All categories

3 years ago

Let A={Φ,{Φ},{Φ,{Φ}}} and B={1,2}. How many relations from A into B?

Mathematics

1 answer:

ahrayia [7]3 years ago

5 0

Answer: There are 64 relations from A to B.

Step-by-step explanation:

Number of relation from set A to B = $2^{\text{ (Number of eklemnts in A x B)}}$

$=2^{\text{(Number of elements in set A ) x (Number of elements in set B)}}$

Since, A={Φ,{Φ},{Φ,{Φ}}} and B={1,2}

i.e. Number of elements in set A = 3

Number of elements in set B = 2

So, Number of relation from set A to B = $2^{3\times 2}$

$= 2^6= 64$

Hence, there are 64 relations from A to B.

You might be interested in

The product of g and 4

ikadub [295]

Answer:

4 grams............

Step-by-step explanation:

4 0

3 years ago

A square is shown below. Which expression can be used to find the area, in square units, of the shaded triangle in the square?

AlexFokin [52]

The area of a square would me multiplying the width times the length and then multiplying by one half would be like splitting the square in half so your answer would be that last one right there. 6*6= area then multiply by 1/2 = half of area Hope this helps! :D

6 0

3 years ago

A rectangular prism has a base area of 2 square feet and a height of 5 feet. What

kvasek [131]

11 cubic feet is the answer

8 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Is 1/4 and 3/8 proportional

Sunny_sXe [5.5K]

Answer:

no it's not because is not

8 0

3 years ago

Read 2 more answers