<span>B. square
</span><span>C. Draw arcs on the circle with the compass set at r units.
</span>
Answer:
44,286
Step-by-step explanation:
The rule is multiplying by -3.
The sequence goes -3, 9, -27, 81, -243, 729, -2187, 6561, -19683, 59049
Add all of those up and you get 44,286
Answer:
10a + 2b
Step-by-step explanation:
Add the lengths of all sides:
3a + b + b + 2a - b + 3a + b + b + 2a - b
Now combine like terms:
10a + 2b
5k3-3k+7-(-2k3+k2-9)
basically rewrite with simple algebra principles
15k-3k+7-(-6k+2k-9)
split up the parenthesis
15k-3k+7+6k-2k+9
sort them
15k-3k+6k-3k+7+9
short down by adding up the similar factors
answer: 18k+18
factorise
18(k+1)
both forms are right
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).
