The final exam is urgent, and the math questions are urgent! Urgent!

1.A car is driving from A to B. If you increase the speed by 20, you can arrive 1 hour earlier than scheduled.

If you drive 120 kilometers at your original speed and then increase your speed by 25, you can arrive 40 minutes earlier than scheduled. How many kilometers is the distance between A and B?

2.There is a highway from City A to City B, which is divided into three sections, of which the first section is twice as long as the third section. On the first stretch of the road, the speed of the car is 40 kilometers per hour; On the second stretch of the road, the speed of the car is 90 kilometers per hour; On the third stretch of the road, the speed of the car is 50 kilometers per hour.

At present, two cars are traveling in opposite directions from City A and City B at the same time, and meet at 1 3 on the second section of the road (1 3 in the direction of City A to City B) after 1 hour and 20 minutes. So how many kilometers are the two cities A and B apart?

3.Vehicles A and B depart from A to B at the same time. After car A traveled the whole 2 3 at the original speed, the speed of the vehicle increased by 1 times, and the result arrived at place B 2 hours earlier than the original planned time; After traveling at the original speed of 30 kilometers per hour, car B increased the speed by 1 times, and as a result, both cars arrived at place B at the same time.

So how many kilometers per hour is A's scheduled travel?

4.There are long-distance buses traveling at a fixed speed between cities A and B. If the speed is 6 kilometers per hour faster than the original speed, then it can be 20 minutes earlier.

If the speed is 5 kilometers per hour slower than the scheduled speed, it will be 24 minutes late. Q: How many kilometers is the distance between cities A and B?

5.A, B and C were racing bicycles, and A arrived at the finish line 24 minutes earlier than B and B 6 minutes earlier than C. He also knows that A's speed is 5 kilometers per hour faster than B's speed, and B's speed is 1 kilometer per hour faster than C's speed.

How many kilometers does the A, B and C race take?

Elementary school? Junior high school? High school ?..Tell me so I can help you find it.

Solution: (1) Let the average percentage of each trouser grinding increase be x, yes.

40（1+x）²=90

Solution: x=1 2=50% or x= -5 2 (unsatisfactory, rounded) Answer: The average percentage of each increase is 50%.

2) During the National Day ** week, the price of each Hu Yidou memory card will be reduced by 2Y yuan, including:

y-3）·（y-6）=0

Solution: y=3 or y=6

When y=3, 2y=6 50-2y=44 40When y=6, 2y=12 50-2y=38 40 (not in line with the topic, discarded) Answer: The price of each memory card was reduced by 6 yuan during the national shortage of Qing ** week.

Just a semi-special corner, wouldn't it be enough to break him down?

^^(1) d=a2-a1=f(d+1)-f(d-1)=d^2-(d-2)^2=2(2d-2)

d=4d-4

d=4 3a1=f(d-1)=f(1 3)=(1 3-1) 2=4 9, so an=a1+(n-1)d=4 9+(n-1)*4 3=4n 3-8 9

q=b2/b2=f(q+1)/f(q-1)=q^2/(q-2)^2q=(q-2)^2

q^2-5q+4=0

q=4 or 1 (rounded).

b1=f(q-1)=f(3)=4

So bn=b1*q (n-1)=4 n

2)a(n+1)=c1/b1+c2/b2+…+cn/bna(n)=c1/b1+c2/b2+…+c(n-1) b(n-1) subtract the two equations, and within a(n+1)-a(n)=d=cn bncn=bn*d=4 n*4 3=4 (n+1) 3 so cn is to.

c1 = 4 2 3 = 16 3 is the first term, qc=cn c(n-1)=4 is the common ratio of the proportional volume series c1 + c2 + c3 + ....cn=c1*(1-qc^n)/(1-qc)=16(4^n-1)/9

Parabola p: y = x

Let AB cross the x-axis at t(t,0)(t≠0).

Then the ab equation is x=t+my and y =x, and the simultaneous elimination of x yields y =t+my, i.e., y-my-t=0

Let a(x1,y1), b(x2,y2).

Then y1+y2=m,y1y2=-t

OA Vertical OB

Vector oa ob=0

i.e. (x1,y2) (x2,y2)=0

x1x2+y1y2=0

x1x2=-y1y2=t

y1y2)²=x1x2

t²=t,t≠0 ∴t=1

That is, AB intersects with the x-axis at the fixed point t(1,0).

The vector oa + ob = the intersection of the vector ococ and ab m m is the midpoint of ab.

Let m(x,y), then 2y=y1+y2=m

2x=x1+x2=1+my1+1+my2=2+m(y1+y2)=2+2my

2x=2+ 2y*2y==>2y+1=x, i.e., the trajectory equation for m is y =(x-1) 2

AOBC is a parallelogram with an area of 2 times the area of δAOB s=2sδAOB=2*1 2|ot| (y1|+|y2|)=|y1-y2|

y1+y2)²-2y1y2

m²+2≥2

When m=0, the AOBC area is 2

Vector oa + ob = vector oc, oc and ab intersection point mm is the midpoint of ab and oc.

Let a(x1,y1),b(x2,y2),m(x,y)x=(x1+x2) 2,y=(y1+y2) 2oa vertical ob,oa*ob=0

x1x2+y1y2=0

y1^2=x1,y2^2=x2

x1x2=(y1y2)^2

y1y2)^2+y1y2=0

y1y2=0 (rounded), y1y2=-1

x1+x2=y1^2+y2^2=(y1+y2)^2-2y1y2=(y1+y2)^2+2

2x=(2y)^2+2

x-1=2y^2

Area of AOBC = |oa||ob|

(x1^2+y1^2)(x2^2+y2^2)=√(x1x2)^2+(y1y2)^2+(x1y2)^2+(x2y1)^2

2+(x1y2)^2+(x2y1)^2

Let a coordinate be (x1,y1),b(x2,y2),m(x,y).

The vector oa+ob=oc, then the vector oc is a diagonal of the parallelogram oacb, and it intersects ab at m, then m is the midpoint of ab, then there is x1+x2=2x, y1+y2=2y

Let the equation for the straight line ab be y=kx+b

Substituting to y 2=x

k^2x^2+(2kb-1)x+b^2=0

x1x2=b^2/k^2

x1+x2=(2kb-1)/k^2

y1y2=(kx1+b)(kx2+b)=k^2x1x2+kb(x1+x2)+b^2=2b^2+b(2kb-1)/k=4b^2-b/k

OA is perpendicular to OB, then there is x1x2+y1y2=0

b^2/k^2+4b^2-b/k=0

Let the coordinates of a be (y1 2, y1) and b (y2 2, y2).

If OA is perpendicular to OB, then there is K(0A)=Y1 Y1 2=1 Y1, and K(OB)=1 Y2

k(oa)*k(ob)=-1=1/(y1y2)

y1y2=-1

ab^2=(y1^2-y2^2)^2+(y1-y2)^2=y1^4+y2^4-2y1^2y2^2+y1^2-2y1y2+y2^2=y1^4+y1^2+y2^4+y2^2

It feels like you're missing something for your topic???

The trajectory of m should be y=0 (x>0) with an area of (xa+xb)ya 2+(xa+xb)yb 2 when xa=xb has the largest area, i.e. xa=1 the area is 2

1,x|x= +2k or x= +2k }

2,x|x=arcsin( +2k or x= -arcsin( +2k }

3,x|x=30° or 330°}

After point A falls, the height of the first bounce of the object is 25 * 2 5 = 10 The height of the second bounce: 10 * 2 5 = 4

The height of the third bounce: 4*2 5=8 5

Or directly bounce up the height for the third time to take the liquid: 25*(2 5) 3=8 5 so away from a: 25-8 5=117 5=

Note: 3 represents a cube.