Please help me explain the exercises about people teaching math eight and I am a middle school stude

One by one, I can't see clearly.

According to parallel.

Triangle ode triangle abc

According to the angle bisector intersects O, so cd db=3 2, cd cb=3 5, be bc=3 5

Set db=2x and cd=3x

This gives CB=5x=20x=4

So cd=12, db=8, be =12, so de=4

de bc = abc perimeter ode perimeter = 4 20 = 1 5od ab = 1 5, through the od point to do om perpendicular to ab to m, on perpendicular to ac to n, of perpendicular to bc to f set om = on = of=h1

Connect the AF, because the triangle ode triangle ABC so the AF is perpendicular to the BC

Let af=h2

1 2ab*h1+1 2ac*h1+1 2sc*h1=1 2bc*h2 (area).

And because h1 h2=3 5 h1= 3 5h2, then 1 2*3 5*h2(ab+ac+bc)=1 2bch2, so ab+ac+bc=5 3bc=100 3, so ode perimeter=1 5*(ab+ac+bc)=20 3

It's very simple, the circumference is 20cm, because the question has only one number, which is absolutely correct.

The two sides of the ODC are isosceles triangles ODC

The circumference is 20

Solution: (1) Set the number of printed copies to x, factory A charges for y A, and factory B charges for y B, according to the title

yA = , i.e. yA = >=1000).

y B = , i.e. y B = > = 1000).

2) When yA&YB:i.e.", the solution inequality gets: x<1800;

When yA=yB: i.e., the solution of inequality gets: x=1800;

When y a, the solution to the inequality yields: x>1800

If you want to print 3,000 copies of the brochure, you should choose Factory A, and the cost is:

yA = yuan) Answer: slightly.

a1, a2, a3 are the three root elimination chains of f(x)-1=0.

So f(x)-1=a*(x-a1)*(x-a2)*(x-a3) and a=0, so f(x)=a*(x-a1)*(x-a2)*(x-a3)+1

f(b)=a*(b-a1)*(b-a2)*(b-a3)+1a*(b-a1)*(b-a2)*(b-a3)!=0, so Nathanoson f(b)=0

Prove to be ambushed.

Proof : Let g(x)=f(x)-1 be known from the known conditions of repentance f(a1)=f(a2)=f(a3)=1 so g(a1)=g(a2)=g(a3)=0

g(x) is also a cubic polynomial with integer coefficients.

So g(x) must be written in this form =a(x-a1)(x-a2)(x-a3) where a is an integer number.

Because b is not equal to a1, a2, a3, a(x-a1)(x-a2)(x-a3) is not equal to 0

i.e. g(b) is not equal to 0 i.e. f(b) is not equal to 1

A school organized 340 teachers and students to travel, with 170 pieces of luggage, and planned to rent a total of 10 cars of two types: A and B. It is understood that each car A can carry up to 40 people and 16 pieces of luggage, and car B can carry up to 30 people and 20 pieces of luggage.

1}Please help you design the car rental plan.

2} For example, the rent of a car with Biguo is 2,000 yuan per car, and the rent of car B is 1,800 yuan per car.

I will first say the answer to the last question, the last question comes out, the first two substitutions give the result 180-1 2 [360-(180-n)], the final result is 90-n 2 in turn, the first question is 75, the second question is 60 hope to adopt.

Based on the triangle inner angle and 180, there is also the nature of the angle bisector is easily calculated.

I will give you the most complete answer, wait patiently.

Column equation bar 1 a:x b:80-x 2400*<=28*x+20(80-x)<=2400*

2.Finding 2 under the condition that 1 is satisfied, the equation is.

400*x*

First question. a+b=80 b=80-a

2400*<=28a+20b<=2400* Substitute b=80-a.

Get 40<=a<=55

The second question is to draw a diagram of linear programming.

z=40<=x<=55

y=80-x

y>=0

Pretty much. Use the following three styles to form a shadow, and use the first one to move the maximum point

Let A have x rooms and B have Y rooms.

x+y=80 。。1）

28x+20y≥2400×

28x+20y≤2400×

Simplify: 1920 28x+20y 2040 480 7x+5y 510 combined with (1) to draw coordinates.

Since the storefront can't be a decimal, the result can be obtained.

1. A can have between 40 and 55 stores.