A math problem for junior high school? A math problem for junior high school

From the question, we can obtain: ( x-6 y)( x+ y)=0 and x>0,y>0

Possibly x=6 y

i.e.: x=36y

Bring in and get the answer.

Summary. A math problem for junior high school

Dear, wait patiently for a <>

Am I right? There are other ways.

2 km for A, 3 km for B.

The engineering team originally planned to repair the project on average every month, B ykm

According to the title, it can be listed, 150 3 (x y).

150 (30-5) [(1+50%) x y] x 3, y 2

So. 2 km for A, 3 km for B.

Teacher, is my method correct? Is there a problem.

Your approach is also correct.

The answer above is correct.

I'm going to be in more detail.

Proven congruence. The ADP area is equal to the trapezoidal ABCD area minus (PAB+PCD).

pab+△pcd）

Because p is the point.

So the height of the PDF is equal to half the height of the trapezoidal ABCD.

According to the area formula.

Let ab=xdc=y

The area of the triangle higher than h is equal to (x+y)*h 2—(x*h 4+y*h 4)=xh 2+yh 2—xh 4—yh 4

xh/4+yh/4=（x+y)h/4

Because (x+y)h2=6

So (x+y)h4=3 square centimeters.

[Wrong question: solve, not verify].

is the midpoint of BC, which proves that PAB PFC (two inner wrong angles, one pair of vertex angles, one corresponding edge, available theorem: AAS, ASA).

2.P is the midpoint of af, and the area of the triangular ADF is equal to the area of trapezoidal ABCD3ADP and ADF are related to the lowest half height.

4.The area of APD is half the area of trapezoidal ABCD, that is, 3 square centimeters.

The parabolic analytic formula y=3 2x 2-6 5x+4 is solved, because ABCD is diamond-shaped, so c(5,4)d(2,0), the substitution is not there.

The width of the solution is xm and the length is 2xm

8x+2x+2x²）*15+2*x²*30+800=2840

x = 4 m wide and 8 m long

Set: If the width of the ground is x, the length is 2x

2x*x (top) + 2 * 2x * 2 (front and rear) + 2 * x (side) } * 15 + 2x * x * 30 + 800 = 2 840

x=4 so 4 meters wide and 8 meters long.

If the bottom surface is x meters wide, it will be 2 x meters long.

800+2x*2*2*15+x*2*15+x*2x*15+x*2x*30=2840

3x²+5x-68=0

x1 = 4 x2 = -17/3 (rounding).

x*2=8 meters.

A: The bottom surface of the garage is 8 meters long and 4 meters wide.

The answer is a bit rough, not very standardized, I forgot how to standardize, you can retouch it yourself, I hope it will be useful to you.

Solution: As shown in the figure below: is one of the cases.

The area of the circle is:

The area of the circumscribed square of the circle is: 4

When the situation is shown in the figure below; The area of the inaccessible part is: (4- ) The square piece of paper has four right angles.

The total inaccessible area is: 4 (4-) = 4-

This is moving along the inside of the square, and if the circle can move arbitrarily, there is only the first case a = 4

s=4-π

Inaccessible parts"Area = 2 * 2 - pi * 1 2 = 4 - pi

The area that cannot be touched is actually the gap when the circle is next to the edge of the square.

The answer is 4-