Senior 1 Math Exercises on Sets, please

Updated on educate 2024-02-08
11 answers
  1. Anonymous users2024-02-05

    Piece; represents the intersection of the straight line 2x-y=1,x+4y=5, and the intersection is found as (1,1) on the straight line y=x, then d is encapsulated by c;

    b=,(1)When a=3,a b=,a b=empty set; (2) When a=1, a b=, a b={1}; (3) When a=4, a b=, a b={4}; (4) When a is not equal to 1, 3, 4, a b = and a b = empty set;

    Since a (cub)=,b=,a=,

  2. Anonymous users2024-02-04

    Scores. k 2 + 4 = (hail with light 2k + 1) line friend 4n = k 4 + 2 = (k + 2) 4

    2k+1 is always an odd number.

    k+2 may be odd or even.

    The source width is contained in m and n

  3. Anonymous users2024-02-03

    8 Yes, (a,b) and (b,a) are different sets, so the ab set can be equal, a b = so there must be 1,3 elements in the ab set.

    a=,b=---2 pairs.

    a=,b=---2 pairs.

    a=,b=---2 pairs.

    a=,b=---2 pairs.

    a=,b=---1 pair.

    It's not right just now.,It's not in line with the topic of a b=.,This is right for you.。

  4. Anonymous users2024-02-02

    a={x |4x+p<0} closed pei lao={x |x<-p 4} and a are contained in the car lift b

    The result is -p 4 less than or equal to -1

    So p is greater than or equal to 4

  5. Anonymous users2024-02-01

    p 4> Kai socks-1 and talk about Sun Bang-p Han talk 4>2

    P<4 and P<-8

    The value range is p<-8

  6. Anonymous users2024-01-31

    The set a=,p= q=, so there is: p=,q=

    Since p=q, then there is: q=

    If q=, then there is: a=0, p=q=

    If q=, then there is: a 2 = a + 1, there is a = (1 + 5) 2

  7. Anonymous users2024-01-30

    cua=,k∈z}

    cub=,k∈z}

    So there is: a is the true subset of cub and b is the true subset of cua.

  8. Anonymous users2024-01-29

    I've just learned, so I hope to communicate more in the future.

    If all elements of set A are also elements of set B, then A is said to be a subset of B;

    If there is an element in b and not in a, and a is a subset of b, then a is said to be a true subset of b.

    a.{Yes {x|.}x root number ten} (false).

    When x=3, x" root number ten, so { is not {x|x root number ten}, not a true subset.

    b.{1,2,3} is a subset (error) of {0,1,2,5}

    Element 3 does not belong to the set of Min homozygous {0,1,2,5}

    c.{x|x-1=0} is {x|(x-1)(x+2)=0} of the Mako and Stool set (correct).

    x-1=0, then x=1

    x-1)(x+2)=0, then x1=1, x2=-2

    So, {x|x=1} is {x|x1=1,x2=-2

    The true subset of the bridge shed.

    d.{-5,7} is {x|x 20} (error).

    Elements -5,7 do not belong to the set {x|x²<20}

    I think this kind of question starts with the concept, I hope it can help you.

  9. Anonymous users2024-01-28

    c subset is that it can have the same as itself, and the true subset is in addition to itself, so don't understand it, and then ask.

  10. Anonymous users2024-01-27

    For example, in analytical and computational physics, it is necessary to have knowledge and skills such as mathematical functions and equation solving. (5) Ability requirements: Of course, I am a mathematics teacher. I have brought back many graduating classes of Gao Liquid World Slow High School, and I am currently teaching the first year of high school. Thank you for looking at me.

  11. Anonymous users2024-01-26

    The square of a3 is less than 10

    The latter does not have a 3 in B, and the former is not a subset of the World Bank of the Zoye latter.

    c Pair. D7 squared Punch pants 20

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