A math problem, urgent, ask for a detailed solution

Updated on educate 2024-04-15
15 answers
  1. Anonymous users2024-02-07

    <> dear, if you are satisfied with the answer, let's do it!! Thank you.

  2. Anonymous users2024-02-06

    It's not good to write up, so I'll give you ideas.

    Divide both sides of the equation by y, it becomes an equation with x y, look at x y as a whole, solve x y equals 4 and look at the problem, divide the numerator and denominator of the problem by y, then it becomes a fraction with x y, and then bring x y=4 in, and we get 6 5

    x cannot be equal to y, if x=y=0 is not in line with the topic.

  3. Anonymous users2024-02-05

    Solution: x- xy-2y=0

    x-2y= xy 0 x 2y

    x^2-4xy+4y^2=xy

    x-4y)(x-y)=0

    x=4y x=y (rounded).

    x=4y substitution gets:

    Because x 0, y 0

    Molecule = 8y-2y = 6y

    Denominator = y + 4y = 5y

    So it is reduced to 6 5.

  4. Anonymous users2024-02-04

    Factoring the equation into ( x-2 y)( x+ y)=0 solves x=2 y or x=- y

    Substituting it into the equation results in 6 5 or 1 3

  5. Anonymous users2024-02-03

    x-root number xy-2y=0

    x-2y = root number xy

    x-2y)^2=xy

    x^2-5xy+4y^2=0

    Multiply with crosses.

    x-y)(x-4y)=0

    x-y=0 or x-4y=0

    x=y or x=4y

    When x=y.

    Original = 1 3

    When x=4y.

    Original = 6 5

  6. Anonymous users2024-02-02

    Let the area of the larger square be x1 and the smaller one be x2, there is x1+x2=468, x1-x2=180, so x1=324, x2=64, so the side length is 18,8

  7. Anonymous users2024-02-01

    So one is 18

    So the other one is 12

  8. Anonymous users2024-01-31

    If the two squares have side lengths a and b respectively, there is.

    a^2+b^2=468

    a^2-b^2=180

    The above two equations are obtained.

    a=18,b=12

  9. Anonymous users2024-01-30

    The big 18 and the small 12 let the side length of the large square be x, and the side length of the small square is y, x*x+y*y=468

    x*x-y*y=180

    The solution is x=18 and y=12

  10. Anonymous users2024-01-29

    Solution: y=f(x)=(2x+5) (x-3)=[2x-6)+11] (x-3)=2+11 (x-3), and the forest collapses x=11 (y-2)+3, i.e., f -1(x)=11 (x-2)+3

    Then f -1 (x+1) = 11 (x-1)+3 gives x=11 (y-3)+1, that is, y=g(x)=11 split key (x-3)+1

    g(4)=11 This circle(4-3)+1=11+1=12

  11. Anonymous users2024-01-28

    丨a+b丨-丨a+c丨-丨b-a丨-丨b-c丨.

    a-b+a+c-b-a-c+b

    a-b

  12. Anonymous users2024-01-27

    Let z = -1 (2 x). Because -3 x 2 and z is a monotonically increasing function of x, -8 z -1 4.

    Then the original formula can be reduced to g(z) = f(x)=z 2+z+1, because a 0, so g(z) opens upward, and the minimum abscissa z=-1 2 is in -8 z -1 4, so the minimum value of g(z) is g(-1 2)=3 4, and the maximum value is g(-8)=57

    Therefore, the maximum and minimum of f(x) can also be obtained.

  13. Anonymous users2024-01-26

    A 1 hour 1 15, B 1 hour 1 12.

    4 (1 15 + 1 12) = 5 hours.

  14. Anonymous users2024-01-25

    Solution :(1) The random variables x and y are independently and identically distributed.

    The possible values of u are 1 and 2;The possible value of v is 1,2p(u=1;v=1)=p(x=1;y=1)=p(x=1)*p(y=1)=2/3*2/3=4/9

    p(u=1;v=2)=0

    p(u=2;v=1)=p(x=2;y=1)+p(x=1;y=2)=1/3*2/3+1/3*2/3=2/9+2/9=4/9

    p(u=2;v=2)=p(x=2;y=2)=1 3*1 3=1 9 and then draw the probability distribution of (u,v), here I omit ah (2) Do you want to ask cov(u,v)

    eu=1*4/9+2*5/9=14/9

    ev=1*8/9+2*1/9=10/9

    EUV = 1 * 4 9 + 2 * 4 9 + 4 * 1 9 = 16 9 (UV distribution omitted) CoV (u, v) = EUV--- EUEV = 16 9---14 9 * 10 9 = 4 81

    If you don't understand, ask me again.

  15. Anonymous users2024-01-24

    The angle PBC is 60°, PB=PC=root number 2, SO triangle PBC is the equilateral triangle to find the height of the triangle PBC, the pyramid A-PBC, through the A perpendicular line, the intersection of the bottom surface is higher than a point D, the D point divides the height into two parts, the ratio is 1:2, find the PD, and then use the Pythagorean to find AD, AD is the height of the pyramid, the bottom equilateral triangle area is (root number 3) 4 sides of the square of the length, and the pyramid volume is 1 3 The base area times the height.

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