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Take a map S from set A to set B, denote S: A ---> B We call A to be our domain, B our codomain. We call, with an small abuse of notation, S(A) our range, that is the set of all maps of elements of A. Or, we call set C the range of S if C = {c | c = S(a) for all a from A} Remark, C is a subset of B. Just for further knowledge, if for all a in A, S(a) is different, or S(a) != S(b) => a != b for all a, b from A (S(a) and S(b) from B), then we say S is one-to-one. It can happen when the "size" of A is smaller or equal than that of B. if the range of S is the same as the codomain. Or for all elements c from B, c = S(a) for some a from A, then we call S to be onto. It can happen when "size" of A is larger or equal to that of B. Further, if S is one-to-one AND onto, it is invertible. I will leave the proof as an exercise. Just two more note: 1. S is linear if it's a map between vector space A, B over field F which also satisfies S(a + b) = S(a) (+) S(b) and S(kb) = k.S(b) where + and x are addition and scalar multiplication from A while (+) and . are for B. 2. S is not necessarily a function.
Use the Hero's formula: Let s = (a + b + c)/2. Then the area of the triangle equals√[s(s - a)(s - b)(s - c)], where a, b, and c denote the sides of the triangle.
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
The answer depends on the information that you do have. Suppose you know all the edge lengths: the three sides of the triangle are a, b and c and the length of the prism is d. Let s = (a + b + c)/2 Then the area of the triangular cross section is sqrt[s*(s-a)*(s-b)*(s-c)] square units. So, surface area = 2*sqrt[s*(s-a)*(s-b)*(s-c)] + d*(a+b+c) square units. Volume = sqrt[s*(s-a)*(s-b)*(s-c)]*d cubic units.
120 sq metres. To see how you get this answer, read on: If the sides are a, b and c, then calculate s = 0.5*(a+b+c) Then the area is sqrt[s*(s-a)*(s-b)*(s-c)]