How do you calculate the 11 questions? I want a step, thanks

11 Let the nth order of x be infinitesimal until it is constant. For example, the numerator and denominator are first conducted, the numerator = 1-1 (1-x 2), x = 0, the numerator is still 0, and the process can be changed to infinitesimal and simple. Do the math yourself.

12 left = e x(exp(xcosx 2-x)-1) e x-1 x, (exp(xcosx 2-x)-1) xcosx 2-x=x(cosx 2-1) also go to cosx.

x+1 x=3, both sides squared at the same time gives x 2+1 x 2+2=9 so x 2+1 x 2=7 x 2 (x 4+x 2+1)=1 (x 2+1 x 2+1)=1 (7+1)=1 8

1) HL or Pythagorean theorem plus SSS (2) AOC congruent BOD (same reason as above) AOB is an isosceles triangle.

, so at least 12 minutes, at this time Dad ran 4 laps and Mom ran 3 laps.

2.Who is the fastest?

3 4 6 is the least common multiple of 12

The trio met at the finish line at least 12 minutes later.

Pro: 1, (1 minute) 2, (mom one circle, dad two circles).

101 is an octal number, which is 65 in decimal, and 65 is an ASCII code with a capital A, so the output is A.

In question 12, the cyclic structure is to swap ab once, and at the same time c minus 1, and only jump out of the loop when a b c, so choose a

128*15

So there are 3 options.

3000-1920=1080 each

The sequence in question 11 is a proportional sequence with a ratio of 1 to 2. This shouldn't be a problem, right? Sum of proportional series = a1*(1-q n) (1-q); a1：

the first line of the proportional series; q: The proportion of the proportional series. That's no problem, right?

Bring a1=1, q=1 2 into the sequence: sum of the sequence =1*[1-(1 2) n] [1-(1 2)] 1-(1 2) n] (1 2) =2-2*[(1 2) n]=2-1 [2 (n+1)].

x approaches positive infinity, and 1 x approaches infinitesimal. The previous part is the bounded function, and the infinitesimal multiplied by the bounded function is equal to the infinitesimal, which is zero.

Day 3 Day 1 Day: Climbed 5 meters, slid down 2 meters at night, actually climbed 3 meters.

The next day, during the day: after climbing 5 meters, it reached 8 meters, and at night it slid down 2 meters, and actually climbed 6 meters.

During the day on the third day, I went out after climbing 4 meters on the basis of 6 meters on the second day, and the descent was already successful before the evening.

5-2 = 3 meters upward on the first day.

On the first day, 5-2 + 3 meters = 6 meters.

On the third day, 5 meters, 6 + 5 = 11 meters, climbed out.