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Ann [662]
3 years ago
11

3. The following sets are not equal - An (BUC) and AU(BnC) Construct a universe U and non-empty sets A, B, and C so that the abo

ve sets are in fact the same. . Construct a universe U and non-empty sets A, B, and C so that the above sets are in fact different.
Mathematics
1 answer:
yarga [219]3 years ago
3 0

Answer:

1.U={1,2,3,4,5}

A={2}

B={2,3}

C={4,5}

2.U={1,2,3,4}

A={1,2}

B={2,3}

C={4}

Step-by-step explanation:

We are given that A\cap (B\cup C) and A\cup (B\cap C)

are different sets

1.We have to construct a universe set U and non empty sets A,B and C so that above set in fact the same

Suppose U={1,2,3,4,5}

A={2}

B={2,3}

C={4,5}

B\cap C=\phi

B\cup C={2,3,4,5}

A\cap (B\cup C)={2}\cap{2,3,4,5}={2}

A\cup (B\cap C)={2}\cup\phi={2}

Hence, A\cap (B\cup C)=A\cup (B\cap C)

2.We have to construct a universe set U and non empty sets A,B and C so that  above sets are in fact different

Suppose U={1,2,3,4}

A={1,2}

B={2,3}

C={4}

B\cap C=\phi

B\cup C={2,3,4}

A\cup (B\cap C)={1,2}\cup \phi={1,2}

A\cap (B\cup C)={1,2}\cap {2,3,4}={2}

Hence, A\cap (B\cup C)\neq A\cup (B\cap C)

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Step-by-step explanation:

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2. Determine the sum of the first 400 ODD numbers.<br><br>​
il63 [147K]

Odd numbers take the form 2n-1, where n\ge1 is an integer. When n=400, the last odd number would be 799. So we're adding

S=1+3+5+\cdots+795+797+799

By reversing the order of terms, we have

S=799+797+795+\cdots+5+3+1

and we can pair up terms in both sums at the same position to write

2S=(1+799)+(3+797)+(5+795)+\cdots(795+5)+(797+3)+(799+1)

so that we are basically adding 400 copies of 800, and from there we can find the value of the sum right away:

2S=400\cdot800\implies S=160,000

###

We could also make use of the formulas,

\displaystyle\sum_{i=1}^n1=n

\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}2

We have

S=\displaystyle\sum_{i=1}^{400}(2i-1)=2\sum_{i=1}^{400}i-\sum_{i=1}^{400}1=400(400+1)-400=400^2=160,000

3 0
4 years ago
Can you help me with this problem please
Over [174]

Remark

I think you want us to do both 3 and 4


Three

QR is 4 points going left from the y axis + 2 points going right from the y axis.

QR = 6 units long.


RT is 4 units above the x axis and 3 units below the x axis

RT = 7 units long


TS = 3 units. For this one you just go from T to S. The graph really helps you. You just need to count.


Problem 4

You need to find the area of the combined figure. You could do it as a trapezoid, which might be the easiest way. I'll do that first.


Area = (b1 + b2)*h/2


Givens

B1 = QR = 6

B2 = UT + TS = 6 + 3 = 9

h = RT = 7


Area

Area = (6 + 9)*7/2

Area = 15 * 7 / 2

Area = 52.5


Comment

You could break this up into a rectangle + a triangle

Find the area of QRTU and add the Area of triangle RTS


<em>Area of the rectangle </em>= L * W

L = RT = 7

W = QR = 6

Area = 7 *6 = 42


<em>Area of the triangle</em> = 1/2 * B * H

B = TS = 3

H = RT =7

Area = 1/2 * 3 * 7

Area = 1/2 * 21

Area = 10.5


Total Area = 42 + 10.5 = 52.5 Both answers agree.

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