problem

Let the line passing(-8,-8) be L1 and another one be L2

Slope of L2 : $${\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Since L1 perpendicular to L2

Slope of L1 x Slope of L2 = -1

Slope of L1 x $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ = -1

Slope of L1 = 2

Equation of L1 : $${\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}}$$=2

y+8=2x+16

2x-y+8=0

Therefore, equation of the line passing through (-8,-8) is 2x-y+8=0

Recipical (you have the wrong spelling) is exchanging the numerator and denominator

or simply saying let the number upside down

for example $${\frac{{\mathtt{7}}}{{\mathtt{8}}}}$$ , its recipical is $${\frac{{\mathtt{8}}}{{\mathtt{7}}}}$$

for an integer , for example 5 , its recipical is $${\frac{{\mathtt{1}}}{{\mathtt{5}}}}$$ ( 5=$${\frac{{\mathtt{5}}}{{\mathtt{1}}}}$$)

f(x)=3x^2 - 5

f(x-1)=3(x-1)^2-5

=3(x^2-2x+1)-5

=3x^2-6x+3-5

=3x^2-6x-2

70=(-7(2x+6)=2

70=(-7(2x+6)

-10=2x+6

x=-16/2

x=-8

(-7(2x+6)=2

2x+6=-2/7

2x=-2/7-6

x=-1/7-3=-22/7=-3.14

10(11n+5)=-10n-10+90n 

10(11n+5)=80n-10

11n+5=8n-1

3n=-6

n=-2

we can factorize it through factor the common factor

for example there is a polynomial 3x^2+x+6y^2+2y

we can have a simple grouping first , put x in a group and y in another

(3x^2+x) (+6y^2+2y)

by extracting the common factor ,

we can see x(3x+1)+2y(3y+1)

as both x(3x+1) and 2y(3y+1) have the same factor(3y+1) still

we need to factorize it also and it will become (x+2y)(3y+1)

factorization is done!

there is another way also 

I think you have some problem while typing

(11.0-12.3)^2+(10.7-12.3)^2+(13.1-12.3)^2+(13.9-12.3)^2+(12.9-12.3)^2+(12.5-12.3)^2+(11.9-12.3)^2+(12.4-12.3)^2

=(-1.3)^2+(-1.6)^2+(0.8)^2+(1.6)^2+(0.6)^2+(0.2)^2+(-0.4)^2+(0.1)^2

=1.69+2.56+0.64+2.56+0.36+0.04+0.16+0.1

=8.11

$${\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

or

$${\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{0.830\: \!127\: \!018\: \!922\: \!193\: \!2}}$$

or the formula from the topic

$${\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{1.464\: \!101\: \!615\: \!137\: \!754\: \!6}}$$

which one is you asking for?

1/2=-1/3u-2/7

1/2+2/7=-1/3u

7/14+4/14=-1/3u

11/14=-1/3u

11/14*-3=u

u=-33/14

It is a reason of proving two triangles that they are congruent.

It means they have the same sides and a same angle which is between the two sides.

For this two triangles, since PN=SQ , MP=RS , angleP=angleS

So these triangles are congruent , with the reason SAS

3+3*3-3+3

=3+9-3+3

=12

It is a complex number.

The answer is 8i.

x^2=-64

X=sqr-64

x=sqr64 x i

x=8i

It is 6.

6x6x6=216

(-5/9)k=2

k=2/(-5/9)

k=2 x  -9/5

k=-18/5

1. x=1/2

2. x=2/3

3. x=-7