Ken=63
sis=75
ken=?+sis
63=?+57
minus 57 both sides
63-57=?+57-57
60+3-50-7=?+0
60-50+3-7=?
10-4=?
6=?
6 more stickers
Answer:
The total surface area of this square pyramid is 
Step-by-step explanation:
we know that
The surface area of a square pyramid is equal to
![SA=b^{2} +4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=SA%3Db%5E%7B2%7D%20%2B4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
we have
----> the length side of the square base
----> the height of the triangular faces
substitute the values
![SA=9^{2} +4[\frac{1}{2}(9)(12)]=297\ mm^{2}](https://tex.z-dn.net/?f=SA%3D9%5E%7B2%7D%20%2B4%5B%5Cfrac%7B1%7D%7B2%7D%289%29%2812%29%5D%3D297%5C%20mm%5E%7B2%7D)
894 mi^2
as 24 x 37.25 = 894
We have that
coordinates are
<span>A(12,7), B(12,−3), C(−2,−3), and D(−2,7)
</span>
i use the formula of midpoint <span>between A and B
</span>Xm=(x1+x2)/2 Ym=(y1+y2)/2
A(12,7) B(12,−3)
Xm=(12+12)/2=12
Ym=(7+(-3))/2=2
the point (12,2) <span>is halfway between A and B</span>
12:15 does not belong, since that is a ratio of 4:5, whilst the others are all 3:4.
