Quick back math problems are best written with ideas

There are a total of 25 + 28 = 53 people who do 1 and 2 questions correctly, and there are only 40 people in total, so 53-40 = 13 people are repeated, that is, they are all the people who do it right, then the remaining 40-13 = 27 people are impossible to answer all of them correctly, assuming that the people who do 1 and 2 questions correctly do the third question correctly (in this case, the least number of people who do the three questions correctly, to meet the "least" requirement), then there are 31-27 = 4 people who do the three questions correctly.

4 people, let's say 15 people get the first question wrong, another 12 people get the second question wrong, and 9 people get the third question wrong, which adds up to a total of 36 people who get the question wrong, that is, at least 4 people get the three questions right.

Here's a simple way to do it.

Treat 40 students as 40 cabbages.

25 seeds were poured on the first water (in order from left to right), 28 seeds were poured on the second time from the 26th place, and 31 seeds were poured on the third time.

After watering, you will notice that at least 4 cabbages have been watered 3 times.

Don't scold me

The number of people who did not get the first question right is: 40-25=15

The number of people who did not get the second question right is: 40-28=12

The number of people who did not get the third question right is: 40-31 = 9

Then if you get at least 3 questions right, it is 40-15-12-9=4

Then at least (4) people have done three questions correctly.

12 is correct at least 28-(40-25) = 13

123 all pairs at least 31-(40-13) = 4 people.

1. These 4 people account for 1 2-1 3 of the original classmates, and the equation 4 (1 2-1 3) = 24 (people).

2. If x ducks are raised, x+50 chickens, x+50+70 geese, and the equation is 2x=x+50+70

3. (24+6) 3+6=16 (person).

The high cross ac of AC over point b is E, ab=10, , ae=6 (isosceles triangle, high bisecting base) so be=8 (Pythagorean theorem).

Now we use the principle of equality of the area of triangles, 12 8 = 10ad

So ad=

The cosine value of angle b = ab square + bc square - ac square 2 * ab * ac = 7 25 the sine value of angle b = under the root number (1-7 25 squared) = 24 25 so the trigonometric area = 1 2 * ab * bc * the sine value of angle b = 48 so ad=48 * 2 100=

According to the Pythagorean theorem, it is easy to find be=8

So the triangle ABC area is 48

So ad=

Make BE vertical AC

ab=bc, isosceles triangle.

So be is also the midline of AC.

ae=ac/2=6

ab=10 by the Pythagorean theorem.

be=√（100-36）=8

So triangle area = ac*be 2 = 48

Triangle area = bc*ad 2 = 48ad=

Evidence: n+2 under the root number - n+1 under the root number "n+1 under the root number - n under the root number; n+2 under the root number + n under the root number "n+1 under the root number;n+2 under the root sign + n) squared under the root sign < squared (n+1 under the root number); n(n+2) under the root number

Proof: ( n+2)-(n+1)<(n+1)- n) [n+2)-(n+1)] [(n+1)+ n]<[n+2)-(n+1)] [(n+1)+ n] (molecular rationalization).

1 [( n+2)+(n+1)]<1 [( n+1)+(n] (n+2)+(n+1)>(n+1)+ n ( n+2)> n is proof.

1. Find the part after the original = wrong formula 1 2a, and then divide by 丨 2a x2a....