Math problems! It s funny, but it s not 20

Suppose A wins, and A's last count is 29, then the number of B before that is 28 or 27,28;If the last number of A is 28,29, then the number of B before that is 27 or, 26,27

Then we know that if A must win, he must count 26 on the penultimate count, so that B can win by counting 28, 29 or 29 on the second count.

In the same way, it can be seen that the number that must be counted in the process of victory is (backwards):

And make sure that B can't count those numbers.

Suppose A counts 2,3, then if B also understands this rule, B will definitely count 4,5, and thus occupy the number 5, so that whether A counts one number 6, or two numbers 6,7, B can count 7,8 or 8, then occupies the number 8, and so on, B must eventually occupy 29 and win, so B must not be allowed to count these numbers.

To sum up, in the case that A and B are equally smart, the first to count wins, and the first number must be 1, 2.

Of course, it is not excluded that the brain is in water, if the person who counts first only counts 1 and counts, then the smart person who counts later counts 2, and the smart one wins.

Count 2 when you reach 29 and 1 when you reach 28.

The first one loses. The number of people counted first is counted by one, and the number of people who count later is counted by two, and vice versa is counted by one. The latter number wins, as long as the sum of the number of the number and the number of the previous number is three.

When the opponent counts two, you also count two, when the opponent counts one, you count three, and when the opponent counts three, you count one.

The first count is 1,2 and then according to the principle that when the opponent counts two, you count one, and when the opponent counts one, you count two, and you will win.

You can cut up to 30 pieces, with the first 5 knives cutting from top to bottom, each knife intersecting the other 4 knives, for a total of 15 pieces.

The 6th knife cuts horizontally from the side and cuts the 15 pieces of cake into 30 pieces.

A: A piece of cake is cut six times, divided into a maximum of 22 pieces.

Solution: (figure at the end).

This is really a problem of "several lines intersecting in a plane".

In order to have the largest number of cuts under the condition of a certain number of cuts, the following two conditions must be met:

1. Ensure that each straight line has an intersection point with any other straight line. 2. Ensure that any intersection does not coincide with any other intersection (that is, only two straight lines can be found along any intersection).

Summary of the law: As can be seen from the figure below, when cutting for the first time, it can be divided into up to 2 parts; The second time can be divided into up to 4 pieces; The third time can be divided into up to 7 pieces; The fourth time can be divided into up to 11 pieces; The fifth can be divided into up to 16 pieces; The sixth can be divided into up to 22 pieces ......The nth time can be divided into up to 1 2 (n 2 + n + 2) graph:

Only 12 pieces can be divided into 6 cuts in this way.

You typed the wrong question! Peaches are more or less than half the size of bananas?

Apples: accounting for three-eighths of the total, that is, 400 * three-eighths = 150 kg I assume you have one more half here, that is, bananas have 1 part, peaches have 1 part + 400-150 = 250 (kg) 250 kg per portion) peaches are 100 * kg).

Banana is 100*1=100 (kg).

Calculation: 150 + 100 + 150 = 400 (kg) 150 100 = 1/2

So bananas are 100 kg, apples are 150 kg, and peaches are 150 kg.

What is the method in 6th grade?

Is a peach a half bigger than a banana?

Forehead... Sixth grade, will you be three dollars at a time?

Questions on summer homework?

According to the conditions you give, the answer to this question should be expressed in an algebraic formula, and there is no need to calculate the value specifically.

The answer is 1500 (2x+35).

The actual number of days spent on building this road is equal to the total length of the road of 1,500 meters divided by the actual daily workload, which was originally planned to be repaired x meters per day, and during the actual construction, it was 35 meters more than twice the original plan, that is, (2x 35) meters. Divide by 1500 (2x 35) and you're good to go.

1500 (2x 35) days.

Received your ** dear you wait a moment I will write the answer on the paper, please be patient, to ensure that all right this is the answer you need Oh dear, you don't understand you can continue to ask me oh pro [than the heart] [than the heart] [than the heart] The key to solving these questions is simplification, and you can calculate it after simplification.

What about math problems that I can't do?

Solution: The original plan was set to use y days longer than the actual plan.

1500/（2x+35）=1500/x+yx=

Solution: Let ad=dc=x, then ab=ac=2x, and let the bottom edge bc=y in the first case:

x+2x=15 ，x+y=6。The solution is x=5 and y=1. The side length of the triangle is 10,10,1

The second scenario:

x+2x=6， x+y=15。The solution yields x=2 and y=13. The side length of the triangle is 4, 4, 13, and the trilateral relationship of the triangle is not satisfied, so the resulting triangle does not exist.

So the three sides of the triangle are 10, 10, 1.

I'm sure you have a problem with the topic, either you wrote it incorrectly or the topic itself is wrong! Take a closer look for yourself.

The title is wrong, right, how can BC be the midline of AC?

B of a + A of B

a^2 + b^2)/ab

(a + b) 2 - 2ab] ab is substituted for a + b = 2, ab = 1, formula = (2 2 - 2*1) 1 = 2