Question

solve this fast !!!!!!!!!​

Answer 1

4³+5⁴

= 64+ 625

= 689

.................

Answer 2

Consider the first one.

We see a square and a triangle at the LHS.

→  No. of sides/vertices of a square is 4.

→  No. of sides/vertices of a triangle is 3.

The sum of nos. of these sides/vertices is 4 + 3 = 7, which is given at the RHS!

Consider the second one.

We see a triangle inscribed in another triangle at the LHS.

We know no. of sides/vertices of a triangle is 3.

If we indicate the no. of sides/vertices of the outer triangle as n, then the no. of sides/vertices of the inner triangle to the nth power will be 3ⁿ = 3³ = 27, which is given at the RHS!

Consider the third one.

We see a triangle inscribed in a square at the LHS.

We know,

→  No. of sides/vertices of a triangle is 3.

→  No. of sides/vertices of a square is 4.

If we indicate the no. of sides/vertices of the square as n, then the no. of sides/vertices of the triangle to the nth power will be 3ⁿ = 3⁴ = 81, which is given at the RHS!

Now, consider the fourth one.

We see a square inscribed in a triangle and a pentagon inscribed in a square at the LHS.

No. of sides/vertices of,

→  a triangle = 3

→  a square = 4

→  a pentagon = 5

If we indicate the no. of sides/vertices of the triangle and the square be m and n respectively, then the sum of nos. of sides/vertices of square to the mth power and pentagon to the nth power will be,

4³ + 5⁴ = 64 + 625 = 689

which should be given at the RHS!