$k:\ y = x-11\ \ \ \Leftrightarrow\ \ \ x-y-11=0\\ and\\ l:\ y = x-7\ \ \ \Leftrightarrow\ \ \ x-y-7=0\\\\the\ distance:\\\\ d(k;l)= \frac{\big{|-11-(-7)|}}{\big{ \sqrt{1^2+1^2} }} =\frac{\big{|-11+7|}}{\big{ \sqrt{2} }} =\frac{\big{|-4|}}{\big{ \sqrt{2} }} =\frac{\big{4\cdot \sqrt{2} }}{\big{ \sqrt{2}\cdot \sqrt{2} }} =\frac{\big{4 \sqrt{2} }}{\big{2 }} =2 \sqrt{2}$

6 0

4 years ago

Read 2 more answers

Can someone actually help me with this question? It's about: Complementary and supplementary angles (visual)

Sophie [7]

90-41 =49
So the answer to x =49 degrees
Complimentary angles are two angles that equal 90 that us why to solve we subtract 41 from 90

4 0

2 years ago

0.3x=15<br> Identify the slope and y-intercept.

masha68 [24]

Steps to solve:

0.3x = 15

~Divide 0.3 to both sides

0.3x/0.3 = 15/0.3

~Simplify

x = 50

This is a vertical line so the slope is undefined.

The line passes through the point (50, 0) and there is no y-intercept since its a straight vertical line.

Best of Luck!

8 0

3 years ago