Verify that |p(a)| = 2^n , if |a| = n for n = 0, 1, 2, 3.

We have to prove that,

|P(A)| = , if |A| = n for n = 0, 1, 2, 3.

For n = 0,

A = {}

P(A) = { {} } = = 1

For n = 1,

A = { a }     ( suppose )

P(A) = { {}, a } = = 2,

For n = 2,

A = { a, b }

P(A) = { {}, {a}, {b}, {a, b} = = 4,

For n = 3,

A = { a, b, c },

P(A) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } = = 8

Thus, it is verified for n = 0, 1, 2, 3.

Now, suppose it is valid for a set B having k elements,

That is, |P(B)| =

Also, there is a set A,

Such that, A = B ∪ {x}

Since, after including the element x in set B,

The element x will be come with every element of set B in the power set of B,

i.e. P(A) = = = =

Hence, by the induction it has been proved,

|P(A)| = , if |A| = n, Where, n∈ N ( set of natural numbers )

u dumb boi

step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq