I have two elementary school 4th grade math problems that I can t do, and I m going to hand them in

Huanhuan departs at 7:40, catches up with Bei Bei at 7:46, takes 6 minutes, and when he turns back, the speed is doubled, and it takes 3 minutes + 6 minutes + 3 minutes = 12 minutes to return to the place where he first met Beibei.

Then after 2 minutes (8:00-7:46=14, 14-12=2) at 8:

00 arrived at the school, and it can be seen that the speed of Huanhuan's bicycle after speeding up is 7 times the speed of Beibei's walking speed (14 2=7).

So Huanhuan takes 3 minutes to ride a bike, and Bei Bei needs to walk 21 minutes.

So Bei Bei departs at 7:46-21=7:25.

Huanhuan from home to school: 8:00-7:40-6 minutes-6 minutes-6 minutes-6 minutes 2=5 minutes.

It took 14 minutes for Bei Bei to walk Huanhuan 5-3=2 minutes.

Huanhuan's 3-minute babe time: 14 2*3=21 minutes.

7:46-21 minutes = 7:25

From 7:46, it took Bei Bei 14 minutes to arrive at school, and Huanhuan arrived from home to school in 14-3-6=5 minutes. The first part of the journey took 3 minutes, and the second part of the journey took (5-3) minutes

Huanhuan 1 minute is equivalent to 7 minutes of Beibei. So Bei Bei shared from home to school.

8 points minus 35 points to 7 points 25 points.

Huanhuan and Bei Bei are classmates, live together, they go to school together, and wear school uniforms together.

Since Huanhuan is twice the original speed after the speed, the time from home to school after the title knows that Huanhuan is 20-6-3-6 = 5 minutes, and it takes 3 minutes from home to the place where he first met Beibei. Therefore, the time it takes for Huanhuan to go to school from the first time she met Bei Bei is 3-2=2 minutes. And Bebe took 20-6=14 minutes.

Therefore, the speed of Huanhuan is 14 2 = 7 times that of Beibei. Therefore, it took 7*3=21 minutes for Beibei to meet Huanhuan for the first time, and it was 7:46 minutes at this time, so Beibei's departure was divided into 46-21=25, that is, 7:

It departs in 25 minutes.

It took 50 minutes between 9:50 and 9 a.m.

The dragon rides for 50-10 = 40 minutes and walks a distance of 250 40 = 10,000 meters.

Xiaohai walked for 50 minutes, walked 60 50 = 3000 meters, because he came back to meet and walked 2 whole journeys, the whole journey = (10000 + 3000) 2 = 6500 meters.

A: The school is 6500 meters away from the bookstore.

1. There are cylindrical containers A and B with an inner diameter of 3 cm and 4 cm respectively and equal depths, first fill A with water, and then pour the water in A into B water, and the depth of water in B is 7 cm lower than the depth of water in A. The depth of water in a is h=?Centimeter?

3/2﹚ ²h=2²π×h－7﹚

h=16 cm.

2. There are cylindrical containers A and B with an inner diameter of 3 cm and 4 cm respectively and equal depths, first fill A with water, and then pour the water in A into B, and the depth of water in B is 3/4 lower than 3/4 of the depth of water in A. The depth of water in a is h=?Centimeter?

3 2 h = 2 3h 4 3 h = 16 cm.

I used to do this question without much effort.

Now I'm in junior high school, but I forgot about it.

Let x minutes catch up, 280x = 840 + 70x solution, x = 4

1. Select A. Put all the elements together.

2. Select B. The common elements of a and b are 2, and the remaining elements are put together after removing 2.

3. Select B. From x=y you can push |x|=|y|, but by |x| = |y|You can't push out x=y.

It's hard to type here, and it should be hard to look at, so I'll give you some ideas.

1) Slightly, this is a basic skill, there is nothing to talk about.

2) My idea is: since cad= dao= abc (the inner wrong angles are equal), then ac=cb.

Since the length of BC is known, and the abscissa of point A is the smaller solution of the zero point of the quadratic function, we can temporarily think that A is a constant, write the solution when the value of the quadratic function is 0 (a set of formulas is OK), obtain the abscissa of A, and combine the ordinate of C to obtain the length of AC according to the Pythagorean theorem, and solve A by AC=CB.

I haven't come up with another solution for a while, but I think there may be more than one solution to this problem.

3) Let the coordinates of e be (x0, y0), the side length of the square is y0, and then the straight line where ab is located can be regarded as an image of the primary function, a and b are both known points, and the primary function can be found, then e satisfies the expression of the primary function and obtains the equation, and then assumes that the coordinates of f are (x1, y0), then y0=x1-x0 (the same square side length) is x1=x0+y0, and the above x1 and y0 satisfy the relationship of the original quadratic function, and substitute the function formula to obtain the equation, and Synthesis, x0 and y0 can be solved, and the side length of the square can be obtained.

4) Consider two cases: ac is the waist, in addition to abc(ac=bc), you can also make ac=ap, so p on the left side of the axis of symmetry image, use a compass to draw an arc with the center of the circle as a and the radius of ac, the parabola is at two points, and the other point in addition to point b is the p-point sought.

AC is the bottom. At this time, the perpendicular bisector of ac intersects the parabola at two points (not to mention that it will not be bisectoral), and both points can be the p-point sought.

I didn't calculate all the numbers carefully, the idea is here, please lz reference.

Solution: (1) The parabolic equation y=ax -5ax+4, when x=0, y=4, the point c coordinate (0,4).

Axis of symmetry: straight line x= -[5a) 2a]=5 2, that is, straight line x=5 2 let y=4, obviously a≠0, solution x1=0, x2=5, point b coordinates (5,4)(2) ab bisect cao

and BC x-axis.

cab=∠bao=∠abc

ac=bc=5

In the right triangle ACO, AC=5, OC=4, by the Pythagorean theorem, AO=3, point A coordinates (-3,0).

The parabolic equation is substituted for point a, and the solution is a= -1 6, and the parabolic equation can also be written (3) from the coordinates of point a and point b, and the linear ab equation is found y=set point e coordinates (m,, side length=eh=

The sides of the square EFGH are on the x-axis.

EF X-axis.

eh=ef=

Point f abscissa: (

Point f ordinate:

The point f coordinates are substituted into the parabolic equation, and the value of m is solved m1 = -3 (rounded), m2 = 3, and the side length =

4) Make the middle perpendicular line of the line segment AC, and intersect the parabola at two points P1 and P2 to form two isosceles triangles.

With A as the center of the circle, AC length as the radius as the circle, and the parabola at C and P3, an isosceles triangle is formed with C as the center of the circle, AC length as the radius as the circle, and the parabola as A and B.