Question

use the following venn diagram to find ( a^c cup b ).
venn diagram with circles a (purple, containing a, b, c, and overlapping with b (green) containing d, e, and bs non-overlapping part f, g, h, i; the universal set has j, k, l, m outside both circles)
options:
○ c,e,i,k,l,s,t,w,z
○ b,c,d,f,i,k,m,r,s,x
○ d,e,f,g,h,i,j,k,l,m
○ c,d,f,g,h,k,m,s,t,w
○ d,g,l,o,p,r,s,u,w,x
○ ∅

Explanation:

Step1: Understand \( A^c \) (Complement of A)

The complement of set \( A \) (\( A^c \)) includes all elements not in \( A \). From the Venn diagram, elements in \( A \) are \( a, b, c, d, e \). So \( A^c \) includes elements outside \( A \): \( f, g, h, i, j, k, l, m \) (and the intersection? Wait, no: \( A^c \) is all not in \( A \), so inside the universal set but not in \( A \). The universal set here has elements: in \( A \) (purple: \( a,b,c \); intersection: \( d,e \)), in \( B \) (green: \( f,g,h,i \); intersection: \( d,e \)), and outside both: \( j,k,l,m \). So \( A^c \) is elements not in \( A \): \( f,g,h,i,j,k,l,m \) (wait, no: intersection is in \( A \), so \( A^c \) is elements not in \( A \), so \( B \)'s non-intersection part (\( f,g,h,i \)) and outside both (\( j,k,l,m \))? Wait, no: \( A^c = \) universal set \( - A \). \( A \) is the purple circle (including intersection). So \( A^c \) is everything not in purple: green circle (including intersection? No, intersection is in \( A \), so green circle's non-intersection is \( f,g,h,i \), and outside both \( j,k,l,m \), and the intersection is in \( A \), so \( A^c \) is \( f,g,h,i,j,k,l,m \)? Wait, no, the intersection \( d,e \) are in \( A \), so \( A^c \) is elements not in \( A \): so \( f,g,h,i,j,k,l,m \) (since \( d,e \) are in \( A \), so not in \( A^c \); \( a,b,c \) are in \( A \), so not in \( A^c \)).

Step2: Understand \( B \) (Set B)

Set \( B \) includes elements in the green circle: intersection (\( d,e \)) and non-intersection (\( f,g,h,i \)). So \( B = \{d,e,f,g,h,i\} \).

Step3: Find \( A^c \cup B \) (Union of \( A^c \) and \( B \))

Union of two sets includes all elements in either set. \( A^c = \{f,g,h,i,j,k,l,m\} \) (wait, no, earlier mistake: \( A^c \) is all not in \( A \), so \( A \) is \( \{a,b,c,d,e\} \), so \( A^c \) is universal set minus \( A \). The universal set elements: \( a,b,c,d,e,f,g,h,i,j,k,l,m \) (from the diagram: purple has \( a,b,c \), intersection \( d,e \), green has \( f,g,h,i \), outside both \( j,k,l,m \)). So \( A^c = \{f,g,h,i,j,k,l,m\} \)? No, wait: \( A \) is the purple circle (including intersection), so \( A = \{a,b,c,d,e\} \). Then \( A^c = \) universal set \( - A = \{f,g,h,i,j,k,l,m\} \)? No, the intersection \( d,e \) are in \( A \), so \( A^c \) is elements not in \( A \): \( f,g,h,i,j,k,l,m \) (since \( d,e \) are in \( A \), so excluded from \( A^c \); \( a,b,c \) are in \( A \), excluded). Then \( B = \{d,e,f,g,h,i\} \) (green circle: intersection \( d,e \) and non-intersection \( f,g,h,i \)). Now, \( A^c \cup B \) is all elements in \( A^c \) or \( B \). So \( A^c \) has \( f,g,h,i,j,k,l,m \), \( B \) has \( d,e,f,g,h,i \). So combining them: \( d,e,f,g,h,i,j,k,l,m \) (since \( f,g,h,i \) are in both, we include them once). So the set is \( \{d,e,f,g,h,i,j,k,l,m\} \), which matches the third option.

Answer:

\( \{d,e,f,g,h,i,j,k,l,m\} \) (the third option: [d,e,f,g,h,i,j,k,l,m])