I d like to ask for two math questions??!!

1, C 2, A Suppose there are n female workers, then the penultimate female worker takes the gift is n-1+(a 10), and the last female worker takes the gift n, because everyone has equal gifts, so n-1+(a 10) = n, and, n = 9 * a 10, we get n = 9. In the shooting competition, since everyone is double-counted, add these rings, divide by two, which is the total number of rings, and divide by 4 to get the average number of rings. The answer is 63.

After thinking about the first question, I think that the remaining tenth of the question refers to the rest including what has been taken, not a copy after taking away two copies, and the rest is set up with a total of gifts, the first female worker took 1+A 10, not 1+(a-1) 10, it should be understandable, according to the latter solution, it is unsolved, and the second female worker took 2+(a-1-a 10) 10, the two should be equal, and the solution is a=90, that is, each person took 10 copies, a total of 9 people.

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Divide by 5 and leave 3, then the number of must be 3 or 8, so the answer to exclude d is divisible by 7, and a and b can be excluded, so that the answer can be determined to be c. Hehe, this kind of question is faster with the elimination method, but since the values of the options are relatively small, it will not be time-consuming to have a substitution method.

I play slowly, and I should also remember it myself.

Divisible features:

2: Even number. 3: The sum of the numbers is divisible by 3.

4: The last two digits are divisible by 4.

5: The last digit is 0,5.

6: divisible by 2,3 at the same time.

7: The last 3 digits are divided from the previous ones, the large number is decrecuted, and the difference is divisible by 7.

8: The last 3 digits are divisible by 8.

9: The sum of the numbers is divisible by 9.

11: The odd digits and even digits are summed separately, and the difference is divisible by 11.

This problem is divisible by 7, and oral arithmetic can be used to exclude a, b

Let's look at the second largest dividend, 6, because there are only two items, you can substitute it, first with d, divide by 2, 3 can be divisible by 6, exclude.

Choose C. If you have extra time, you can substitute it for verification. Accord with.

ps: The remainder of 238 divided by 4 is the same as the remainder of 38 divided by 4 and is determined by the divisibility at the top.

This kind of problem.

It's called the congruence problem.

Here is a formula for this topic: the difference is the same as the difference, and the least common multiple of the divisor is the period. If a number is divided by 4 and left with 1, divided by 5 and left with 2, divided by 6 and left with 3, then -3 is taken and expressed as 60n-3

Back to the question, let's say this number is 3

The least common multiple of 6 is 60n-2, which is the minimum 3-digit number when n.

That's it. Hope you understand. In this question, n takes 2 to be 23 one.

Minimum 3 digits 118

The odd term is n

Even terms are n-1

Then [a1+a(2n-1)]n 2=36

a1+a(2n-1)=72/n

a2+a(2n-2)]*n-1)/2=30a2+a(2n-2)=60/(n-1)

The difference is a1+a(2n-1)=a2+a(2n-2), so 72 n=60 (n-1).

60n=72n-72

n=62) bought 298 bottles of beer and drank 298 bottles left, of which 294 bottles could be exchanged for 42 bottles of beer.

After drinking 42 bottles of beer, there are 42 bottles left, which can be exchanged for 6 bottles of beer.

After drinking 6 bottles of beer, plus the extra bottles in front, you can change 1 bottle for a total of 298+42+6+1

3) We can all analyze the internal graphs as 1 2 of the outer graphs. Using recursive reasoning, we can calculate that 802 25 = 200 (square centimeters). But in this problem, we can completely bypass the calculation, by judging that the side length is 80, and 8 is the 3rd power of 2, rough estimate, and quickly judge that the area sought is an integer hundred, so you can directly choose 200

, a series of equal differences has 2n-1 terms, the sum of all odd terms is 36, and the sum of all even terms is 30, then the value of n is ( ) a, 5 b, 6 c, 10 d, 11 2, "Red Star" beer to carry out "7 empty bottles for 1 bottle of beer" preferential ** activity. Now that Mr. Zhang is known to have drunk a total of 347 bottles of "Red Star" beer during the event**, Mr. Zhang was asked how many bottles of beer he bought with the least money? a.

296 bottles b298 bottles c300 bottles d

302 bottles 3, a square with a side length of 80 cm, connect the midpoint of the four sides in turn to get the second square, so that the third, fourth, fifth, and sixth squares can be obtained by continuing, and the area of the sixth square is how many square centimeters? A, 128 square centimeters B, 162 square centimeters C, 200 square centimeters D, 242 square centimeters.

A, B, and C sugars, all of which cost equally, ie.

Weight ratio A:B:C = (6* :

The ratio of costs will be 1:1:1

Total cost = (6* = [all three are the same, 3 by A] Total weight = 6* +=

Cost per kilogram = = yuan.

Hello, first of all, there is a formula for assorted candy questions for you to remember

The average price a=n {(1 a1)+(1 a2)+(1 a3)+(1 an)}

Substituting the formula yields 3 (1

Hope it helps.

This question is undoubtedly the total ** divided by the total mass to get the unit price, with a little trick to use the common multiple. Assuming that a single cost is RMB, then the total price is RMB, the total mass is 6* and the unit price is RMB.

Careless, turned the plus sign into a minus, the upstairs reminded that!

Thank you! 1. Let f(x)=ax +bx=x(ax+b), when f(x)=0, there are two solutions: x=0 and x=-b a.

2. From a>0, we know that y=ax +bx is a parabola that opens upwards and intersects with the x-axis at (0,0)(-b a,0);

a>0,b<0, -b a>0,y=ax +bx pass the first.

One, two, four quadrants.

I knew it before, I forgot now, I'm sorry for the math teacher.

The quadratic function y=ax +bx, when a>0, b<0, 1,2,4 quadrants pass through.

When the axis of symmetry is y-axis, b=0

Chose d 33 and summed 3,6,12,24

According to this law, the next wave of smart pants should be 3*16=48

So the next number is a few Sen Bi Xing.