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## Here's the Solution to this Question

Solution:

(A):

1: $\{ 1,-1\}$

2: $\{ -3,-2,-1,0,1,3\}$

(B):

1: {x| x is a vowel }

2: {x| x is an integer and $-2\le x\le2$ }

(C):

1: $A\cup C=\{ a,b,c,0,1\}$

2: $C\times B=\{ (0,x),(0,y),(1,x),(1,y)\}$

3: $B-A=\{ x,y\}$

4: $(A\cap C)\cup B=\{\phi\}\cup \{x,y\}=\{x,y\}$

(D):

$A_n=2(3)^n+5$

(1): Put n = 0

$A_0=2(3)^0+5=2(1)+5=7$

(2): Put n = 5

$A_5=2(3)^5+5=2(243)+5=491$

(3): Put n = 3

$A_3=2(3)^3+5=2(27)+5=59$

(4): For 8th term, put n = 8

$A_8=2(3)^8+5=2(6561)+5=13127$

(5): For 2nd term, put n = 2

$A_2=2(3)^2+5=2(9)+5=23$

(6): $S_n=\sum [2(3)^n+5]=2\sum 3^n+5\sum 1$

$=2(3^0+3^1+...+3^n)+5n \\=2[\dfrac{1(3^{n-1}-1)}{3-1}]+5n \ [\text{Using GP}] \\=3^{n-1}+5n-1$

(E):

(F): R = {(-1,-1),(-1,0),(-1,1),(-1,2),(-1,3),(-1,4),(-1,5),(0,0),(0,1),(0,2),(0,3),(0,4),(0,5),(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5)}

(G): Domain of R = {-1, 0, 1, 2, 3, 4, 5}

(H): Range of R = {-1, 0, 1, 2, 3, 4, 5}

(I): Digraph:

(J): This relation is reflexive and transitive but not symmetric as-

Reflexive: $(x\le x)$, this is true

Transitive: $\\(x\le y)\ \& (y\le z)\Rightarrow (x\le z)$, this is true

Symmetric: $(x\le y)\Rightarrow (y\le x)$, this is not true.