Answer:
Step-by-step explanation:
The surface area of a square pyramid is the sum of the area of the squared base + 4 times the area of each triangular face, therefore:
where:
is the area of the base, where
L is the length of the base
is the area of each triangular face, where
h is the height of the face
Substituting,
For the model in this problem,
L = 12
h = 8
Therefore, the surface area here is:
In a rhombus opposite angles are equal and all 4 add up to 360
assuming one angle is 53, the opposite angle will also be 53 and then adjacent angles will be (360 - (53 x 2))/2 = 254/2 = 127*(deg)
I think the answer you're looking for is 127
Answer:
d. 11
Step-by-step explanation:
The outlier is D. 11 because it is the furthest amount from any other set.
Answer:- a.The given expression is equivalent to 
Given expression:- ![[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%283xy%5E%7B-5%7D%29%5E3%7D%7B%28x%5E%7B-2%7Dy%5E2%29%5E%7B-4%7D%7D%5D%5E%7B-2%7D)
![=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B%283%29%5E3x%5E3y%5E%7B-5%5Ctimes3%7D%7D%7Bx%5E%7B-2%5Ctimes-4%7Dy%5E%7B2%5Ctimes-4%7D%7D%5D%5E%7B-2%7D.........%28a%5Em%29%5En%3Da%5E%7Bmn%7D)
![=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B27x%5E3y%5E%7B-15%7D%7D%7Bx%5E8y%5E%7B-8%7D%7D%5D%5E%7B-2%7D)
![=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}](https://tex.z-dn.net/?f=%3D%5B27x%5E%7B3-8%7Dy%5E%7B-15-%28-8%29%7D%5D%5E%7B-2%7D............%5Cfrac%7Ba%5Em%7D%7Ba%5En%7D%3Da%5E%7Bm-n%7D)
![=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}](https://tex.z-dn.net/?f=%3D%5B27x%5E%7B-5%7Dy%5E%7B-7%7D%5D%5E%7B-2%7D%3D%2827%29%5E%7B-2%7D%28x%5E%7B-5%7D%29%5E%7B-2%7D%28y%5E%7B-7%7D%29%5E%7B-2%7D.........%28a%5Em%29%5En%3Da%5E%7Bmn%7D)
Thus a. is the right answer.
