The math problems in the second year of junior high school, seeking math experts, need the process..

1. Left = y 2 + (2a + 2) y + 2 + 3a = y 2 + 2by +4b So 2a + 2 = b, 3a = 4b, a = -8 5, b = -6 5

2. Let the radius of the original cylinder be r, and the small cylinder is r, r 2*10=2 r 2*10, r = (2 of 2) r

3. Yes. Draw two perpendicular lines in the circle through the center of the circle, and the area connected by the four points that intersect the circle is 32

sina+5cosa / 6sina+2cosa=4/3+(7/3cosa)/( 6sina+2cosa)=4/3+(7/3)/( 6tana+2)=69/42

If there is an error in the calculation, please understand...

1. y^2+2y+1+2ay+3a=y^2+(2+2a)y+(1+3a)= y^2 + 2by +4b

2+2a=2b

1+3a=4b

a=-3 b=-2

2.Let the original radius be r, and the radius of the small cylinder is r(2 r 2 10) (2 10) = r 2 = 2 r 2 r = 2·r 2

3.The length of the hypotenuse of the square = (32 + 32) = 8 cm = the diameter of the circle To draw the square in the circle, that is, the length of the hypotenuse of the square cannot be greater than the diameter of the circle. That's why I can draw it in a circle.

4.(8sina+5cosa) (6sina+2cosa) divided by cosa

8sina/cosa+5)/(6sina/cosa+2)

8tana+5)/(6tana+2)

ad is a common edge, and am=an since ab=ac, and am=2bm, an=2cn

AD is the angle bisector of angle A.

So, according to the corner edge theorem, two triangles are congruent.

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It's really troublesome, this vote labor and management don't want it.

a+b=kc;

b+c=ak;

a+c=bk;

The sum of the three forms is obtained:

2(a+b+c)=k(a+b+c)

a+b+c≠0;

k=2;Hello, I'm glad to answer for you, skyhunter002 for you to answer your questions, if you don't understand anything about this question, you can ask, if you are satisfied, remember to adopt if there are other questions, click on me to ask for help after this question, it's not easy to answer the question, please understand, thank you.

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Distance traveled less after deceleration = 50

Less traveled = 30 reduced speed traveled in the same time.

You can calculate the travel time = 50 30

Time has then speed = original distance Time = (400 + 50) (50 30) = 270 km h

Using the equation, let the original velocity be x km h time t hx-30) t=400

xt=400+50

Just solve this system of equations.

The result is the same.

Speed before deceleration = 400 (50 30) = 400 (5 3) = 240 kmh.

Hello, it's a pleasure to answer your questions, skyhunter002 to answer your questions.

If you don't understand anything about this question, you can ask it, and if you are satisfied, remember.

If you have other questions, click to ask me for help after this question, it is not easy to answer the question, please understand, thank you.

Good luck with your studies.

The 1st floor is concise and correct, but it may be a little unclear.

It's better to use the method of solving equations.

The speed before deceleration is x km h

So there is. 400 x=(400-50) (x-30) gives x=240

Let the speed before deceleration be x km/h, then there is:

x-(400-50)÷400/x=30;

x-350x/400=30;

x/8=30;

x=240;

The speed is 240 km/h.

Let the speed of the deceleration train be V

Time t = 400 V

v-30)t=350

400(v-30)/v=350

v=240 km/h

Let the average velocity before deceleration be x, then.

400（x-30)/x=(400-50)

Solve x=240