Ask the question of the 6th grade. I want to ask a 6th grade math problem

Because |2x-y-2|Greater than or equal to 0, (3x+5y+10) squared greater than or equal to 0 and |2x-y-2|+(3x+5y+10) squared = 0 so |2x-y-2|=0, (3x+5y+10) squared = 0, so 2x-y-2=0, 3x+5y+10=0 solution gives x=0 y=-2

So the original formula = (0-1) to the power of 2005 + (-2+1) to the power of 2006 (-1) to the power of 2005 + (-1) to the power of 2006.

I thought it was a question asked by the game, but this math problem is so difficult.

So 2x-y-2=0 3x+5y+10=0

Then it is a system of binary equations, and after calculating the solution (x=0y=-2), we can see that x-1) to the power of 2005 + (y+1) to the power of 2006 = 1

2x-y-2|+(3x+5y+10) squared = 0 to know that 2x-y-2 = 0

3x+5y+10=0

To solve x and y, substitute (x-1) to the power of 2005 + (y+1) to the power of 2006.

The main point of this question is |a|+(b) 2=0 then a=0 and b=0

Salinity = mass of salt (mass of salt + mass of water).

If you say that the shaded part is the area of the overlapping part of the intersection of four small circles, it should be:

1.Half of the area of the small circle is: 4 2 2 = 1 (the radius of the small circle). 1*1*square decimeter.

2.With the diameter of the small circle as the base and the center of the large circle as the vertex, a triangle is formed, and its area is 2*1 2=1 square decimeter, 3The area of the shaded area is: (square decimeter.

The speed of the ship is x kilometers per hour.

Then there is: (x+8)*13=520

x = 32 returns as retrograde, and the return time is equal to the distance divided by the speed, ie.

520 (32-6) = 20, i.e. 20 hours to return to the original location.

Answer: 20 hours.

1.First calculate the speed of the ship (520-8*13) 13=32km h2Calculate the retrograde speed of the ship 32-6 = 26km h

Hehe, this classmate, I read it for a long time before I understood the question. This question involves the problem of going with the flow and going against the water.

v shun = v ship + v water, v reverse = v ship - v water; Of course, the question involves the middle of the river and the bank of the river, and there is a bit of a difference between the water. Now let's get started:

1. A boat goes down the river in the middle of the river, travels 520 kilometers in 13 hours, and gets v shun = 520 13 = 40 = v boat + 8 v boat = 32;

2. How many hours does it take to get the boat back to the same place along the coast? Since it is against the water, then v water = 32-6 = 26km h;

3. Time t=520 26=20h. That is, the boat is back along the coast, and it takes 20 hours.

Look, hehe!!

Answer: 20 hours.

Step 1: The boat travels along the water, 520 km, divided by 13 hours equals 40 km, which is the speed of the ship plus the speed of the water in between. So 40-8 = 32 kilometers is the speed of the ship in still water.

Step 2: When the boat is driving against the current, the actual speed = 32-6 = 26 km, the distance traveled is the same as the original. So time is equal to 520 divided by 26, which is equal to 20 hours.

1. First calculate the number of pages with a multiple of 6: 200 6=, so there are only 33 pages with a multiple of 6, and the probability is 33 200

2.If the sum is even, then the two cards taken cannot be one odd and one even, any two cards have 9x8 = 72 combinations, two odd numbers 5x4 = 20 kinds, and two even numbers 4x3 = 12 kinds.

So the probability of an even number is: (20+12) 72=4 93 It is best to convert dividing by five-twelveth to multiply by twelve-fifths, and then use the multiplicative distributive law to multiply one-fifth of twelve into parentheses, and then add and subtract, which is relatively simple, and finally calculates as four-fifths.

1. The page number should be a multiple of 6, 200 6 33 ......3, so there are 33 numbers, so the probability is: 33 200

2. There are 5 odd numbers and 4 even numbers.

If the first card is an odd number, then the second card has 8 cases, of which there are 4 odd numbers, so there are 40 situations in total, of which 20 are eligible;

If the first card is found to be an even number, then there are still 8 cases in the second card, of which 3 are even, so there are 32 cases in total, of which 12 are eligible.

So there are a total of 72 cases, and 32 are eligible, so the probability that the sum is even is: 32 72 = 4 9

3. [5/6 (2/6 1/6)] 5/12 [5/6 3/6] 5/12 [2/6] 10/5 4/5.

÷﹙2×3﹚

Answer: The probability is 1 out of 33.

2. Odd numbers plus odd numbers or even numbers plus even numbers can get even numbers, so 1 can only be encountered by , and 3 can only be encountered by . Then the odd number plus the odd number has 4 3 2 1 10Similarly, the probability of an even number plus an even number is 3 2 1 6, 10 6 16.

Thus the probability that the sum of numbers is even is 1 16.

and 3 to the least common multiple, is 6,200 6 = 33. The probability is 33 200 100% = 66%.

2. Only one even number, a singular number may not be an even number. The probability that the first is an odd number is 5 9, the probability of an even number is 4 9, and the probability of simultaneous occurrence is 20 81, so the probability that the sum sought is an even number is 61 91

3. [5/6 (1/3 6)] 5/12 (5 6-(1 2)) 5 12

Eldest brother! How easy is it!! Do it yourself.

That's a mistake:

If you look at the AOB COD, then its area ratio is equal to the square of the similarity ratio, the similarity ratio (high): 4 2=2, the area ratio: 14x4 (6x2)=14 3,2 ≠14 3.

The problem is that the length of the upper and lower bottom of the trapezoid is not the same as the height of the above 2 triangles, there is a conflict, if you have to do this problem, then, you can't use the height of the above 2 triangles, or you can't use the bottom of the trapezoid, anyway, you can't do it, you can lie to yourself and think that the first method is right. You can also lie to yourself about the second one.

If you are not convinced, please feel free to ask questions. Wha...

Calculate the area of the two blank triangles first, and then decelerate the area of the trapezoid.

o point is not in the position of 2, so it can only be done according to the first algorithm, which is absolutely correct.

Will there be an unknown number in the primary school?

x+y=360

2x+4y=890

The solution is x=275

y=85, so there are 275 robbers.

There are 85 dogs.

Hahaha, count it yourself.