What is the stoper of a vertical line

Mathematics

2 answers:

kirza4 [7]3 years ago

7 0

HORIZONTAL LINE WHICH IS THE AT WHICH ALL VERTICAL LINES MUST STOP

Sphinxa [80]3 years ago

6 0

The slope of a vertical line is 0

Hope this helps!

You might be interested in

What is three quarters of £42 and a quarter of £42

maksim [4K]

$three \ quarters \to \frac{3}{4} \\\\ \frac{3}{4} \ of \ 42\£ = \\\\ =\frac{3}{4}*42\£= \\\\ =\frac{3}{2}*21\£= \\\\ =\frac{63\£}{2} = \\\\= \boxed{31.5\£} \\\\\\\ a \ quarter \to \frac{1}{4} \\\\ \frac{1}{4} \ of \ 42\£= \\\\ =\frac{1}{4}*42\£= \\\\ =\frac{21\£}{2}= \\\\ =\boxed{10.5\£}$

3 0

3 years ago

Read 2 more answers

Terry pays £480 per month in rent.

Sholpan [36]

Answer: £516

Step-by-step explanation:

7.5 increase = 100 + 7.5 = 107.5

107.5/100 = 1.075 (multiplier)

480 x 1.075 = 516

8 0

3 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$