I have a math problem that I don t know how to solve

Because, de bc, ef ab, therefore, the quadrilateral defb is a parallelogram, so, de=bf; Again, E is the midpoint of AC and EF AB, so EF is a median line of the triangle cab, so F is the midpoint of BC; So, de=bf, bf=cf; So, de=cf

E is the midpoint of AC, EF AB

f is the midpoint of BC.

de∥bc,ef∥ab

The quadrilateral DBFE is a parallelogram.

de=bf and f are the midpoints of bc (i.e., bf=cf).

de=cf

The solution is as follows: Solution: Because de bc, ef ab so dcbf is a parallelogram;

And because E is the midpoint of AC, DE is the median line of BC; So de=1 2bc;

In the same way, EF is the median line of AB;

So, f is the midpoint of ac; So, fc=1 2bc;

So, de=fc;

I hope you can understand.,If you don't understand, you can ask.。。

EF ab and e is the ac midpoint, f is the bc midpoint, bf=fc and write the rest.

From the inscription f is the midpoint of the bottom edge, de bc

So, de=bf=1 2bc=fc

It is sufficient to prove the parallelogram first, and then the congruence of the two triangles (ade efc).

Start the conversation with little time.

C 8 min.

A 8+10=18 minutes.

B 8 + 10 + 12 = 30 minutes.

A + B + C = 56 minutes.

Happy fairy78, hello :

Knowledge points: It is always the first to come with less time, and the last to have more time.

Conversation order: C, A, B.

Minimum required: 8 (8 10) (8 10 12) 56 (minutes).

Solution: The final result is to arrive home 20 minutes earlier than usual, which means that Reiko walks in 30 minutes after arriving at the nearest station, and Dad only needs 20 2 = 10 minutes to complete by car.

In other words, if Reiko arrives at the nearest station, if she waits at the station, her father's car will arrive at the station in 10 minutes.

So Reiko arrived at the station 10 minutes earlier than usual.

Because it was 20 minutes earlier, Reiko's 30-minute walk away from Dad only took 20 minutes to drive a round trip, which means that Reiko's 30-minute Dad only needed 10 minutes to finish.

Suppose Reiko arrived at the station 30 minutes early, and Reiko passed by her father and didn't see her father, then 30 minutes later, her father arrived at the station, which means that her father had to go backwards for 10 minutes, 30+10=40 minutes, 40 minutes earlier.

1.The price has been reduced by 20%.

Looking at it this way, the original price of a dress is 5 yuan, 1000 yuan can buy 200 sets, and now a piece of clothing is 4 yuan. Understandable...

2.a, b, m are all non-zero natural numbers, and if b a, then b is smaller; If B a, then B out of a A is greater.

If there are no other hints for the question, and it is a fill-in-the-blank question or something, then it will generally default to b a).

3.A circle is 360°, and if you remove 45°, you still have 315°, so the remaining part of the area is 315 45=7 times the subtraction.

Uphill distance: 20 1 = 20 thousand pounds before touching the rice regret pants.

Let Xiao Li's speed be x kilometers per hour.

3/2x - x = 18/6

x = 6

Solution 1: When two people meet, the distance traveled is 3x, 2x; 3x-2x=18；x=18km, Xiaojing walked 54km, Xiao Li 36km, when the two met, it took the same time, the same 6 hours, so Xiao Li's speed is 36 6=6km h

Solution 2: Let the velocity of Xiaoli be xkm h, then the small mirror is 3 2xkm h, then 6*(3 2x-x)=18 x=6km h

Let's assume that "B fourth" is right for E, then "B for 3rd" is not right for A, "C is fifth" for "true", and "B second" for C is not right and "C first" is correct, but this is a contradiction. So the assumption is not valid.

Therefore, "A third" said by E is correct, and then judged on this basis, since A is third, then B third said by A is not correct, then "C fifth" is correct;

Therefore, b says that d is fifth, and "e fourth" is correct;

From C fifth: C says that C is the first is wrong, and "B is second" is right;

Based on the above rankings, it can be concluded that: d first.

Therefore, the rankings of these five students are: D first, B second, A third, E fourth, C fifth.

A says: B is third, C is fifth.

B says: E fourth, D fifth.

C says: A first place, E fourth place.

D says: C first place, B second place.

E says: A third place, B fourth place.

Someone guessed correctly in every position. , then:

D must be fifth, because only one person mentions D;

At the same time, E is the fourth place that has been established (both say E is fourth);

Then C must be the first (and there is another person who says that C is fifth, which contradicts D fifth);

Then A must be the third (and there is another person who says that A is the first, which contradicts the first C);

Of course, the remaining B can only be second.

The rankings are: C, B, A, E, D.

C first, B second, A third, E fourth, D fifth.

According to what the teacher said, it can be inferred that C is the first and B is the second, so what A says is wrong, then E is half right, that is, A is the third.

In the end, the only thing left is what B said

1st place: c; 2nd place: b; 3rd place: a;

4th place: e; 5th place: d;

Don't fool me, the answer will come out when it's listed, are you kidding???

C first, B second, A third, E fourth, D fifth.

a, third, b, second, c, fifth, e, fourth, d, first.

Solution: One foot of air, take half of it every day, then, take 1 2 on the first day, and take half half on the second day, that is, (1 2) 2,。。

Take (1 2) n on day n

1) Then, after day 5 there is 1-[(1 2)+(1 2) 2+.1/2)^5]=1/32

Day 6 remaining: 1 64

Remaining after day 10: 1 1024

2）1-[（1/2）+（1/2）^2+..1/2)^n]=(1/2)^n

3）（1/2）^n<1/1000000

2 n>10 6, take the natural logarithm on both sides at the same time:

nln2>6ln10

n>6ln10/ln2=

n = After 2020 days, less than one millionth of a foot of air remains.

Essentially, it is to examine the summation of proportional sequences.

1-1 2=1 2 left on the first day

On the fifth day, 1 2 (n=5)=1 32 remained

6 1/ 2ⁿ （n=6）=1/64

10 1/ 2ⁿ （n=10）=1/1024n 1/ 2ⁿ

n=7, less than 1 100

Solution: (1) After the fifth day, a total of 1 2+(1 2) 2+(1 2) 3+(1 2) 4+(1 2) 5=1-(1 2) 5 was taken

So there is 1-1+(1 2) 5=(1 2) 5 feet.

After the sixth day, a total of 1 2 + (1 2) 2 + (1 2) 3 + (1 2) 4 + (1 2) 5 + (1 2) 6 = 1-(1 2) 6

So there is 1-1+(1 2) 6=(1 2) 6 feet.

After the tenth day, a total of 1 2+(1 2) 2+(1 2) 3+ ...... were taken1/2)^9+(1/2)^10=1-(1/2)^10

So there is still 1-1+(1 2) 10 feet left.

2) The length of one foot of air after the nth day of the same method is (1 2) n feet.

3) i.e. (1 2) n<1 10 6 solution gives n>=20, so after about 20 days, one foot of air is less than one millionth of a foot.

Let the length of the ruler be 1

Then the first day is 1x1 2

The next day was 1x1 2x1 2

And so on -

1) Fifth day (1 2) 5

Day 6 (1 2) 6

Day 10 (1 2) 10

2) Day n (1 2) n

3) After X days, the remaining part of a foot of air is less than one millionth.

Column inequality (1 2) x<1 10000001 1000000=(1 10) 6=(1 2x1 5) 6, so the original inequality is (1 2) x small is equal to (1 2x1 5) 6 to get 2 (x-6) large equal to 5 6

After that I used a calculator to do the math. 5 to the 6th power is equal to 15625.

2 to the 13th power is equal to 8192

2 to the 14th power is equal to 16384

So x-6 is greater than or equal to 14, i.e. x>20

Because the 14th day is already insufficient, it is after the 13th day.

Day 1: 1 2 = 1 (2 1)......n=1

Day 2: 1 2*1 2=1 (2*2)=1 (2 2)......n=2 Day 3: 1 2*1 2*1 2=1 (2*2*2)=1 (2 3)......n=3

Day n: 1 (2*2*2*......=1 (2 n) so: 1) fifth day: 1 (2 5).

Day 6 (1 2) 6

Day 10 (1 2) 10

2) Day n (1 2) n

3）1/（2^n）<1/(10^6)

2^n>10^6

n=20 (2 19=524288 2 20=1048576) So: after about 20 days, less than one millionth of a foot of air remains.

Half of it is taken every day"It's multiplied by 1 2 per day.

2)(1/2)^n

3) Let x days after meet the requirements, there are (1 2) x<1 1000000, and the solution x is 20

The length remaining on day n is: x=(1 2) n

The equation is given to you, and the rest is done by yourself.

This is a proportional number series problem, first list the equation, and directly bring in the number of days to calculate.

a ba 9 with |b－1|are inverse numbers to each other.

a²－ba＋9=0，|b－1|=0

b = 1 is substituted for b = 1, a ba 9.

Decomposed (a+3)(a-1)=0

A1=-3 and a2=1

Substituting a1=-3 yields (a, b, a) (ab)=4 3a2=1, and (a, b, a) (ab)=0

A ba 9 and the second term is 6a, so that he is a non-negative number. a=3，b=1

So the original formula = (3-1 3) (3+1) = 2 3