Applied Mathematics Textbook answers compiled by Yan Suzhen

Updated on educate 2024-02-09
17 answers
  1. Anonymous users2024-02-06

    Textbooks can be used:

    Mathematical Analysis (Liang Zao, Fudan University Press).

    Advanced Algebra (Peking University).

  2. Anonymous users2024-02-05

    Since I am a math major, I have limited time during the summer vacation.

    I suggest you take a look.

    Mathematical Analysis (North Grinding University Press) is fast and stupid.

    Advanced Algebra (Xianke Zhang).

    The second one is difficult.,But since it's a math major.,Don't read the math books of the acre of engineering.。。。

  3. Anonymous users2024-02-04

    I am a 08 student majoring in mathematics and applied mathematics, and the mathematical analysis I use is from Fudan University, and the advanced algebra is from Peking University.

  4. Anonymous users2024-02-03

    55+5=60 The greatest common factor of 36 and 60 is 12, so there are a maximum of 12 students with three good students.

    30=15+15=14+16=13+17……Calculated (15*12)*4=720 square centimeters.

    Minimum, so the area of this label paper should be at least 720 square centimeters.

    The bottom area is 50 * 30 = 1500 square centimeters 3000 1500 = 2 10-2 = 8, so the water surface is 8 cm from the mouth of the water tank.

    4.There were 21 insect specimens and 14 butterfly specimens, the number of butterfly specimens accounted for 14 21 of the total number of insect specimens, and the number of simplified butterfly specimens accounted for two-thirds of the total number of insect specimens.

  5. Anonymous users2024-02-02

    Mathematical analysis at East China Normal University, ordinary differential equations at Northeast Normal University, and advanced algebra.

  6. Anonymous users2024-02-01

    Solution: If the length of the side perpendicular to the wall is x meters, the area of this hut is y square meters, and the length of the other side is (20-2x) meters.

    y=x(20-2x) from the title

    2x2+20x

    2(x2-10x)

    2(x2-10x+25-25)

    2﹝(x-5)2-25﹞

    2(x-5)2+50

    So, when x = 5, y has a maximum, y max = 50 is obtained by x = 5, 20-2x = 20-10 = 10 (m) Therefore, a rectangle with a length of 10 meters and a width of 5 meters can make the hut the largest area, and the maximum area is 50 square meters.

  7. Anonymous users2024-01-31

    Let the length of one side of the rectangle be x, then the length of the other side is (20-2*x) 2f(x)=x*(20-2x) 2, that is, f(x)=-x 2+10x parabola, the opening is downward, the axis of symmetry is x=5, and in its defined domain, therefore, when x=5, the area of the rectangle is the largest, that is, it is enclosed into a square.

  8. Anonymous users2024-01-30

    The largest area is enclosed into a square, 20m against the wall only encloses 3 sides, the side connected to the wall is x, and the other side is 20 2x

    Area s x * (20 2x).

    2x^2+20x

    2(x^2-10x)

    2(x^2-10x+25-25)

    2(x^2-10x+25)+50

    2(x-5)^2+50

    When (x 5) 2 0, s is maximum, i.e., when x 5, s is maximum 50

  9. Anonymous users2024-01-29

    Let the width of the hut be x meters, then the length (10-x) meters, and the area x(10-x) is the maximum area of the hut, that is, when the maximum value of the quadratic function y=x(10-x) is found x=5, y gets the maximum value of 50

    A: When the length is 10 meters and the width is 5 meters, the enclosed hut has the largest area of 50 square meters.

  10. Anonymous users2024-01-28

    Let the width be x, then the length is (20-2x), and the area function can be expressed as.

    y=x(20-2x)=-2xsquared+20x

    When x = 20 [(-2)(-2)] = 5 meters, the length is 10 meters, and the area is the largest.

  11. Anonymous users2024-01-27

    Let the width be x, the length be (20-2x), and the area be y

    So: y=x(20-2x)=-2x squared + 20xx=20 [(-2)(-2)].

    x=20/4

    x=5 so:

    Length=10The area is the largest.

  12. Anonymous users2024-01-26

    If the length is x and the width is 20-x, then the area is s=x*(20-x).

    s=-(x-10)²+100

    When x=10, s is maximum! The maximum is 100 square meters!

  13. Anonymous users2024-01-25

    The length and width of the hut are x, y

    Then there is 2x+y=20, find the maximum value of xy.

    Bring y=20-x into xy to get 20x-2x.

    i.e. -2 (x-squared -10x).

    According to the graph, the maximum value is 50, where x=5 and y=5

  14. Anonymous users2024-01-24

    Using the original wall, the square with a side length of 10 meters has the largest area.

    10 10 100 square meters.

  15. Anonymous users2024-01-23

    Anshi University is a very rubbish school, don't take the test, if you want to take the test, just take the mathematics test of the University of Science and Technology of China or Anhui University.

  16. Anonymous users2024-01-22

    If the road is not very far, it is recommended that you go directly to the school to inquire, many students have finished the course, the books are lost, and it is not convenient to collect them directly? Or go to the faculty of the department.

  17. Anonymous users2024-01-21

    Mathematical Analysis Advanced Algebra Spatial Analytic Geometry Probability Theory and Mathematical Statistics Complex Variable Functions Real Variable Functions Functional Analysis Mathematical Physical Equations Ordinary Differential Equations Abstract Algebra Matrix Analysis Topology (Elective).

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