Find the surface area of the regular pyramid. round to the nearest tenth if necessary.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

3 0

the answer is a

hope this helps

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3 years ago

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g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and

allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, $[ab]_K$ is equal to $[ba]_K$ , thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, $[ab]_H = [ba]_H$ . Since a and b were generic elements of H, then H/G is abelian.