Two math problems, the best with a prize!

1. At the same time, brother and sister walked 180 * 2 = 360 meters.

The elder brother walks 90 meters per minute, the younger sister walks 60 meters per minute, and the brother walks 90-60=30 meters more than the younger sister per minute.

The time for brother to walk is t=360 30=12 points.

Their home is s=90*12-180=900 meters from the school.

2. According to the title, the truck has traveled 108 * 2 = 216 kilometers more than the bus.

Passenger cars travel 54 kilometers per hour, trucks travel 84 kilometers per hour Travel time t=216 (84-54)=hours.

So the total trip of the truck s=84* = km.

In this hour, the train traveled a distance of one and a half kilometers and 108 kilometers more, and the distance between A and B is s=(km.

1) Their velocity ratio is 90:60=3:2, so when they meet, the ratio of distance they have traveled is also 3:2.

So the distance sought = 180 2 (3-2) (2+1 2) = 900 meters.

2) The first time they met, they walked a total of one course, and the second time they met, they traveled a total of three times, and their velocity ratio was 54:84=9:14, so in the same time, the distance ratio they traveled was also 9:

14 , so the whole journey = 108 2 (14-9) (14+9) 3 = 331,2 km.

1) Set the distance between the home and the school is x meters.

x-180）/60=（x+180）/90

90x-180×90=60x+180×6030x=27000

x=900A: Their home is 800 meters away from the school.

2) The distance between A and B is x kilometers.

x+x/2-108）/54=（x+x/2+108)/84x=

Answer: The distance between A and B is kilometers.

1. Solution: Let their home be x meters away from the school, and list the equation according to the question

x+180）/90=(x-180)/60

Solve the equation to obtain: x=900

A: Their home is 900 meters from the school.

2. Solution: Let the distance between A and B be x kilometers, and the equation is listed according to the question

x+x 2-108) 54=(x+x 2+108) 84 solves the equation to obtain: x=

Answer: The distance between A and B is kilometers.

1. When we met, my brother walked more than my sister: 180*2 360 (m) My brother walked more than my sister per minute: 90 60 30 (m) The brother and sister walked together:

360 30 12 (min) They traveled twice as much as from home to school, a total of (90 60) * 12 1800 (m).

Distance to school: 1800 2 900 (m).

2. When we met, the truck traveled more than the bus: 108*2 216 (km) (the truck exceeded the midpoint by 108 km, and the bus was 108 km away from the midpoint).

The two cars traveled: 216 (84 54) hours) 54 84) * km) (equivalent to 3 times the distance between two places) The distance between A and B: km).

1) 180*2 (90-60)=120 (points).

The distance from home to school is 120*60+180=900 meters.

1-3/8 - 2/5 = 1/5 Answer: . . For the first time, pure milk made up two-thirds of the total, and water accounted for one-thirdThe second pure milk accounts for 2/3 of the total 1/2 = 1/3 of water 1/3 1/2 + 1-6 -1/3 -1 = 2/3 of the total 3rd pure milk 1/6 = 1/18 2/3 of water 1/6 + 1-1-18/1 - 1-9/18 = 17/18

Eating 1/3 of water + 2/3 + 17/18 = 35/18 Drinking only one glass of milk 35/18"1 Answer: . .

1-3 8-2 5=7 20 400 kg of fruit is a superfluous condition, of course, multiplication of fractions can also be used.

1-(1 3+1 2+1 6)=0 Drink as much.

1. Solution: Let A and B be separated by x kilometers.

Then: x-40%x-20%x=96

Solve the equation to obtain: x=240

A: A and B are 240 kilometers apart.

2. Solution: Medical compensation: 400 + (8200-400) * 45% = 400 + 3510 = 3910

The out-of-pocket payment is 8200-3910=4290

Answer: According to the regulations, Uncle Wang has to pay 4,290 yuan by himself.

1 solution: Let the total distance be x, then we can get:

80%x+60%x=x+96 solution: x=240 (km).

2 solution: 400 + (8200-400) (1-45%) = 400 + 4290 = 4690 (yuan).

(1) If the curve is an ellipse, then there is k+1≠0

k/(k+1)>0 （1）

4-k)/(k+1)>0 （2）

The solution to this inequality group is: for (1) there is k>0 and k+1>0 or k<0 and k+1<0

There is k>0 or k<-1

For (2) there is 4-k>0 and k+1>0 or 4-k<0 and k+1<0

There are -10(k+1) (k-4)>0

then there is k>4 or k<-1

If the focus is on the y-axis, then there is.

k+1)/(4-k)>0

k+1)/k>0

Then k has no solution at this time.

Therefore the focus must be on the x-axis, and the angle of inclination of an asymptote line is 60 degrees, i.e., the slope is: tan60 = 3

i.e. b a = 3

where a= (k+1) k), b= (k+1) (k-4)).

Substituting the solution yields k=-2 (satisfying k>4 or k<-1).

Substituting k = -2 gives us the hyperbolic equation is: 2x 2-6y 2=1 (3).

3) Assuming that there are two points p,q, then the straight line l is the perpendicular bisector of pq, then the slope of the straight line pq and the slope of the straight line l are negative reciprocal to each other, that is, the slope of the straight line pq is -1, then let the straight line pq be y=-x+b (4).

Then the straight line pq and hyperbola have two intersections.

4x^2-12bx+6b^2+1=0

then (12b) 2-4*4*(6b 2+1)>0, i.e., 3b 2-1>0

b> 3 3 or b<-3 3 (5).

then x1=(12b+ (3b 2-1)) 8,x2=(12b- (3b 2-1)) 8

Then we know from equation (4): y1=(-4b- (3b 2-1)) 8, y2=(-4b+ (3b 2-1)) 8

Then the midpoint h(x3,y3) of p(x1,y1),q(x2,y2) must be on y=x-1, x3=(x1+x2) 2=3b 2, y3=(y1+y2) 2=-b 2

x3, y3) substituting y=x-1.

b 2 = 3 b 2-1, then there is b = 1 2 does not satisfy the equation (5), so the contradiction, such p, q points do not exist.

refers to the secondary root number.

1.Transporting a batch of goods with two kinds of large and small cars, the load capacity of 2 large cars is exactly equal to the load capacity of 3 small cars, and 6 large cars and 10 small cars transport a total of 76 tons at a time. The carrying capacity of the cart is (6) tons.

76 (6*3 2+10) = 4 tons -- trolley.

4*3 4=6 tons -- cart.

2.Easy to calculate.

Large cars = 9 small cars.

Therefore, the weight of the small car is 4 tons, and the large car is 6 tons.

2。[(1+9)+(2+8)+.4+6)+5]x10+[（

The two straight lines are parallel, and the slope of the two straight lines is equal.

2m=5m=5/2

5m=10m=2

Proof: If the number written on the front of the 6 cards is a1, a2, a3, a4, a5, a6, and the number written on the reverse side corresponds to b1, b2, b3, b4, b5, b6, then the absolute values of the number written on the front of the 6 cards containing hail and the number written on the back are |a1-b1|，|a2-b2|，|a3-b3|，|a4-b4|，|a5-b5|，|a6-b6|If these 6 numbers are not equal to each other, then they can only take the 6 values of 0, 1, 2, 3, 4, and 5

On the other hand, |ai-bi|Same parity as ai-bi (i=1,2,3,4,5,6) So|a1-b1|+|a2-b2|+|a3-b3|+|a4-b4|+|a5-b5|+|a6-b6|It is the same as the parity of (a1-b1) + (a2-b2) + (a3-b3) + (a4-b4) + (a5-b5)+ (a6-b6) = (a1 + a2 + a3 + a4 + a5 + a6) one (b1 + b2 + b3 + b4 + b5 + b6) = (1 + 2 + 3 + 4 + 5 + 6) one (1 + 2 + 3 + 4 + 5 + 6) = o, while 0 is an even number and 15 is the odd number of the pies, and the two are contradictory

So, |a1-b1|，|a2-b2|，|a3-b3|，|a4-b4|，|a5-b5|，|a6-b6|At least two of these 6 numbers are identical