A junior high school math problem, please speed

Year 1 x = 1; Maintenance fee y=20, then substitute the equation: 20 = a + b in the second year x = 2; Maintenance fee y = 40 + 20, then substitute the equation: 60 = a * 4 + b * 2

Solve the system of equations: a = 10 b = 10

So the expression for y is: y = 10 x +10 x2) to the cost of year x: 1000 + 10 x +10 x column equation:

1000 + 10 x +10 x < = 330 x solution inequality x should be 4 years.

It's hard to read the y expression in your question clearly!

1) Solution: Because when x=1, y=20 when x=2, y=60, substitute y=ax +bx to get:

20=a+b

60=4a+2b

Solution: a=10

b=10y=10x²+10x

2) Solution: If it takes z years to recover the cost, then it can be obtained in the z year.

330z-y=1000

Solution: Z = After it is put into production, the enterprise will be able to recover its investment in the first year.

1)20=a+b

40=a2²+b2

Solve the system of equations, and the expression y=5x +10x2) lets the cost recover in z years, then it can be obtained in year z.

1000+5z²+10z=330z

You will always solve a quadratic equation, there is a formula in the book).

Let the shadow of the flagpole be x long, then there is.

Solve x = connect the highest point of the flagpole to the end of its shadow, which has an intersection with the building. Let the distance between this intersection and the ground be y, and compare the length of y and 5m to draw conclusions. y>5m, then the shadow length can reach the window on the second floor of the library. On the contrary, no.

Well, according to the similarity triangle theorem, there is.

The solution is y=>5m

So you can get to the window on the second floor.

1) Bring in the rubber scatter b coordinates, a=1. The quadratic function is y=x circular pose.

Let the linear solution beam cavity analysis formula y=kx+b, bring in b and c coordinates k+b=1,-2k+b=4 solution, k=-1, b=, and the intersection point with the y-axis is (0, 2), which may be recorded as e. Intersection with the x-axis a coordinate (2,0).

2)s△obc=s△oce+s△obe=1/2*oe×2+1/2*oe×1=3

s△oad=1/2×2×|Point d ordinate |The ordinate of the =3 d point is 3, and since the minimum value of the quadratic function is 0, the ordinate is only 3. Bring y=3 into y=x, x= point coordinates (- 3,3) or.

As shown in the figure, the middle line on the hypotenuse of the right triangle is equal to half of the hypotenuse, it can be seen that the line of the bamboo pole is an arc with a radius equal to 5, and the central angle of the pin is equal to 30 degrees

Proof: Let the ten digits of this three-digit number be b, then the three-digit number can be expressed as: 100a+10b+c

The resulting new three-digit number can be expressed as: 100c+10b+a, so: (100a+10b+c)-(100c+10b+a)=100a+10b+c-100c-10b-a=99a-99c

99(a-c)

So the difference between the original three-digit number and the new three-digit number must be divisible by 99.

Don't know what you're asking?

100a+10b+c-100c-10b-a=99a-99c=99(a-c)

Solution: 1) Replace b(2m,m) to y=8 x, get, m=8 2m, solve m2=4, solve m= 2, because b is in the first quadrant, so m=2,2) solve the system of equations:

y=2x,y=8 x,get,x1=-2,x2=2,so a(-2,-4),e(2,4).

Let the line ab:y=kx+b, and change a(-2,-4)b(4,2) generation, get, -2k+b=-4,4k+b=2, solution, k=1, b=-2

So the straight line ab:y=x-2

3) Let the straight line eb:y=kx+b, replace e(2,4),b(4,2) for people, obtain, 2k+b=4,4k+b=2, and solve k=-1, b=6

So the straight line eb:y=-x+6, and the straight line intersects the x-axis at (6,0).

So bof area = (12)*of*ordinate of point b = (1 2)*6*2=6

4) There are 3 (-4, -2), (0, -6), (8, 10) p

1, b( 1,0) is substituted into a straight line, and there is b=1;So if the linear equation is y=x 1 and the ordinate of point c is 4, then d(3,4) is obtained;

2、o(0，0)、d(3，4)。Let p(t,0), then if od=pd, the solution is p(6,0); If op=pd, then p(25 6,0). 3. Under the condition of 2, only p(6,0) can, and the center of the circle p is (6,0) and the radius is 5, then the radius of the circle o is 1.

Solution: (1) Because point b is symmetrical with point A with respect to the origin, the coordinates of point A are (1,0), so the coordinates of point B are (-1,0).

Again, the straight line y=x+b (b is a constant) passes through the point b

0=-1+b b=1

cm：y=4

Point d is the intersection of cm and the straight line y=x+1.

4=x+1 x=3

D point coordinates (3, 4).

3) The pod is an isosceles triangle, there are two cases, od=op=5 p-point coordinates (5,0) pd=2 radius of 5 circle o =| pd-op |=5-2 5 od=pd=5 p-point coordinates (6 ,0) op=6 radius of circle o =| pd-op |=1

Clever and quick calculations, right?

Let's take a look at it first, 7 -7 2 -7 (2*3) -7 (2*3*4) -7 (2*3*4*5) -7 (2*3*4*5*6) -7 (2*3*4*5*6*7).

x-1 2 x=(1-1 2) x=1 2 x, 1 2 x-1 3 1 2 x=2 3 1 2 x=1 3x, and then 6 ......1/2009。

1 2009 x=1 (x is 2009, easy to write.) ）