Let the sequence {an} satisfy A0 = a, a (n + 1) = can + 1-C, where a and C are real numbers, and C is not equal to 0

Let the sequence {an} satisfy A0 = a, a (n + 1) = can + 1-C, where a and C are real numbers, and C is not equal to 0

0

It is known that SN is the sum of the first n terms of the sequence {an}, A1 = 1 / 2, 2Sn = sn-1 - (1 / 2) n minus the first power + 2 to find the general term formula of an

2S=Sn-(1/2)^n+2,
Subtracting 2A = an + (1 / 2) ^ n,
Divide both sides by (1 / 2) ^ n to get
2^（n+1)an=2^n*an+1=…… =2a1+1=2,
∴an=1/2^n.

In the sequence {an}, if A1 = 2, an = 3A (n-1) + 5 (n ≥ 2, n belongs to n *), then an=______ .

an+x=3a(n-1)+5+x
an+x=3[a(n-1)+5/3+x/3]
Let x = 5 / 3 + X / 3
x=5/2
an+5/2=3[a(n-1)+5/2]
So an + 5 / 2 is equal, q = 3
So an + 5 / 2 = (a1 + 5 / 2) * 3 ^ (n-1) = 9 / 2 * * 3 ^ (n-1) = 1 / 2 * 3 ^ (n + 1)
an=-5/2+1/2**3^(n+1)

Parachutists perform low altitude parachuting. When the plane flies 224m above the ground, the parachutists leave the plane to do free fall in the vertical direction
After a period of exercise, open the parachute immediately. After the parachute is deployed, the athlete will decelerate and descend with an average acceleration of 12.5m/s ^ 2. For the safety of the athlete, the maximum landing speed of the athlete is required to be no more than 5m / s, What is the height from the ground at least? What is the height of free fall when landing? (2) what is the shortest time of athletes in the air

What is the minimum height of the parachute from the ground
(1) Let the height of the parachute be h, the speed be V0, and the landing speed be VT = 5m / s,
h0 = 224m
Vt ^ 2-v0 ^ 2 = 2ah, substituting a = - 12.5m/s2,
So h = 99m,
How high does landing mean falling freely
（2）h'=V^2/2g=5^2/20=1.25m
It is equivalent to falling freely at 1.25 meters
What is the shortest time an athlete can spend in the air
(3) The free falling time T1 = 5 s can be obtained
The average deceleration time after deployment T2 = 3.6s, Tmin = 8.6s

There are two two two digit numbers, the number on the ten digit is n, the number on the one digit is a, B respectively, and the sum of a and B is equal to 10
What rule do you find in the results? Can you explain this rule with what you learned in the chapter multiplication of integers?
Note: what we have learned in Grade 7 and 8

Correct answer:
If two two digit numbers are x and Y respectively, then
x=10n+a
y=10n+b
So there's a product of two two digits
x*y=（10n+a）（10n+b）=100n2+10n（a+b）+ab
And a + B = 10, so
x*y=100n*n+10n*10+ab=100n*n+100n+ab=100n（n+1）+ab
That is to say, the result of multiplication of two two digit numbers satisfying the conditions in the question is equal to 100 times of the product of two digit ten digit number multiplied by (ten digit number plus 1), plus the product of these two digit numbers. It is used for quick calculation. It's not easy to understand the words. It's easy to understand if I give you a few examples
Like 52 and 58,
Then 52 * 58 = 5 * (5 + 1) * 100 + 2 * 8 = 3016
11*19=1*（1+1）*100+1*9=209
After proficiency, you can directly write the product of any two digits that meet the conditions mentioned in the question: the first two digits are n * (n + 1) (if the result is less than two digits, take one digit), and the last two digits are ab (if the result is less than two digits, fill in a 0)
For example, 52 * 58, the first two digits of the result are 5 * (5 + 1) = 30, and the last two digits are 2 * 8 = 16, so the result is 3016
Another example is 11 * 19. The first two digits of the result are 1 * (1 + 1) = 2, and the last two digits are 1 * 9 = 9 = 09 (less than two digits are 0), so the result is 209
Let's take another example that satisfies the following conditions:
For example, calculate the product of 98 and 92, 98 * 92, the first two numbers of the result are 9 * (9 + 1) = 90, the last two numbers are 8 * 2 = 16, so the result is 9016, you might as well verify it

What is the relationship between the maximum static friction and sliding friction? Why?

In general calculation, the sliding friction can be regarded as equal to the maximum static friction unless it is specified. But in practice, the sliding friction is smaller than the maximum static friction
This theoretical proof is actually very cumbersome and unnecessary (actually I don't know, just know that this proof is very cumbersome), and the conclusion drawn from the experiment

(1) Two times of X is equal to the difference between half of X and 4. Finding the product of X (2) x and - 3 is equal to the sum of three times of X and 2, finding X

(1)2x=1/2x-4
4x=x-8
3x=-8
x=-8/3
(2)x*(-3)=3x+2
6x=-2
x=-1/3

The distance between the two places is 320 km, with 45 km per hour for passenger cars and 35 km per hour for freight cars. When the two cars meet, how many kilometers more do passenger cars travel than freight cars?

320 ÷ (45 + 35) = 320 △ 80 = 4 (hours) 45 × 4-35 × 4 = 180-140 = 40 (kilometers) a: when two cars meet, the bus travels 40 kilometers more than the truck

1999 2000/2001*1 2/1999
2000 out of 1999 and 2001 times 1 out of 1999

2000 out of 1999 and 2001 times 1 out of 1999
=(2000 out of 1999 + 2001) * 2001 out of 1999
=1999 * 1999:2001 + 2001:2000 * 1999:2001
=2001 + 2000 / 1999
=2002 and 1999

99+100+101+.+999=?

=448*1100+550+99+100=493549