It s a very realistic math problem in life I m afraid it s a bit difficult .

Categorical discussions. 1) If you sell 250w game coins, you can get 20 yuan, pay a handling fee of 5 yuan, and the actual income is 15 yuan.

Then 100w game currency = 6 yuan.

2) Sell 800w game coins, charge 50, procedures 10, and actual income 40.

Then 100w game currency = 5 yuan.

3) Sell 1000w game coins, income = 78-10 = 68, then 100w game coins = yuan.

4) If you sell 1700w game coins and the handling fee is 10 (or 15) yuan, the actual income is 90 (or 85) yuan.

Then 100w game coins = or yuan.

5) Sell 2000w game currency, the actual income is 120 yuan.

Then 100w game currency = 6 yuan.

To sum up, in comparison, if you sell 1000w game coins to get the highest million ratio (per 100w game currency yuan), then you should buy and sell in units of 1000w game coins.

Let's say you have 2000w

2000w = 135 yuan.

Actual get: 135-15 = 120 yuan.

Each game currency is 120 2000W

Let's say you have 1700w

1700w = 100 yuan.

The actual amount of 100-10 = 90 yuan.

Each game currency is 90 17000w

And so on, to figure out which one will have the most benefits.

The conditions provided are non-linear, and if you want to get an accurate answer, I'm afraid you have to solve it with students who specialize in mathematics!

The original system of equations can be sorted as, x+2y=50......①

3y-2x=5……②

Changeable, - got.

20, substituting x into can obtain, y=15

So, x=20

y=15

Multiply the two sides of the second equation by 5 to get 115y-10x 25, and then add the two equations to get 35y 525, y 15, and substitute y 15 into 3y-2x 5 to get 3 5-2x 5, a 2x 10, and get x 5. Thank you.

After A is done as AC BF to C, in RT ABC, ABC=90°-60°=30°, AC=1 2AB=160<200

City A will be affected by the typhoon.

With A as the center of the circle and 200 as the radius, draw the arc to cross BF at D and E, and in RT ACD, CD= (AD -AC)=120

de=2×120=240

240 60 = 4, A: The time affected by the typhoon is 4 hours.

Answer: (1) City A will be affected by the typhoon, and if it crosses the perpendicular line of A as BF and intersects with point O, then AO is the shortest distance between the center of the typhoon and City A. Because the angle ABO=30 degrees, AO=160, City A will be affected by the typhoon.

2) This question needs to pass a to make a circle with a radius of 200, and at bf intersect two points, you can set it to m,n, and then find the length of mn, according to the Pythagorean theorem, fm=120, and then multiply by 2, that is, you can calculate mn=240, Mn is the process of the typhoon affecting city A, so the required time is divided by 240 by 60 to get 4, so the typhoon will affect city A for 4 hours.

It will be 4 hours.

1) The perpendicular line of BF is 160 km <200 km) if the perpendicular point D is 160 km 2) the side of the right triangle with AD as the right angle, if it is not affected by the typhoon, the side corresponding to the right angle is at least 200 km.

According to the Pythagorean definition, the other side on the BF is at least 120 km, note that both sides of point C can be triangular shapes.

The landlord can make his own drawings to make it easier to understand.

(1) ac:bc=4:3 , ac:ab=4：5， bc:ab=3：5

ac=8， bc=6

2) The time for Q to reach the end point is (6+8) 2=7 seconds, and the time for point P to reach the end point is 10 seconds, so point Q reaches the end point first, when Q moves on BC, 0y=pb*h 2=4x(10-x) 5, when Q moves on Ca, 3y=(10-x)*3(7-x) 5=3(10-x)(7-x) 5 4 so y=4x(10-x) 5, 0y=3(10-x)(7-x) 5 3(3) when pq ab, h=pq=3(14-2x) 5

Pa=X, so at the right angle Paq right angle abc

pa/aq=ac/ab x/(14-2x)=4:5

x=56/13

pq=42/13, pb=74/13

pq pb = 42 74 = 21 37≠bc ac, so the two are not similar.

4) When x=5, we can find that p is the midpoint of ac, q is the midpoint of ab, connecting am, obviously, pq is the perpendicular line of ac, so mc=ma, and the perimeter of the triangle bcm = bc + mc + mb = bc + mb + ma = 6 + mb + ma

In mab, there is ma+mb ab, and m, a, and b are equal to each other when they are in a straight line.

Therefore, when the M point coincides with the Q point, the circumference of the triangle BCM is the smallest, L=BC+AB=16cm

If you have any questions, please feel free to ask, if you are satisfied, thank you for adopting.

Ok, help you answer, the first question, I won't talk about this at the beginning, according to the Pythagorean theorem, it's very simple ac=8, bc=6. This question is not difficult, but it is important because it gives you a very important hint that when one point reaches the end, the other point also stops. You look at their speed, that is to say, P moves up to 10s, point Q moves up to 7s, that is, 7 seconds is over.

The second question, two schemes, I use direct hair, first determine that PBQ is a triangle, but the difference is that Q has to pass through the point C, which has an impact on the high of PBQ, so it should be segmented, the first section Q moves on BC, that is, within 3S (the range of independent variables) is based on PB=10-X, and the height is SinB*BQ, Sinb=AC AB=4 5, BQ=2X, bring in the triangle area formula (calculate it yourself) When Q moves to AC, The same thing (omit a lot of words here, hehe).

The third question is that at this time, it is all hypothetical and assumes existence. Then the time at this time must be greater than 3s, less than or equal to 7s. At this time, ap=x, and aq=14-2x, there must be cosa=ap qp=4 5, (aqp is similar to the big triangle) bring in and find x=64 13, so that the time of the movement is determined, so as to find pb and qp to see if their ratio meets the requirements, because there are right angles, if the proportion meets the requirements, then similar, otherwise, not similar.

The last question, to be honest, feels a bit biased and difficult to understand. At 5 seconds, QP is exactly the median line, and the area of the triangle MBC is determined, which is equivalent to finding a case that makes the triangle have the shortest circumference while the area is unchanged. That is, when the case of bm=cm (that is, the case of isosceles, if this needs to be proved, then I have to helplessly say, no way, hehe) set pm=x, use the Pythagorean theorem to solve it, and draw a diagram by yourself to see it.

If you don't understand, just ask.

aqab=qh bc, i.e.: 14-2x10= qh 6, solution: qh = 35(14-2x)cm, y= 12pb qh = 12(10-x) 35(14-2x)= 35x2- 515x+42(3 x 7);

The functional relationship between y and x is: y= -45x2+8x (0 x 3)35x2-515x+42 (3 x 7);

3) ap=xcm, aq=(14-2x)cm, pq ab, apq acb, apac=aqab= pqbc, i.e.: x8=14-2x10= pq6, solution: x= 5613, pq= 4213, pb=10-x= 7413cm, pqpb= 42137413=≠ bcab, when the point q moves on the ca, so that pq ab, the triangle with points b, p, q as the fixed point is not similar to abc;

4) Reason for existence: AQ=14-2X=14-10=4cm, AP=X=5cm, AC=8cm, AB=10cm, PQ is the median line of ABC, PQ BC, PQ AC, PQ is the perpendicular bisector of AC, PC=AP=5cm, AP=CP, AP+BP=AB, AM+BM=AB, when the point M coincides with P, the circumference of BCM is the smallest, and the circumference of BCM is: MB+BC+MC=PB+BC+PC=5+6+5=16cm

The minimum circumference of BCM is 16cm Comments: This question examines the determination and properties of similar triangles, the Pythagorean theorem, and the shortest distance problem This question is very comprehensive and difficult, and the key to solving the problem is the application of the idea of combining equations and numbers You should be able to understand it.

Of course it's hard.

The conclusion is wrong!

Let the side length of the square be 3

Then you can find:

ag = root number five.

af = two root number two.

fc = root number two.

The square of ag = af times fc is obviously not true.

I have a headache when I look at this question... Find a master to do it for you