A 4th grade math problem, 4th grade math problem?

Remember the depth of the well l meters, then.

The length of the rope is (2l+2 8).

After the trifold, (2l+16) 3-2=l

Solution, l = 10, the length of the rope is (2l + 2 8) = 36 Therefore, the depth of the well is 10 meters, and the length of the rope is 36 meters.

If the well is x meters deep, the rope is (2x+16) meters long.

is obtained from (2x+16) 3-2=x, x=10

The well is 10 meters deep and the rope is 36 meters long.

Set the depth of the well to be x and the length of the rope to be y

2x+8=y

3x+2=y

x=6 y=20

It should be done this way, I don't know if I ever learned such a complex equation in the fourth grade.

Set x meters deep. 2x+2*8=3x+3*2

The solution is x=10

then 2x+2*8=36

Answer: 10 meters of well, 36 meters of rope.

Set the depth of the well x2(x+8)=3(x+2).

x=10 I didn't learn the equation, look at me.

Let's meet next.

The depth of the well is 10 meters, and the length of the rope is 36 meters. Because there are 8 meters outside the folding well, that is, there are 16 meters of rope, and there are 2 meters outside the three-fold well, that is, there are 6 meters outside the three-fold well, and the three-fold is longer than the rope length of one well depth more than the two-fold, so the depth of the well is 10 meters.

There are h meters of wells.

2h+16=3h+6

Get h=10, so the rope is 36 meters.

a÷36=17……Mouth.

a 36 x 17 + port.

a 612+ port.

a 36 has a remainder and that remainder must be less than 36

The maximum number of a is 647 36x17 + 35 647, and the dividend and divisor are enlarged tenfold at the same time.

647x10)÷(36x10)

The remainder is 350 A B 23

A 20) (B 20).

It is still equal to 23

A 20 B x 20

A B x20 20

A B 23

Ax25) (B4).

Equal to 2300

A x25 B x4

A B x25x4

23x25x4

a 36 = 17, a = 17 36 = 612, a does not have a maximum and minimum problem, a is always 612. After the dividend and the divisor are enlarged tenfold at the same time, the number in the box is still 17. A B = 23, (A 20) (B 20) is also equal to 23.

A 25) (B 4) = 100 (A B ) = 100 23 = 2300

Because when the numbers in the first row and the first column are no matter how many (as if it is a), the numbers in the first row and the third column are 5 8 a (7 a) 6, and the numbers in the third row and the first column are 5 8 a (a 3) 10, so according to the sum of the numbers in the diagonal from the upper right corner to the lower left corner, the sum of the numbers in any row, column, and diagonal is: 6 8 10 24

So the numbers in the first row and the first column are: 24 8 5 11 The numbers in the second row and the third column are: 24 3 8 13 The numbers in the third row and the second column are: 24 10 5 9 So the complete chart is:

11 7 6

The sum of the numbers for each horizontal, vertical, and diagonal line is 24

From top to bottom, from left to right:

Nowadays, the questions in elementary school are so difficult++

1.After a little bit, you can draw 1 parallel line, and the first question is wrong.

2.The second question and the first question have the same meaning, wrong, wrong, wrong!

3.Square. Trapezium. Rectangles and parallelograms, both quadrilaterals, this right!

It's your homework! Be sure to keep the rolls neat and tidy!

Draw a parallel line of a known straight line over a point outside the straight line, you can draw an infinite number of lines (x) Draw a parallel line over a point outside the straight line, you can draw an infinite number of lines (x) The trapezoid is a quadrilateral, which is a special quadrilateral (, ) (check).

Wrong, wrong, right, one or two questions are all tangible, and there is only one point.

Draw the parallel lines of the known straight line at a point outside the straight line, you can draw countless lines (right) Draw the parallel lines of this straight line at a point outside the straight line, you can draw countless lines (wrong) The trapezoid is a quadrilateral, it is a special quadrilateral (false) True/False question

According to the inscription, 5-3=2, so the length is larger than the width 2, through the drawing, it can be found that the part of the square that is more than the original rectangle can be divided into four parts: the area 1=3 5=15, the area 2=2 5=10, 3, 4 parts of the rectangle have the same length on one side, that is, the width of the original rectangle = (65-15-10) (3 + 5) = 5

It's easy to set x, but I'll think about that kind of thing.

The approximate number of Kaibei for arguing a three-digit decimal place is, and the largest decimal is ( ) and the smallest is ( ).

Let the original number be x

x/52=6……40

So x=352

So the original number was 352

Then with 352 25

This gives the correct quotient of 352 25=14......2