Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0, and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, The first term of sequence {BN} (BN > 0) is C, and the first n terms and Sn satisfy SN-S (n-1) = √ Sn + √ s (n + 1) (n ≥ 2) 1. Find the general term formula of sequence {an} and {BN} 2. If the sum of the first n terms of the sequence {1 / BNB (n + 1)} is TN, what is the minimum positive integer n with TN > 1000 / 2009

Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0, and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, The first term of sequence {BN} (BN > 0) is C, and the first n terms and Sn satisfy SN-S (n-1) = √ Sn + √ s (n + 1) (n ≥ 2) 1. Find the general term formula of sequence {an} and {BN} 2. If the sum of the first n terms of the sequence {1 / BNB (n + 1)} is TN, what is the minimum positive integer n with TN > 1000 / 2009

The sum of the first n terms of 1) a = 1 / 3, an = fn-c - (f (n-1) - C) = fn-f (n-1) = - 2 / 3 * (1 / 3) ^ (n-1) ｜ an is (1 / 3) ^ n - 1 ｜ C = 1 and ∵ SN-S (n-1) = {Sn + √ s (n-1)}}} {Sn - √ sn-1 = 1}} {Sn = n, Sn = n ^ 2} BN = sn-sn-1 = 2n-12) BN to get 1 / bnbn BN + 1

The taxi charge standard of a city is as follows: 5 yuan for less than 3 kilometers, 1.5 yuan for more than 3000 meters per kilometer, how much do you need to pay for 5 kilometers?
How many kilometers can I take if I pay 23 yuan

5 yuan for 3 km,
Charge for the rest of the journey: (5-3) × 1.5 = 3 (yuan)
5 km: 3 + 3 = 6 (yuan)
For the fare of 23 yuan, you can take: (23-3) △ 1.5 = 40 / 3 (km)

Draw the rectangle nailed with wooden strips into parallelogram and compare their perimeter and area

Perimeter unchanged
Smaller area

The maximum and minimum values of Q function z = x ^ 2 + y ^ 2 in circular field [x - (2's root)] ^ 2 + [y - (2's root)] ^ 2 ≤ 9
The questions you asked before,

Let x = √ 2 + ρ cos α, y = √ 2 + ρ sin α (0 ≤ ρ ≤ 3), then z = x ^ 2 + y ^ 2 = (√ 2 + ρ cos α) ^ 2 + (√ 2 + ρ sin α) ^ 2 = 4 + ρ ^ 2 + 2 √ 2 ρ (COS α + sin α) = 4 + ρ ^ 2 + 4 ρ sin (α + π / 4). Therefore, the maximum value: Z ≤ 4 + ρ ^ 2 + 4 ρ = (ρ + 2) ^ 2 ≤ (3 + 2) ^ 2 = 25

Math problem, please hurry up
1. On the afternoon of June 20, 2013, the shenzhou-10 astronauts conducted a special space lecture on Tiangong-1, which is about 340km away from the earth. They demonstrated and explained interesting physics experiments in weightlessness environment for teenagers all over the country, Xiaohong weighs 8.98kg less than Xiaoming. If Xiaohong weighs 40kg on the ground, her weight will be reduced by about 3% for every 100km from the earth, can you calculate the approximate weight of Xiaoming on the ground?
2. If the solutions of the equations 5x + k = 20 and 6x + 3 = 0 are reciprocal to each other, then the value of K is?

When the weight is reduced by 3% for every 100 km rise, the weight is reduced by 340 km rise: 340 / 100 * 3% = 10.2%;
It weighs 8.98 kg on Tiangong-1 and x kg on the ground
　　X*（1－10.2%）＝8.98
X = 10 (kg)
The weight of Xiaoming on the ground is: 40 + 10 = 50 (kg)

In a parallelogram, if the perimeter is equal to 48 and the length of one side is known to be 12, the length of each side shall be calculated. If AB = 2BC, the length of each side shall be calculated

① 48-2 * 12 / 2 = 12, each side is 12. ② 48 / 6 = 8.8 * 2 = 16, side length is 8.8.16.16

A passenger car and a freight car are running in the same direction on the parallel track. The passenger car is behind the freight car. The length of the passenger car is 200 meters and the length of the freight car is 280 meters. The ratio of the speed of the passenger car and the speed of the freight car is 5:3. The crossing time for the passenger car to catch up with the freight car is 1 minute. The speed of each car is calculated. If the two cars are running in opposite directions, what is their crossing time?

Suppose the speed of the passenger car is 5x M / min and that of the freight car is 3x M / min, then there is: 5x = 3x + 280 + 200, and the solution is x = 240. Therefore, the speed of the passenger car is 5 × 240 = 1200 (M / min), and that of the freight car is 3 × 240 = 720 (M / min). If two cars are running in opposite directions, then the crossing time is (280 + 200) / (1200 + 720) = 0.25 (min) = 15 seconds It's 15 seconds

A math problem,
Master Wang used to take five minutes to make a part, but now his work efficiency has been improved, and the time to make a part has been shortened by 20%. How many parts can he make when he used to make 500 parts a day

It used to be 500 parts a day, but now it can be x parts
500*5=5*（1-20%）*X
X=625

In a square piece of paper with a circumference of 100cm, cut the largest circle. What is the radius of the circle?

A: the radius of this circle is 12.5cm

A (A-1) - (the square of a-b) = - 2, ab = 5. Find [(the square of a + the square of B) / 2] - 2Ab
The answer is as follows:
Because a (A-1) - (the square of a-b) = - 2, A-B = 2
also
[(square of a + square of B) / 2] - 2Ab
=[(1 / 2) * (a-b) square] - AB
=[(1 / 2) * 2 squared] - 5
=-3
But I can't understand this step = [(1 / 2) * (a-b) square] - AB how can 2Ab become AB?
Please know!

Because one AB is taken to the front to make up (1 / 2) (a-b) ^ 2, you can expand the formula yourself
(1/2)(a-b)^2-ab
=(1/2)(a^2-2ab+b^2)-ab
=(1/2)(a^2+b^2)-ab-ab