All categories

4 years ago

A = (1, 3, 5, 7, 9) B= 2, 4, 6, 8, 10) C = (1.5.6.7.9) A U(BOC) =

Mathematics

1 answer:

Artemon [7]4 years ago

8 0

Answer:

$A \cup (B \cap C) = (1, 3, 5, 6, 7, 9)$

Step-by-step explanation:

Set A = (1, 3, 5, 7, 9)

Set B = (2, 4, 6, 8, 10)

Set C = (1, 5, 6, 7, 9)

We have to evaluate: A U (B O C) i.e. A union (B intersection C)

First we need to evaluate B intersection C. Remember that intersection of two sets only results in common elements of both the sets. Union of two sets result in all the elements of the two sets combined.

So,

$A \cup (B \cap C) = (1, 3, 5, 7, 9) \cup [(2, 4, 6, 8, 10) \cap (1, 5, 6, 7, 9)]\\\\ A \cup (B \cap C) = (1, 3, 5, 7, 9) \cup (6)\\\\ A \cup (B \cap C) = (1, 3, 5, 6, 7, 9)$

You might be interested in

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

No links i will give you brainliest. please

CaHeK987 [17]

Answer:

n=radical of 28

Step-by-step explanation:

n^2=bc

n^2=28

n=radical of 28

8 0

3 years ago

Find equation of linear function represented by the table below in slope intercept form

icang [17]

Answer:

y=4x+3

Step-by-step explanation:

The slope of the line is (13-7)/(2-1)=4. The line equation is hence y=4x+3

7 0

3 years ago

Find the sum of all multiples of 4 between 25 and 100

Mademuasel [1]

Answer:

Sum of all multiples of 4 between 25 and 100 is 1116.

Step-by-step explanation:

Multiples of 4 between 25 and 100 = 28,32,36,40..............,96

let's check the series whether it is an arithmetic progression

a1 = 28 , a2 = 32 , a3 = 36 , a4 = 40

Common difference (d)=a2-a1=32-28 = 4

=a3-a2=36-32 = 4

=a4-a3=40-36 = 4

Common difference exists then the series of numbers is an Arithmetic progression.

To find number of terms :

a(a1) = 28 , d = 4 , l( last term ) = 96

l = a + (n-1)d

96 = 28 + (n-1)4

96 - 28 = (n-1)4

68 = (n-1)4

68/4 = (n-1)

17 = n - 1

17 + 1 = n

18 = n

Sum of all terms:

a = 28 , d = 4 , n = 18 , l = 96

Sn = n/2 ( a + l )

S18 = 18 / 2 (28 + 96)

S18 = 9 (124)

S18 = 1116

Therefore,Sum of all multiples of 4 between 25 and 100 is 1116.

NOTE:

Here 100 is also a multiple of 4 but question is asked between 25 and 100 so 100 is not considered in multiples.

7 0

3 years ago

Is (1, 4) a solution to the equation y = 7x?

VashaNatasha [74]

Answer:

Step-by-step explanation:

y=7x

4=7(1)

4=7

This is not a true statement.

4 0

4 years ago