The more detailed the math puzzle, the better

Each person can bring x kg of luggage for free, and pay 1 yuan for every y kg of excess weight.

90-3x=15y

90-x=35y

x = 20 kg.

When one person takes the luggage of 3 people, the overweight part is the excess part of the three people plus the free part of the other two, and the free part of each person has to pay: (35-3-5-7) 2=10 (yuan), that is to say, if you pay the weight of the free part according to the overweight payment rate, you have to pay 10 yuan, so the baggage weight ratio of the three people is 13:15:

17, the total weight of 90 kilograms, the luggage of the three people weighed respectively:

90 * 13 (13 + 15 + 17) = 26 (kg) 90 * 15 (13 + 15 + 17) = 30 (kg) 90 * 17 (13 + 15 + 17) = 34 (kg) free part is 90 * 10 (13 + 15 + 17) = 20 (kg), and the overweight payment rate is 3 (26-20) = yuan kg).

It has nothing to do with the weight of the baggage.

If each person can bring luggage for free x yuan, there is.

x+3+x+5+x+7=x+35

Therefore, x=10 yuan.

1) Let the number B be x

Then: x(1+1 4)=20

Solution: x==16

2) Let the average vehicle transport cargo yield x tons per hour, then: (3x) 3 4=45 solution: x=20

3) If the school buys x soccer balls, then: 21 5 7=1521+x) 5 6=(15+x).

Solution: x=75

Looks like you're trying to use the junior high school method.

The arrow above ignores, finds.

From the meaning of the title, g and o are both on ac, let the radius of the circle o be r, and the radius of the arc ef is r, then r + r + root number 2 * r = ac = 23 root number 2, and because it is a cone, so the arc ef = the circumference of the circle o, that is.

1 4 (2 R) = 2 R, through the above two formulas can obtain R = 5 root No. 2-2, R = 20 root No. 2-8 bus is R = 20 root No. 2-8, surface area = sector AEF area + circle O area = R 2 4 + R 2 = 10 (27-10 root No. 2).

This question may seem troublesome, but it's actually okay.

Just analyze.

Child 1, subtract 2 pieces of sugar, child 2, get 2 pieces of candy, give 4 pieces of sugar, child 3, get 4 pieces of sugar, give 6 pieces of sugar,。。 It can be seen that every child loses 2 pieces of sugar.

When he can't give the candy according to the regulations, he must only have 1 piece of sugar in his hand. (If there are 2 pieces of sugar, it is possible to continue to give).

Only the children who did not give candy and the children who gave candy last could have a ratio of 13:1.

To list it simply, the initial sugar number is 13,11,9,7,5,3,1 and there are 7 children.

It is also possible to list equations.

There are x children.

2 (x-1)+1)/1=13/1

Solution: There are a total of n children, and the last child before the start has 2k+1 candy. Then the second child has 2k+1+(n-2) 2 at the beginning, i.e

2 (n+k)-3 pieces of sugar. The game can only be played in K round, and after K round, the second child is left with 2 (n+k)-3-2k 2n-3 pieces. The first person starts with 2k+1+(n-1) 2, i.e.

2k+2n-1 block. The first person in each round will increase 2n-2 pieces, and K round will increase K (2n-2) blocks, and the first person after K round will have 2k+2n-1+ k(2n-2), that is: 2n(k+1)-1 pieces, and the ratio of the number of candies according to the adjacent children is 13:

1, then there are 2n(k+1)-1 13(2n-3), simplified:

n（12-k）＝19。And 19 19 1 1 19, so n 19, k 11, n has a maximum of 19 people.

Let's take a picture of this function, and you can see that it's about x=2 symmetry, and you'll see that for a y value, it will have two x values, and especially when y=1, there will be three values. If you judge it, you will know that as the question is set, f(x) +af(x)+b=3 can only have equal roots f(x)=1 or one of them is f(x)=1 and the other is less than 0. So x1 = 1 x2 = 2 x3 = 3, choose d.

f(x) af(x) b=3 has three roots, according to the root of the unary quadratic equation, let f(x)=t, then the case of the root of the equation about t either has one root or two roots (the case without roots does not need to be studied in this problem), that is, f(x)=t (t has been solved) should have three solutions, considering the image of the function f(x), if there are three solutions, this t can only be 1. Thus of the three roots, x2=1, and the other two are symmetrical with respect to the line x=2 (the axis of symmetry of the function f(x) is x=2), and the solution is x1=1 and x3=3. Now it's easy to do.

The equation f(x) +af(x)+b=3 has two f(x) solutions because the quadratic coefficient is not zero.

And because there are three solutions to x, and f(x) itself is a piecewise function, there are three different solutions when f(x)=1, respectively, x=1, 2, and 3;

Therefore, x 2 + 2 x 2 + 3 x 2 = 14 is correct, choose d

You can buy up to 35,480 bottles.

A total of 62,090 bottles of water when I change it!

Who finished drinking--

...What the hell the topic, copied in **