B\right) \cap \left(A \cup C\right)$

#=>color(red)(Auu(BnnC)sube(AuuB)nn(AuuC)#

Let,
$\text{ } y \in \left(A \cup B\right) \cap \left(A \cup C\right)$

$\implies y \in \left(A \cup B\right) \wedge y \in \left(A \cup C\right)$

$\implies \left(y \in A \vee y \in B\right) \wedge \left(y \in A \vee y \in C\right)$

$\implies y \in A \vee \left(y \in B \wedge y \in C\right)$

$\implies y \in A \vee y \in \left(B \cap C\right)$

$\implies y \in A \cup \left(B \cap C\right)$

• $x \in \left(A \cup B\right) \cap \left(A \cup C\right) \implies x \in A \cup \left(B \cap C\right)$

#=>color(red)((AuuB)nn(AuuC)subeAuu(BnnC)#

From the both red part , we get by using the rule of equal set,

#color(red)(ul(bar(|color(green)(Auu(BnnC)=(AuuB)nn(AuuC)))|#

Hope it helps...
Thank you...

## How do you solve #8x^5 + 10x^4 = 4x^3 + 5x^2#?

Ananda Dasgupta
Featured 5 days ago

$x = 0 , \pm \frac{1}{\sqrt{2}} , - \frac{5}{4}$

#### Explanation:

We can easily rewrite $8 {x}^{5} + 10 {x}^{4} = 4 {x}^{3} + 5 {x}^{2}$ in the form

$2 {x}^{4} \left(4 x + 5\right) = {x}^{2} \left(4 x + 5\right)$

Subtracting ${x}^{2} \left(4 x + 5\right)$ from both sides, leads to

$2 {x}^{4} \left(4 x + 5\right) - {x}^{2} \left(4 x + 5\right) = 0$

which means that
'
$\left(2 {x}^{4} - {x}^{2}\right) \left(4 x + 5\right) = 0$,

i.e.

${x}^{2} \left(2 {x}^{2} - 1\right) \left(4 x + 5\right) = 0$

Now, for a polynomial to vanish, one of its factors have to be zero. So, either ${x}^{2} = 0$, or $2 {x}^{2} - 1 = 0$ or $4 x + 5 = 0$.

If ${x}^{2} = 0$, we must have $x = 0$

or if $2 {x}^{2} - 1 = 0$, we get ${x}^{2} = \frac{1}{2}$, i.e. $x = \pm \frac{1}{\sqrt{2}}$

or, finally, $4 x + 5 = 0$ which implies that $x = - \frac{5}{4}$

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

CountryGirl
Featured 5 days 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:

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