The first two binary quadratic equations, the first two binary quadratic equations

1. Let the length of the train be x meters and the speed be y meters and seconds.

1000+x=1×60y

1000-x=40y

The solution is x=200 y=20

So the train is 200m long and the speed is 20ms

2. Let the leftmost number of the six-digit number not x, and the last five digits form a new number y, x should meet the integer number within 1 to 9, and y should meet the condition that it is a five-digit number.

The original six-digit number: 100000x+y

After transposition: 10y+x

List equation 3 (100000x+y)=10y+x, according to which x satisfies integers from 1 to 9, y satisfies the condition that it is a five-digit number, i.e., 1 x 9 and x is a positive integer, and 10000 y 99999 and y is a positive integer.

The solution is x=2, y=85714 (solution: start substituting from x=1 to see if y meets the conditions, y 99999 when x 2, so there is no need to consider the situation when x 2, that is, as long as x=1 and x=2 are generated, the solution is solved).

Supplementary question: Substitute x and y: 1 2=a+2b+c=93 3=-3a+3b+c=6

0×1=b+c=2

The solution yields a=2, b=5, and c=-3

1.Let the speed be x and the length of the car be y

60x=1000+y

40x=1000-y

So x=20 y=200

So the train speed is 20m s the length of the train is 200m

2.Let the first digit of the original number be a, and the remaining digit after removing the first digit is x, then according to the meaning of the question: 3 (a 100000+x)=10x+a x=142857 a

x=142857*2=285714

1.Let the commander be L

Speed: (1000-l) 40=(1000+l) 60

get l, and then get velocity v

1.Set the length of the train L, the speed V

1000+l)/v=1min=60s

1000-l)/v=40s

The solution yields v=20m s l=200m

2.Lack of conditions.

1.Let the length of the train be x and the speed be y.

60y=1000+x

40y=1000-x, solution. x=200,y=202.Let the leftmost digit of the original number be x, and the last 5 digits of the original number be y, then.

3(100000x+y)=10y+x where x and y are natural numbers, and 1<=x<=9.

Find out the right solution. You can ask for this yourself!

3. a + 2b + c = 9

3a + 3b + c = 6

b + c = 2 , find the values of a, b, c are 2, 5, -3 respectively, so the value of the original formula is -34

1、(1000+y)=60x；(1000-y)=40x

2. Multiply by 3 and the first digit is not carried, so the first digit is , push down to get two numbers 142857, 285714

3、a+2b+c=9；3b-3a+c=6；b+c=2, get a=2, b=5, c=-3, answer -34

Solution: Let's buy x bottles for the first time and Y bottles for the second time.

x+y=70

Find y=22 x=48

x>y, in line with the requirements.

Other: x+y=70

Find y=46 x=34

A: The class bought a bottle of water for the first and second time.

Substituting x=1, we get a+b-3=0, a+b=3

(a-b) 2+4ab=a 2+2ab+b 2=(a+b) 2 is substituted into a+b

Original = 3 2 = 9

In order to improve the housing conditions of residents, a new batch of houses is built every year in a development zone, and the per capita housing area increases year by year (per capita housing area = total housing area of the district Total population of the district, unit: m people).From 1999 to 2001, the statistics of the total population and per capita housing area at the end of each year are shown in Figures (1) and (2) respectively.

Answer the following questions based on the information provided in the two figures:

1) In the two years of 2000 and 2001, the area of housing increased in 2001 compared with the previous year, and increased by 10,000 square meters.

2) Due to the needs of economic development, the total population of the district is expected to increase by 20,000 by the end of 2003 compared to the end of 2001. In order to achieve an average annual increase of 11 square metres in the total housing area of the district by the end of 2003, the average annual growth rate of the total housing area in the district in 2002 and 2003 should be reached.

k=(11-10)/（22-20）=

Average annual growth rate =

In 2002 and 2003, the average annual growth rate of the total housing area in the district should reach 50 per cent

A gets a wrong in (1) and gets the equation solution as x -3, and y -1 brings x -3, y -1 into (2) to get b = 10 B gets b in equation (2) wrong, and gets the equation solution x 5, and y 4 gets a = -1 in x 5, y 4 (1).

x+5y=15 （1）

4x-10y=-2 （2）

The solution is x=14, y=

A is wrong about A, then the answer he gets should be substituted in (2) to be true, so B = 10

B is wrong about B, then the answer he gets should be substituted in (1) to be true, so a=-1

Using this pair of a and b to solve, we get x=14, y=29 5

1、(x-3y)(x-4y)=0

x=3y or x=4y

x y=1 3 or x y=1 4

2. (x 2+y 2) 2-(x 2+y 2)-6=0(x 2+y 2+2)(x 2+y 2-3)=0x 2+y 2=-2 or x 2+y 2=3

Because x 2+y 2>0

So x 2 + y 2 = 3

3、x^2y^2-18xy+81+x^2+y^2-2xy=0(xy-9)^2+(x-y)^2=0

xy=9 and x=y

So x=3, y=3 or x=-3, y=-3

Because a is a root of x 2-5x+1=0, so a 2-5a+1=0, so a-5+1 a=0, a+1 a=5

1.The perpendicular line from A to OB is denoted as point C, and it can be seen from the figure that AOC is 30°, and since AO is 400 meters, AC is 200 meters defined by a right triangle. Since 200 250, the noise of tractors affects schools.

2.According to the conclusion of the first sub-problem, the Pythagorean theorem can be obtained to ob about 346 meters, and then the physical formula time = distance speed can be obtained, and the noise has the greatest impact on the school after the second can be obtained.

3.Do AD at ae (the two points d and e are located on both sides of point c), let them both be 250 meters, and use the cosine theorem of the triangle: a2=b2+c2-2accos a, respectively, into the obtained equation:

2502=4002+x2-2 400·x·cos30°, using the formula method: [-b + root number (b2-4·a·c)] 2a) and [-b-root number(b2-4·a·c)] 2a), you can obtain: ad=196, ae=496, ae-ad=300, and use the physical formula again:

Time = distance speed, the time is 30 seconds.

I'm so tired Is math in the first year of junior high school so hard?! The tractor turned out to be so fast

Friend, what are you trying to solve?