There are 3 math problems, one by one, and 3 math problems

1), vertex p(5,25 4), axis of symmetry line x=5,y=ax +bx+c passes through the origin o(0,0),e(10,0),c=0,100a+10b=0,25a+5b=25 4,b=5 2,a=-1 4.

The analytic formula of the parabola: y=-1 4x +5 2x

2) ab = 8, a, b symmetry about x = 5, a(2, 0), y = -1 4 * 2 + 5 2 * 2 = -1 + 5 = 4, ad=4.

3) Let ab=m, then a((5-m 2,0), ad=y=-1 4*(5-m 2) +5 2*(5-m 2)=-1 16 m +25 4

l=2ab+2ad)=-1/8m²+2m+25/2.

When m=-2 [2*(-1 4)*]=4, the maximum value of the circumference l of the rectangular ABCD is: I=-2+8+50=56

1)x1*x2=c/a=0

c=0x1+x2=(-b/a)=10

10a=b4ac-b*b)/4a=25/4a=-1/4 b=5/2

ab=6 ao=2

Substitute 2 into the function.

That is, we get AD and set ao=x

l=2*y+（10-2x）*2

The maximum value of l can be obtained.

In the third year of junior high school, I don't take my homework seriously!

(1) b= The function is b, and the argument is a

2) b= maximum number of heartbeats.

3) b= higher than normal heart rate, dangerous.

1) y=160-80 x (2) Value range: 0(1)q= (2)Value range: 0(3)q=10, t=40

One. , a is an independent variable and b is a function.

3。The maximum number of heartbeats is b=<148, so it is dangerous.

Two. y=160-80x, and the value of x can range from 0 x 2.

Three. q=, the value range of the independent variable t is 0 t 60, and when q=10, t=40.

Question 1. 1 b=175-(a-1)* a is an independent variable and b is a function.

3 Not normal.

Question 2. y=160-x*80 0<=x<=2 q=

0<+=t<=60

20 minutes.

1. Sorting out the algebraic formula yields 3x+1 6

1) Positive: 3x+1 6>0;i.e. x>-1 182) less than -2: 3x+1 6<-2; i.e. 3x<-13 6; i.e. x<-18 13

2. According to the title, add the two numbers to show that 3 cats and 3 dogs cost 120 yuan, so 1 cat and 1 dog cost 40 yuan.

70-40 = 30 yuan, that is, 1 dog costs 30 yuan.

40-30 = 10 yuan, that is, 1 cat costs 10 yuan.

3. Set up a large truck to install x each time, and a small truck to load y each time, and the equation can be listed according to the title:

6x+15y=360

8x+10y=440

Both equations are reduced to the following points

2x+5y=120

4x+5y=220

The second equation is subtracted from the first equation.

2x=100, so x=50

Substituting the original equation, y=4 can be solved

So, large trucks carry 50 tons each time, and small trucks carry 4 tons each time.

1. Set the side length of the square vegetable garden to be x meters.

x*x*15+30*(4x)=3600, and the solution is x=12 or -20 (rounded) (m).

2. There are a total of n teams, each team has to play 2 games against another (n-1) team, that is, 2*(n-1), A and B play 2 games, and in turn B and A play 2 games, in fact, there are only 2 games between them.

So there is a total of n*2*(n-1) divided by 2, and the column formula: n*2*(n-1) 2=n*(n-1)=90, which gives n=10, or -9 (rounded).

3. (1) The ball stops, the final velocity is 0m s, and the speed is uniformly decelerated, the formula: x=(v0+vt)t 2 25=20 2*t, and the solution is t=

2) v=at, 20=a*a=8m s (squared), so the velocity of the ball decreases by 8m s per second on average

3) x=v0t+, substituting data to get t=

Just set an equation and you'll know.

If the side length of the square vegetable garden is x, then the square vegetable garden area is x squared.

15x²+4x=3600x=

Hahahaha LZ handshake.

I'm also writing about these three questions, so it's a good fate.

1.The sum of the outer angles of the polygon is 360°, so the sum of the inner angles is 1080° 360° = 720° The sum of the inner angles of the n-side is equal to 180°(n-2).

So 180°(n-2) = 720°

The solution is n=6, i.e. the number of sides is 6

2.Because the point m is on the x-axis, the m ordinate is zero, i.e., 5-t=0

We get t=5 abscissa d: t-3=2, so the m coordinate is (2,0).

3.If both lines are parallel to the other, then the two lines are parallel.

The first question is D, because only the sum of the internal angles is a multiple of 360° can be used for plane mosaicing, and nothing else.

The third question is c and d, because the sum of the two sides is greater than the third side, but c is not eligible, and the sum of the three sides in d is not equal to 20

1d2b c adds up to 0 or 180

3c (the sum of the two sides is greater than the third side) d (the sum is not 20).

1,d2,b(a) This is a multiple-choice question, because the condition does not say whether it is in the same direction or in the opposite direction as the original direction. I draw it on the second floor, but if you go in the opposite direction, you should choose A. 3,c

,8,10,12)

3.There are questions about the question, there is no answer.

1.Let the triangle ABC be the straight line parallel to the bottom edge is EF, intersect with AH and D, and let the length of AD be X, then the perimeter of the trapezoidal BEFC is.

zl＝bc+ef+be+fc

1＋2x×（1／cos60）＋2（1×sin60－x）×（1／sin60）

Trapezoidal area mj =1 2(BC+EF) DH

1/2×[1+2x×(1/cos60)]×1×sin60－x)

s=(zl) square mj=(bc+ef+be+fc) squared [ 1 2(bc+ef) dh ].

As a result, let's ask for yourself!

2.Let a1, a2, a3, a4, a5....ak is the series after the high score is removed.

am,a(m+1)..an is the series after removing the low score.

Then: 1 n(a1+a2+..an)>1/k(a1+a2+..ak)

1/n(a1+a2+..an)<1/(n-m+1)(am+a(m+1)+.an)

3。It's one thing.

The square of n 3) The square of n is 1 (3n 3) The square of n.

1. As shown in the figure, trapezoidal perimeter = 1 + (1-x) + 2x = 2 + x, trapezoidal area = [(1-x) + 1] * (3 2) x 2 = 3 4 (2-x) x = 3 4 (2x-x squared).

The circumference of the trapezoidal is squared = (2+x)2=x squared + 2x+4, and the derivative of s is found at the extreme value of (0,1).