Answer: 665
explanation:
it follows this equation 3^n – 2^n where n is integer starting from 0

thus
3^0 – 2^0 = 1 – 1 = 0
3^1 – 2^1 = 3 – 2 = 1
3^2 – 2^2 = 9 – 4 = 5
3^3 – 2^3 = 27 – 8 = 19
3^4 – 2^4 = 65
3^5 – 2^5 = 211
3^6 – 2^6 = 729 – 64 = 665

There is one more way to solve this

3*previous number + 2^(n-1)

0x3+1=1;
1×3+2=5;
5×3+4=19;
19×3+8=65;
65×3+16=211;
211×3+32=665;

wondering how it is possible?

lets solve 3*previous number + 2^(n-1)

=> 3*(3^(n-1) – 2^(n-1)) + 2^(n-1)
=> 3^n – (2+1)*2^(n-1) + 2^(n-1)
=> 3^n – 2^n – 2^(n-1) + 2^(n-1)
=> 3^n – 2^n i.e. our first equation

Answer: X = 3
Solution
in first row next number is a multiplication of 3 and the previous number
second row, next number is a multiplication of 5 and the previous number
in third row 3rd number (147) is a multiplication of 7 and previous number, i.e. 21, thus X*7 = 21 => x = 21/7 = 3

Answer: 130
Explanation:

1=(1x1x1)+1=2
2=(2x2x2)+2=10
1=(3x3x3)+3=30
4=(4x4x4)+4=68
5=(5x5x5)+5=130

answered by Mohammad khalid!!

Answer:

3.5	+	4.5	=	8
+		+
9.5	–	3.5	=	6
||		||
13		8

Explanation:
We can make four equations
A+B = 8
C-D = 6
A+C = 13
B+D = 8

now we have four variables with four equations so it should be easy to solve. let’s solve these four equations

if we add the four equations
A+B+C-D+A+C+B+D = 8 + 6 + 13 + 8
2(A+B) + 2C = 35
2C = 35 – 2*8 —- by replacing A+B
2C = 35 – 16 = 19
C = 9.5

from 2nd Equation C – D = 6
D = C – 6
D = 9.5 – 6
D = 3.5

from 3rd equation
A+C = 13
A = 13 – C
A = 13 – 9.5
A = 3.5

from 4th equation
B + D = 8
B = 8 – D
B = 8 – 3.5
B = 4.5