Math problems in the first year of junior high school. Understood, please. Urgent!!

1. 1. The law is 3n*(-1)n+1

Both n and n+1 are exponents.

n*(-1)n+1-2 3n*(-1)n 33, bring in: -6561-6561+2+2187=-10933 two, 1, 4+ 27+

2. Up to 27+4+ on the 18th

28000-27000-1105=-5 5 yuan.

4. An integer whose absolute value is not greater than 3 means that the number with an absolute value less than or equal to three is -3, -2, -1, 1, 2, 3

The single digit to the power of 2009 is 2.

Because the power of 2 single digit is , at a time.

Every four rounds, 2009 4 remainder is 1, so is the first one, with a single digit of 2

The first column is arranged in order to the nth power of 3; The first column minus two is the second column; The first column multiplied by minus one-third is the third column.

April 19, per share. The highest is the 18th, and the lowest is 26 yuan. Mr. Zhang lost 5 yuan.

An integer with an absolute value of no more than 3 means that numbers with an absolute value less than or equal to three are -3, -2, -1, 1, 2, 3

The single digit of 2 to the power of 2009 is 2.

Solution: Let the gold x grams contain 250-x grams of silver, so 1 19x+1 10(250-x)=16

The solution is x=so 250-x=

Answer: If you encounter this kind of problem in the future, you can set one of them as an unknown number first, and then find the relationship between the two, and then you can list the equation.

Solution: If the gold weighs x grams, then the silver weighs (250-x) grams.

1/19）x+(1/10)(250-x)=16∴10x+4750-19x=3040

9x=1710

x=190250-x=60

A: The alloy contains 190 grams of gold and 60 grams of silver.

Let the gold content in the alloy be x, the silver content is 250-x grams, x 19+(250-x) 10=16, x=190, and the gold content in the alloy is 190, and the silver content is 60 grams

Solution: Let the gold content be x grams.

So the silver content is (250-x) grams.

There is a question to know. 1 19x+1 10*(250-x)=16, simplified to obtain 25-(1 19-1 10)x=16, and x=190

So the gold content is 190 grams, and the silver content is 250-190 = 60 grams.

If gold x grams, then silver (250-x) grams.

x/19+(250-x)/10=16

x = 190 so 190 grams of gold, 60 grams of silver.

Hands up, hope it helps!

(1) If the passenger is transported back and forth by car alone.

It must be sent three times, and 4 people must be sent each time.

40 5 60 = 3 and 1 3 (hours) > 3 hours If the passenger is sent back and forth by car alone, try to illustrate that the probability of getting all 12 passengers to catch the train is 0 (2) If the rest of the people can walk for a while by car, it is possible to catch the train for a while, and the speed of walking is a kilometer per hour, and the car will send each person x kilometers, then each person will walk (40-x) kilometers.

x-(40-x)]/60=[(40-x）/2]/a (1)x/60+(40-x)/a<=3 (2)

The solution gives a>=2,5

That is, you can catch a train by walking at a speed of kilometers per hour or more.

If the walking speed is 5 kilometers per hour, the time required is 1 hour and 35 minutes.

Four people took the first ride to 30 kilometers, which took 30 minutes, and the next two and a half hours the five people walked to the station by themselves.

Then drive back, of course the remaining 8 people will walk 3 km (because it takes 27 minutes to drive back). Then the four of them were hitchhiking, it took 31 minutes to get to 34 kilometers, and the remaining 6 kilometers took the next 1 and a half hours, walking on their own.

Then in the drive back, the last 4 people had already walked about 8 kilometers (because it took 26 minutes to get back) and then used the rest of the time to get them to the station.

(1) Assuming that a passenger is transported by car, it will take 3 hours and 20 minutes to pull the passenger if the passenger does not move at all.

But if these passengers are in a hurry, I think they will definitely walk on their own while the car starts, then the time required should be: 1 hour and 15 minutes + 1 hour 5 minutes and seconds + 30 minutes and seconds, obviously, the sad passengers have caught the train... At the same time, this is also the answer to the second question.

Let it be divided into right angles starting at 3 points through x. (At right angles, the difference between the hour and minute hands is 15 minutes).

x-(15+(5/60)x)┃=15

then x-(15+(5 60)x)=+15 or =x-(15+(5 60)x)=-15

Get x=0 or x=360 11 x=0 rounded x=360 11

Let's start at 3 o'clock and pass through x to split into flat angles. (The difference between the hour and minute hands at a flat angle is 30 minutes).

x-(15+(5/60)x)┃=30

then x-(15+(5 60)x)=+30 or =x-(15+(5 60)x)=-30

Get x=-180 11 or x=540 11 x=-180 11 rounded x=540 11

Let's start at 3 o'clock and coincide with x points. (The overlapping hour and minute hands are 0 minutes apart, that is, equal).

x=15+(5/60)x

Get x = 180 11

5 60 is the number of degrees the hour hand travels.

Isn't the hour hand at 3 o'clock and the minute hand at 12, that is, the hour hand and the minute hand are at 90°?

,m=5,n-5=1,n=6,m^n=156252.The length of the second side is a+b-(5-a)=2a+b-5, and the third side length is 2a+b-5+a+b=3a+2b-5, then the perimeter is a+b+2a+b-5+3a+2b-5=6a+4b-10, and any two sides of the triangle are longer than the third side, but the first side + second edge = third side, this triangle should not exist.