This is a trick question. First just understand that any number over itself is 1 so

is just

so it is just
2
radians = degrees x PI/180
radians = 60 x pi/180 = 1.04719 radians
round answer as needed
if you need it in terms of pi it would be PI/3 radians
0.6666 (repeating) x 0.75 = 0.5
0.6666 (repeating) / 0.75 = 0.8888888 (repeating)
Step-by-step explanation:
Hey there!
The points of line AB are; (-1,-4) and (2,11).
Note:
- Use double point formula and simplify it to get two eqaution.
- Use condition of parallel lines, perpendicular lines to know whether the lines are parallel or perpendicular or nothing.
~ Use double point formula.

~ Keep all values.

~ Simplify it.



Therefore this is the equation of line AB.
Now, Finding the equation of line CD.
Given;
The points of line CD are; (1,1) and (4,10).
~ Using formula.

~ Keep all values.

~ Simplify it.


Therefore, 3x - y- 2 = 0 is the eqaution of line CD.
Use condition of parallel lines.
m1= m2
Slope of equation (i)


Therefore, m1 = 5
Slope of second equation.


Therefore, m2 = 3.
Now, m1≠m2.
So, the lies are not parallel.
Check for perpendicular.
m1*m2= -1
3*5≠-1.
Therefore, they aren't perpendicular too.
So, they are neither.
<em><u>Hope </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
<h3>The lateral area for the pyramid with the equilateral base is 144 square units</h3>
<em><u>Solution:</u></em>
The given pyramid has 3 lateral triangular side
The figure is attached below
Base of triangle = 12 unit
<em><u>Find the perpendicular</u></em>
By Pythagoras theorem

Therefore,

<em><u>Find the lateral surface area of 1 triangle</u></em>


<em><u>Thus, lateral surface area of 3 triangle is:</u></em>
3 x 48 = 144
Thus lateral area for the pyramid with the equilateral base is 144 square units