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Black_prince [1.1K]
3 years ago
9

Polynomial function F(x)=2-11x+2x²​

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0

Answer:

should be 4

Step-by-step explanation:

You might be interested in
Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are
Svetlanka [38]

Answer:

(a) Number of sets B given that

  • A⊆B⊆C: 2¹⁰.  (That is: A is a subset of B, B is a subset of C. B might be equal to C)
  • A⊂B⊂C: 2¹⁰ - 2.  (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let x_1, x_2, \cdots, x_{20} denote the 20 elements of set X.

Let x_1, x_2, \cdots, x_{10} denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain x_1, x_2, \cdots, x_{10}.

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from x_1, x_2, \cdots, x_{20}.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 2^{10} -2 = 1024 - 2 = 1022 possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's 2^{20} = 1048576 possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 2^{20} - 2^{10} possibilities left.

5 0
2 years ago
Q # 19 The data below shows the. number of hours a week on average group of students spend volunteering for community service pr
alisha [4.7K]
If those two are your only choices then the answer is none of the above. 

<em>Attached is a cumulative frequency table for your data.</em>

If you take a look at the two tables given, the frequencies were not tallied properly. If the frequency column is wrong, then the cumulative frequency will be wrong. 

The answer is then none of the above or find one that matches the table attached. 

7 0
3 years ago
PLEASE HELP ME!!!!!!
Sonja [21]

Answer:

\huge \red{ c_3 = - 1}

Step-by-step explanation:

c_1 = 1 \\  \\ c_n= - 2c_{n-1}+5 \\  \\ c_2 = - 2c_{2-1}+5 \\  \\ c_2 = - 2c_{1}+5 \\  \\ c_2 = - 2(1)+5 \\  \\ c_2 = - 2+5 \\  \\  \huge \purple{c_2 =3} \\  \\ c_3 = - 2c_{3-1}+5 \\  \\ c_3 = - 2c_{2}+5 \\  \\ c_3 = - 2(3)+5 \\  \\ c_3= - 6+5 \\  \\ \huge \red{ c_3 = - 1}

6 0
3 years ago
Connie bought 172 tulip bulbs. She wants to plant an equal number of bulbs in each of 4 flower beds. How many bulbs should Conni
Levart [38]

Answer:

43

Step-by-step explanation:

172/4= 43

4 0
3 years ago
What does 14x - 200y = when x = 12 and y = 3.2?
myrzilka [38]
14(12)-200(3.2)
168-640= -472
5 0
2 years ago
Read 2 more answers
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