A math problem, ask for a high finger!

It took 20 days.

The process is as follows. If A plants x trees a day, then B and C can plant 2x, 3x together can plant 6x trees a day, and 1 3 can be completed in five days

All tasks: 6x*5*3=90x.

Set the start to the end of the sharing of y days.

Efficiency, time, and completions.

A x .y. .xy

B 2x y-3. 2x*(y-3)

C 3x y-8. 3x(y-8)

The column equation xy+2x(y-3)+3x(y-8)=90xminus x. y+2(y-3)+3(y-8)=906y=120

y=20

The time of 1 tree of A is 2 for B and 3 for C, and it can be set for A to plant x trees in 1 day, then B 2x and C 3x.

Work together for 5 days to complete a third of the total, 5(x+2x+3x)=1 3, x=1 90

There are 2 3 left, C takes 8 days off, B takes 3 days off, A does not rest, and finally completes it together, setting up y days.

xy+2x(y-3)+3x(y-8)=2/3，y/90+(y-3)*2/90+(y-8)*3/90=2/3，y=15

From start to finish, use y+5=15+5=20 days.

v A = 1 v B = 2 v C = 3 then the total workload is 90 There are 60 left in the 8 days of C's rest 5 * 2 + 1 * 8 = 18 42 left The remaining workload of 3 people is 42 (1 + 2 + 3) = 7 days The total number of days is 5 + 8 + 7 = 20 days.

If A is 1 day x , then B and C are 2x and 3x respectively

Working together for 5 days: 5(x+2x+3x)=1 3, get:x=1 90 leaving 2 3, set up for a day.

a 90 + (a-3) * 2 90 + (a-8) * 3 90 = 2 3 that is, a + 2 a - 6 + 3 a - 24 = 60

6a=90a=15

Therefore, from start to finish, a total of 5+15=20 days.

Wait a minute, if your problem is not solved, wait for me to add to you after watching the movie!

Solution: (1) Because, OAB=90°, OA=AB, therefore, OB is a right triangle of the waist of the equal Changli calendar, that is, AOB=45°, according to the nature of rotation, the distance from the corresponding point to the center of rotation is equal, that is, OA1=OA=6, the corresponding angle A1OB1= AOB=45°, and the rotation angle AOA1=90°, therefore, the degree of AOB1 is 90°+45°=135°

2) AOA1= OA1B1=90°, OA1B1, and OA=AB=A1B1, the quadrilateral OAA1B1 is a parallel resistant search quadrilateral

3) Area of oaa1b1 = 6 6 = 36

Substituting the points (1,3) into the function yields: k=3

Draw a picture, set the point a(a, b), then e(a+b 2, b 2), from the title:

b=3/a;

b/2=3/(a+b/2)

So a=(root number 6) 2

b = (root number 6).

So the coordinates of e are: (root number 6, (root number 6) 2).

Let a(x1,yi) d(x2,y1) e((x1+x2) 2,y1 2) k=3

The image of y=k x(x>0) passes through the points a and e to obtain x1*y1=3 (x1+x2) 2*y1 2=3

From the square we get y1=x2-x1

x1 = (root number 6) 2 x2 = (3 root number 6) 2 y1 = root number 6 then e (root number 6, root number 6 2).

Solution: 3=k 1 k=3 then y=3 x

Let a(x,3 x) then b(x,0) side length 3 x, e(x+3 (2x),3 (2x)),x+3 (2x))*3 (2x))=33 2+9 (4x)=3

x²=3/2

x>0, then x = root 6 2

e(Gen6, Gen6 2).

From the known point (1,3), it can be seen that the equation is y=3 x, let the coordinates of point A be (a,b), then the coordinates of point e are (a+b 2,b 2), and from point A and point e on the image of the function y=3 x, then b=3 a and b 2=3 (a+b 2), solve this system of equations, take a positive solution, and substitute the coordinates of point e as (a+b 2, b 2) to solve the problem.

Solution: The graph of this function has no intersection point with the x-axis.

Because the vertex of this function is (-83,2274) and the quadratic coefficient is 17 > 0, i.e., the opening is upward, so the vertex (-83,2274) is the lowest point.

Therefore, the graph of the function y 17(x 83) 2 2274 does not intersect with the x-axis.

Let y=0 find δ, δ0 means that there are 2 intersections, and by analogy, =0 1, 0 0.

How can this function y be equal to 0?

Add two positive numbers.

This type of general observation or letting y=0 look at the positive and negative cases of the discriminant δ.