Answer:
4 1/6, 25/6, 4.167
Step-by-step explanation:
I'm dumb
Answer:
382 cm²
Step-by-step explanation:
Front face + Back face:
A = 2(a + b)h/2
A = 2(14 cm + 8 cm)(7 cm)/2
A = 154 cm²
Left face:
A = 7 cm × 6 cm = 42 cm²
Right face:
A = 9 cm × 6 cm = 54 cm²
Bottom face:
A = 14 cm × 6 cm = 84 cm²
Top face:
A = 6 cm × 8 cm = 48 cm²
Total surface area =
= (154 + 42 + 54 + 84 + 40) cm²
= 382 cm²
Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
Answer: 12
Step-by-step explanation:
12 times 12 is 144 so 12.in each row
Answer:
<em>C</em><em>o</em><em>r</em><em>r</em><em>e</em><em>c</em><em>t</em><em> </em><em>answer</em><em> </em><em>of</em><em> </em><em>464657474+64736825483</em><em> </em><em>is</em><em> </em><em>65201482957</em>
