Math problems, come, hurry!!!! 30

A and B are 180 kilometers apart, it takes 3 hours for a car to drive from A to B, and 5 hours for a truck to travel from B to A.

Solution: 180 (180 3+180 5) 180 (60+36) 15 8 hours.

If you remove "A and B are 180 kilometers apart", then.

1 (1 3+1 5) 1 (8 15) 15 8 hours.

180 (180 5+180 3) = hours.

When you don't know 180 kilometers, first set the distance between the two places x kilometers, and the equation is listed.

x (x 5+x 3) = 1 (1 5+1 3) = that is, assuming that the two places are x kilometers apart, the equation can be obtained by approximating x.

The hour is 1 hour, 52 minutes, and 30 seconds.

180 3 is the speed of the car, 180 5 is the speed of the truck, and 180 (60+36) is the result of the first question, which is about 8/15

The second question, if A and B are separated by x kilometers, then the speed of the car is x 3 kilometers per hour, and the truck is x 5, then their total speed is x 3 + x 5, and the time they meet is x (x 3 + x 5), and then the denominator is divided to get x (3x 15 + 5x 15).Get x (8x 15) about 15 8 hours I hope you can understand.

This guy, how to read the book, I've been graduating for ten years, I'll answer it! Let's say they meet in x hours:

The speed of A is: 1800 km/h in 3 hours.

B's speed is 1800 km/h for 5 hours.

The equation is as follows:

1800 km (3 hours) x hour + (1800 km 5 hours) x hour = 1800 km.

600 x km + 360 x km = 1800 km.

960x km = 1800 km.

x = (hours) = 1 hour minute.

The answer is: The time it takes for the two cars to meet is 1 hour and 1 minute.

Car speed 180 3 = 60

Car B speed 180 5 = 36

Encounter time: 180 (36+60) = hours.

If there is no distance of 180 km.

then set the total distance as the unit "1".

A car speed 1 3

B car speed 1 5

Encounter time 1 (1 3 + 1 5) = hours.

1. Because a and b are reciprocal to each other, ab=1

Since the absolute value of m is 3, the square of m is 9

Because c and d are opposite to each other, 3 parts of 5 m, c + 3d is 0, and the whole equation becomes 1-9-0=-8

2. (7-2) + (5-2b) = a? Is it a typo?

x+y4, because a is less than 0, and a of b is less than 0, a can only be a positive number.

So b-a+1 is a positive number, and the absolute value of the segment is equal to itself.

a-b-5 is a negative number, and the absolute value is equal to its opposite number of grips, which is 5+b-a, so the absolute value of b-a+1 plus the absolute value of a-b-5 is b-a+1+5+b-a=-2a+2b+6

5, -2x to the fifth power + 9x + 2

1.ab = 1 m = 9 -3 d3 d/5 c + 3d = 0 original formula = -82A is whispering**? Void?

3. 100x+y

4.b 0 original qi zhi eliminate = b-a+1 + a-b-5 = -45I'm sorry, I don't know.

As an auxiliary line AE, it can be known from DE bisecting AB perpendicularly, AE=BE=5cm, because AEC= B+ EAB

b=∠eab

So AEC=30°

So ac=

Link AE

From the knowledge of the perpendicular bisector, ae=5

Because the angle AEC = 2 angle b = 30 degrees.

The side of a right triangle with a 30-degree angle is equal to the half-known AC = half of AE = from the hypotenuse

Equal to five out of two.

Drawing, linking AE, there is a perpendicular bisector, we can get AE=BE=5, the angle AEC is equal to 30 degrees, so AC is equal to five-twoths.

Master Wang pays tax every month (2000-1600) * 5% = 20

Master Wang can actually get 2000-20=1980 per month

1. (2000 1600) 5 20 (yuan) Answer: Master Wang should pay 20 yuan of tax every month.

20 1980 (yuan) Answer: Master Wang can actually get 1980 yuan per month.

Tax Payable: 20 Actual Available: 1980

(2000-1600)*5%=20 The tax payable is 20 yuan, and the actual tax is 2000-20=1980 yuan.

Solution: Let the total number be x

Class 61 + Class 62 = 1 2x

9+1/5x=2/1x

3/10+1/5=2/1x

The 61 class accounted for 3 10 of the total

Total = 9 (3 10) = 30

A: The total number of June 1st entertainment programs organized by the sixth grade is 30 programs.

Class 61 9 programs.

Class 63 is two-thirds more than Class 61 9 + 9 * 2 3 = 15 programs Class 62 is one-fifth of the total, then 61 + 63 is four-fifths 15 + 9 = 4 5 * total So the total is 30

Class 62 is 30-15-9 = 6 programs.

Please give me the portion, thank you.

9 in 1 class.

Our six (3) classes have two-thirds more programs than six (1) classes--- 3 classes 15 Six (1) and six (2) classes perform exactly half of the total number of programs, and our six (2) classes perform one-fifth of the total number of programs --- 1 class 3 10

There are 6 classes of 302 in total.

Let the total number of programs be x, then.

9+1/5·x=1/2·x

Solution x=30

In fact, the message "Xiaohua: We have two-thirds more programs in Class Six (3) than Class Six (1)" is not necessary.

Root number (8a-b) + root number (b-4) = 0

Because the number in the root number cannot be less than 0

So the root number 8a-b = 0

8a-b=0

Root number b-4 = 0

b-4=0b=4

8a-b=0

8a-4=0

8a=4a=1/2

2a*root(ab)*(root(ba) root(1b))2*(1 2)*root(root(8) (1:2))2*(1 2)*root(root(8) (12)*(12)*(12)*(12)*(root(root(8))2*1*(root(1))).

..Root number (8a-b) + root number (b-4) = 0

Description 8a-b=0, b-4=0

So a=, b=4

And then the one you ask for can be brought in.。。

Root number (8a-b) + root number (b-4) = 0

Description 8a-b=0, b-4=0

So a=, b=4

Wheel circumference = 2 r = d = meters, number of laps = distance Perimeter = 1000 laps.

Wheel circumference = 2 r = d =

RPM r=distance circumference=1000 turns.