Ask a math problem that needs to be solved urgently!!!!

The main problem in solving this problem is that it is used to exchange yuan.

Formula 1: 2a+b+5c=400

Formula 2: 4a-b-c=100

Formula 1 + 2 yields: 6a+4c=500 is represented by a: c=(500-6a) 4=125-3a 2

c should be less than 100 and greater than 0 to get: 50 3 a 250 3 substitute the above c value into formula 2 to get the b value represented by a: b = 11a 2-225 and b should be less than 100 and greater than 0 to get:

450 11 a 750 11 The range of a obtained from above is: 450 11 a 750 11 then n=a+b+c=a+11 2a-225++125-3 2a=5a-100

So the minimum value is a=450 11 and n=1150 11

The upstairs is really ingenious, I have nothing to say, strong

2a＋b＋5c＝400 ..

4a－b－c＝100 ..

Yes c (250 3a) 2 0, a 84 *5 there is b (11a 450) 2 0, a 40a (40, 84), but considering that b is an integer, a can only take 42 at least when a 42, b 6, c 62

a＋b＋c＝5a－100≥5*42－100＝110

There is a conceptual error on the first floor! As a result, the scope of A is wide, imagine how B and C can be zero at the same time? a=200》100。

The fourth floor is more rigorous, not bad!

I didn't see the question on the fifth floor! Who said a, b, c are integers?

2a+b+5c=400 is denoted as m

4a-b-c=100 is denoted k

m+k: 6a+4c=500

m+5k: 22a-4b=900

So c = 125-3a 2

b=11a/2-225

So n=a+11a 2-225+125-3a 25a-100

a, b, c are non-negative numbers that are not greater than 100.

and 0 b + c = 4a - 100 200

So 25 a 75

Therefore, the minimum value of n is 25

1. Xiao Ming walked at a speed of 10 kilometers and hours (hours, Xiao Qiang set off from school and rode a motorcycle according to the original road for 24 minutes (24 60 hours) to catch up with Xiao Ming, so the motorcycle speed = 10 (kilometers and hours.

2. The school is located x kilometers away from the shopping mall. Awareness according to the topic (x 3)-(x 5)=1-12 60=4 5

Solution(x 3)-(x 5)=4 5 gives x = 6 km.

3. A certain unit marched, walked for an hour, traveled 20 * kilometers more than an hour by car, and walked (150-90) * [ kilometers.

Traveled a kilometer by car.

4. Set a plan to dig x square per hour. Also in t hours, (x-3)*t=51 was dug in the morning and (x+6)*t=51+27=78 was dug in the afternoon

The solution (x-3)*t=51, (x+6)*t=78 gives x=20 squares.

5 Let the distance between A and B be s, and the speed V from B to A.

According to the title, there are s 40+s v=(2s) 48

Solution s 40+s v=(2s) 48 yields v=60

6. Set the distance s between A and B, the speed of Li and Zhang (v+2).According to the title, there are 2v+2*(v+2)+36=s, 4v+4(v+2)-36=s

The solution is 2v+2*(v+2)+36=s, and 4v+4(v+2)-36=s gives s=108 km.

7. (1) Divide 36 into the sum of two numbers divisible by 8 and 4: one by 4 + 4 by 8. 3 cars for 4 people + 3 cars for 8 people.

5 cars for 4 people + 2 cars for 8 people. 7 cars for 4 people + 1 car for 8 people. 9 cars for 4 people.

There are 5 combinations in total.

2) Among the above 5 combinations, the cost of one car by 4 people + 4 cars by 8 people = 200 + 4 * 300 = 1400 yuan is the most cost-effective.

The cost of 9 cars by 4 people = 9 * 200 = 1800 yuan is the most expensive.

8. According to the topic, the walking frequency ratio of A and B is 5:6A and B step distance 4:7. Therefore, the walking speed ratio of A and B is (5*4):(6*7)=10:21

Now A runs 55 meters first, and B starts to chase A. After the t time, 55+10t=21tThe solution yields t=5So A runs another 10*5=50 meters, and B can catch up with A.

9. When a seventh-grade student was doing his homework, he accidentally knocked over the ink, so that only the following words were seen in a homework question: "A and B are 40 kilometers apart, when the speed of a motorcycle is 45 kilometers, and when the speed of a freight truck is 35 kilometers, (Q: How long does it take for a motorcycle and a cargo truck to meet when they start from A and B in the opposite direction?) The parentheses indicate some of the text covered in ink) Complete this assignment and list the equations.

Set to meet when you were a child. According to the meaning of the title, there are 45t + 35t = 40 to solve t = hours.

This is probably because the person who lazy did the homework left by the teacher and did it yourself, and then posted it on the Internet to ask for help with your homework.

I think. You've got enough time to type so many words for you to finish.

i) Proof: aa1c1c is a square, aa1 ac

Plane abc plane aa1c1c, plane abc plane aa1c1c=ac, aa1 plane abc

ii) Solution: by ac=4, bc=5, ab=3

ac2+ab2=bc2

ab ac establishes a spatial Cartesian coordinate system with A as the origin, then A1(0,0,4),B(0,3,0),B1(0,3,4),C1(4,0,4).

bc1＝(4，−3，4)，ba1＝(0，−3，4)，bb1＝(0，0，4)

Let the normal vector of plane a1bc1 be n1 (x1,y1,z1) and the normal vector of plane b1bc1 be n2=(x2,y2,z2).

then n1 (0,4,3).

n2＝(3，4，0)

cos＜n1，n2＞=16/25