Solving. I don t know how to do math. Math problems, no

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.

The rope folded in half is twice as small as the original.

It turns out that the rope has: (2+2)*2=8

It turns out that wood has: 8-2=6

1.Because a m+a n=4 and a (m+n)=2 (a m+a n) 2=a 2m +a 2n +2a (m+n)=a 2m +a 2n +4=16

So a 2m +a 2n=12 So a (-2m)+a(-2n)=1 a 2m +1 a 2n=[a 2m +a 2n] [a (m+n)] 2=12 (2 2)=3

(2009 444+1)+2x2009 222] Because 2009 111+2009 333 2x2009 222 so a b 1 so a b

This index,It's too troublesome to knock on.。。。

Let the m power of a = x, the n power of a = y, then there is x+y = 4,,,xy = 2, the square of the original formula = (1 x) + the square of (1 y)...

1 x + 1 y) squared -2 (xy).

(x+y) (xy)) squared -2 (xy) is substituted into the known available formula = 3

a^-2m+a^-2n

1/a^2m+1/a^2n

a^2m+a^2n)/a^(2m+2n)=[(a^m)^2+(a^n)^2]/[a^(m+n)]^2=[(a^m+a^n)^2-2a^(m+n)]/[a^(m+n)]^2

3. The detailed steps are almost the above, and you can ask if you don't understand.

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.

The volume of the stone = 300 * 2 = 600 cubic centimeters.

Sleepwalking in the sky and leaving farewell (Li Bai).

Since there are two intersections, y=ax 2+bx+c=0 has two different solutions, set to x|and x2, then x1 then ab two point coordinates are a(x1,0), b(x2,0), then x1=[-b- (b 2-4ac)] (2a), x2=[-b+ (b 2-4ac)] (2a), the length of the line segment ab = x2-x1= (b 2-4ac)] a.