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Tju [1.3M]
3 years ago
6

Given A = {(1, 3)(-1, 5)(6, 4)}, B = {(2, 0)(4, 6)(-4, 5)(0, 0)} and C = {(1, 1)(0, 2)(0, 3)(0, 4)(-3, 5)}, answer the following

multiple choice question: From the list of sets A, B, and C, state the domain of set B.
1.Domain of set B: {0, 6, 5}

2.Domain of set B: {2, 4, -4, 0}

3.Set B does not have a domain
what number is?
Mathematics
1 answer:
ANEK [815]3 years ago
8 0

B=\{(\boxed{2}\ 0);\ (\boxed{4},\ 6);\ (\boxed{-4},\ 5);\ (\boxed{0},\ 0)\}\\\\(x,\ y)-domain\ is\ "x"\\\\Answer:\ 2.\ Domain\ of\ set\ B:\{2,\ 4,\ -4,\ 0\}

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catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts. A banquet committee is to select 7 ​appetizers, 8 main​
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Answer:  The required number of ways is 46200.

Step-by-step explanation:  Given that a catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts.

A banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts.

We are to find the number of ways in which this can be done.

We know that

From n different things, we can choose r things at a time in ^nC_r ways.

So,

the number of ways in which 7 appetizers can be chosen from 8 appetizers is

n_1=^8C_7=\dfrac{8!}{7!(8-7)!}=\dfrac{8\times7!}{7!\times1}=8,

the number of ways in which 8 main courses can be chosen from 11 main courses is

n_2=^{11}C_8=\dfrac{11!}{8!(11-8)!}=\dfrac{11\times10\times9\times8!}{8!\times3\times2\times1}=165

and the number of ways in which 4 desserts can be chosen from 7 desserts is

n_3=^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!\times3\times2\times1}=35.

Therefore, the number of ways in which the banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts is given by

n=n_1\times n_2\times n_3=8\times165\times35=46200.

Thus, the required number of ways is 46200.

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