Z=2/3
decimal z=0.66667
do you need a graph?
The answer is around 79, but if you want to round it, you could round it up to 80.
it's 72°, since move two point
Answer:

Step-by-step explanation:
Given


Solving (a) and (b):
See attachment for plot and labelled vertices
Solving (c): The area
This is calculated using:

Where:
-- 
-- 
-- 
This gives:



This gives


We have been given that side length of pyramid the square base is 12 cm and its slant height is 10 cm. We are asked to find the height of the pyramid.
We will use slant height of a pyramid formula to solve our given problem.
,where,
s = Slant height,
h = Height,
a = Each side of square base.
Upon substituting our given values in above formula, we will get:



Switch sides:

Let us square both sides:




Now we will take positive square root of both sides.


Therefore, the height of the pyramid is 8 cm and option 'b' is the correct choice.