I have a math problem that I won t ask for advice, thank you

If the number of people in class A is x, the number of people in class B is 93-x 10*2 3*x+5*1 3*x+10*1 3*(93-x)+5*2 3*(93-x)=695 x=45 trees donated by class A: 1 3*45*1+2 3*45*2=75 trees The number of people in class B is 93-45=48 trees donated by class B: 1 3*48*2+2 3*48*1=64 trees.

The second algorithm: the number of trees in class A is x the number of trees in class B is 93-x the total number of trees donated in the two classes is: 695 5 = 139.

Then 2 3*x*2+1 3*x*1+2 3*(93-x)*1+1 3*(93-x)*2=139 x=45 people Class A donated trees are: 2 3*45*2+1 3*45*1=75 Class B = 139-75=64 trees.

Person A x person, Person B Y.

x+y=93 (1)

2/3x*10+1/3x*5+1/3y*10+2/3y*5=695>>>25/3x+20/3y=695 (2)

1) Substitution (2).

This gives x=45, y=48

A tree = 45 * 2 3 * 2 + 45 * 1 3 * 1 = 75 B tree = 48 * 1 3 * 2 + 48 * 2 3 * 1 = 64

Answer: The area of the plastic film is actually a rectangular area.

The length of this rectangle is 15 meters, and the width is "the circumference of a semicircle with a radius of 2 meters".

Circumference of a circle = d

Obviously, if the radius is 2, then the diameter d should be 4, so the circumference of the circle is 4, and the circumference of the semicircle is half of the circumference of the circle, so the circumference of the semicircle is 2 Therefore, the area of the rectangle to be calculated at the end = 15 2

The final answer can be directly written 30 square meters, or you can count it as square meters.

Calculate the area of the cross-section first, since it is 2 semicircles, the area is equal to one circle, s= *r 2=4.

Counting the area of the rectangle, the length is already known to be 15, and the width is half the circumference of the circle. So width = 2

So the area of the rectangle is s'=15*2 =30 So plastic film is needed.

I don't know if I need to add the area of two circles to this, so if I don't use it, it is:

If you want to, it is:

Calculate the area of the two circles first, (2 2

Then calculate the area of the side, 2 2

Then add the two together

A works alone for 18 days and cooperates for 12 days, then B's solo working time is.

To minimize the cost, the time required to complete the work is exactly 31 days, and B works as long as possible, then it can be assumed that B's working time is x, and the rest of the time is completed by A, then you can get.

x*1/36+1/18*(31-x)=1

Solving the equation can obtain x=26, that is, A works for 5 days, B works for 26 days, and the total cost is 2000*5+1400*26=46400 yuan.

Solution: Let the functional relationship between the daily sales volume y and the pricing x be y=kx+b, and substitute x=50, y=100, x=80, y=40 into y=kx+b to get {100=50k+b

40=80k+b

The solution yields k = -2 and b = 200

The functional relationship between daily sales volume y and pricing x is y=-2x+200

Let the function relation be y=kx+b

then 100=50k+b

40=80k+b

The solution yields k = -2 and b = 200

So the function relation is y=-2x+200

The profit is y(x-30)=(200-2x)(x-30)=-2(x-65) 2+2450

Therefore there is a maximum profit when x=65. (The maximum profit is 2450 yuan).

There is a functional relationship between the daily sales volume of goods y pieces and the price of x yuan, so let y=a*x+b

100=a*50+b

40=a*80+b

The solution is a=-2

b=200, so y=-2*x+200

Profit xy=-2*x 2+200*x=-2*(x 2-100*x+250-250)=-2*(x-50) 2+500

So when x=50 is the biggest profit.

1) The functions y pass through the points (0,100) and (100,0) with the intercept formula:

y/100+x/100=1

> y=-x+100

2) Let the profit be z

z=(x-30)y

x-30)(100-x)

x^2+130x-3000

(x-65)^2+65*65-3000=-(x-65)^2+1225

So when x=65, profit z has a maximum value of 1225

Let y=ax+b

Bring 100, 50 and 40, 80 in.

100=50a+b

40=80a+b

Get a=-2 b=200

So y=-2x+200

Let the analytical formula of the primary function be y=kx+b

From this we can list the system of equations: 50k+b=100

80k+b=40

The solution yields k = -2 and b = 200

The analytic formula of the primary function is y=-2x+200

Let y=kx+b

100=50k+b

40=80k+b Just solve this system of equations.

Solution: Let y=kx+b

From the meaning of the title: (simultaneous equations).

50k+b=100 ①

80k+b=40 ②

Solution: k=-2 b=200

The functional relationship between sales volume y and pricing x is:

y=-2x+200

There is a functional relationship between the daily sales volume y pieces and the pricing x yuan, y=kx+b, bring in (50,100), (80,40), and solve it.

2.When the price is 65 yuan, the daily sales profit of the product is the largest.

First of all, you can solve the primary function y=-2x+200Do you have a second question?

AE is known to bidivide DAB and BC

FAE Raid BAE (ASA).

fe=bece:be=1:2

fe:ce=1:3

CD ab again

fdc∽△fab

s△fdc：s△fab＝1：9

S fdc:s quadrilateral abcd 1:8

Fae didn't laugh at bae

s△fdc:s△abe=2:9

S ABE:S quadrilateral plexus with ABCD 9:16 Turn left|Turn right.

Let the total project be 1, the amount of work done by A every day is A, and the amount of work done by B every day is B (A + B) * 12 = 1

A*8+B*18=1

If A = 1 20 and B = 1 30, then it takes 20 days for A to do it alone and 30 days for B.

A: A completes it alone, you need to pay 2000 * 20 = 40000 yuan B: B completes it alone, you need to pay 1400 * 30 = 42000 yuan C:

After the cooperation is completed, it needs to pay 2000 * 12 + 1400 * 12 = 40800 yuan, which is less expensive than plan A.

Let A complete the project x every day, and B completes the project y every day has 12(x+y)=8x+18y

4x=6yx=12(x+y)=1

x=1 20 y=1 30 The number of days that engineering team A works alone is 20 days B 30 days engineering team A works alone 2000 * 20 = 40000 engineering team B works alone 1400 * 30 = 42000 cooperation completed: 3400 * 12 = 40800

Collaboration is done with less money.

5 3 8 (edged).

2) What is more than 200, the maximum is a flat angle, 180, what does 200 mean?

is a regular dodecagon, with an inner angle and 12 150 1800 4,

1. (1) The sum of the inner angles of the quadrilateral is 360°, which can be pushed, and it can be composed of 3 quadrilaterals, so there are 12 edges, except for the four sides, which coincide in pairs! The middle line segment is equivalent to the auxiliary line! You can go: the result is 8 sides.

2) You made a mistake: it should be 20°. Let the outer angle be n, then 3n+20°+n=180°, and the outer angle n=40°, then the inner angle is 140°, so the center angle of the opposite side is 40°. The number of edges a=360° 40°=9

2. The internal error angle is equal: c= b=51° so aob=180°- c- b=87°

3. The center angle of the side is equal to 30° and the inner angle is equal to 150°, the number of sides n=360° 30°=12 Launch: The inner angle has 12 inner angles, so the sum of the inner angles is equal to 1800°

4. The conditions seem to be a little problematic, you see!

Proof: If you connect an OE, then OE AB

Make of cd at point F

ab=cdof=oe

i.e. the distance from the point o to cd is equal to the radius of the small circle.

The cd is tangent to the small circle.

Connecting ao, bo, co, and do, that is, the triangle abog and the triangle cdo congruence, and the perpendicular lines of ab and cd are made by o, which are oe and of, that is, oe is equal to of, so cd is tangent to the small circle.