40 - 4 = 36
average = 36/9 = 4
answer
average number of members who joined the math club is 4 per month
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:4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1
Step-by-step explanation:
Factor 4a^2-4a+1. 4a2 − 4a + 1 4 a 2 - 4 a + 1. Rewrite 4a2 4 a 2 as (2a)2 ( 2 a) 2. (2a)2 − 4a+1 ( 2 a) 2 - 4 a + 1. Rewrite 1 1 as 12 1 2. (2a)2 − 4a+12 ( 2 a) 2 - 4 a + 1 2. Check that the middle term is two times the product of the numbers being squared in the first term and third term. 4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1.
Step-by-step explanation:
23. area of rect,r = 11 x 3.5 = 38.5cmsq
area of tri, t = 6 x3.5/2 = 10.5 cmsq
area of shaded region, sr = r-t
=> sr = 38.5- 10.5 = 28.5cmsq
24. area of square, s = 10x10 = 100mmsq
area of circle, c = pi x 5^2 = 78.5mmsq
area of shaded region, sr = s - c
=> sr = 100- 78.5 = 22.5mmsq
25. area of circle, c = pi x 6^2 = 113.04insq
area of tri, t = 6x12/2 = 36insq
area of shaded region, sr = c - t
=> sr = 113.04 - 36 = 77.04insq
(7x6x5)/(1x2x3)=35. 35 groups :)
