Second year of junior high school questions! Ace enter! Junior 2 Math Problem !! Urgent!! Ace enter!

Let BC be x and AC be y, then in the triangle ABC x 2+y 2=2 2+8 2In the triangle ACD and BCD x 2-2 2 = y 2-8 2, the solution can be found to find cd

Your math learning logic is fundamentally wrong. There's a good way to do it, why not use it? Don't say anything that is not in the grade, the teacher asks. Our teachers have always liked to self-study students. I hope you can think about how exactly you want to learn.

For this problem, there is no point in using the valley theorem, it is nothing more than an algebraic problem, let a variable x, list all the equations, and solve it.

1. According to the diagram, the triangle DFO and BOE are translated to the position of the triangle KCH and the triangle AGK respectively, so that we can get that the triangle OGH is an equilateral triangle, then S AOC + S BOE + S DOF S OGH, and the triangle OGH is an equilateral triangle with a side length of 2, there is S OGH = 3 under the root number, then S AOC + S Boe + S DOF is 3 under the quadratic root number

S AOC + S Boe + S DoF + S ACK = 3) 2, according to Zu Xuan's theorem (equal area theorem) we can get that the area of the special-shaped quadrilateral bedf = square ABFE, then the sum of the area of the shaded part = 1 square meter.

The second one is very simple, the two parts of BFE and FCD are half of ABCD, AD=2, AB=1. Then the interception area should be 1.

Solution: Suppose a bus departs every x minutes at a speed of V1 and a taxi at a speed of V2, then:

5（v1+v2）=v1*x

10（v2-v1）=v1*x

Multiply Equation 1 by 2 and subtract Equation 2 to get: 20v1=v1*xx=20.

Answer: Slightly.

The farthest distance between the longitudinal functions of A and B is 1 hour, that is, during the game, the farthest distance between teams A and B is 1 hour.

n^5-5n^3+4n

n(n^4-5n^2+4)

n(n^2-4)(n^2-1)

n(n+2)(n-2)(n+1)(n-1)=(n-2)(n-1)n(n+1)(n+2) can be seen to be the multiplication of 5 consecutive natural numbers.

1 of 5 consecutive natural numbers must be a multiple of 3, 1 must be a multiple of 5, and 2 even numbers must be 1 of 4.

So the greatest common divisor is 2*3*4*5=120

The farthest at 1 hour is derived from the image.

Otherwise, you can calculate the analytic formula of the function: A y=kx through the dot (1,20) y=20x

x "1) B over the point (1,16)."

y②=16x

x" and then bring in the calculations yourself.

1 hour is the farthest, and the function graph represents the relationship between time and distance. So in 1 hour is 20-16km apart is the farthest.

Solution: Let the speed of A be 3x km/h and the speed of B be 4xkm/h.

20 minutes = 1 3 hours.

According to the title, it is called:

10/(4x)-6/(3x)=1/3

1 3x=x=so the speed of the first is km/h).

B speed: kilometers per hour).

Answer: The first speed is km/h, and the B speed is 6kmh.

Let the speed of A be 3x kmh, then the speed of B is 4xkmh, and the equation is established.

6/（3x）+1//3=10/（4x）

Solving the equation gives x=3 2, so.

The speed of A is 9 2 km/h, and the speed of B is 6 km/h.

Kao, this one is simple, A is, and the other is 6

Let the speed of A and B be 3x and 4x. respectively

10000/4x-6000/3x=20

x = meters per minute.

A speed B speed.

Two numbers are squared at the same time.

The root number 15 is 15; The root number is.

Because 15 >, so.

Root number 15> root number.

Colleague Squared: Ah, no need to estimate.

Because dg is parallel to ba, 1= 3

Since AD and EF are both perpendicular to BC, AD is parallel to EF, so 3 = 2

So 1= 2

a*a+2a+b*b-6b+10=0

a*a+2a+1+b*b-6b+9=0 (split 10 into 1 and 9) a+1) + (b-3) squared = 0 is obtained by talking about the non-negative property.

a+1=0 b-3=0

So, a=-1 b=3