If the difference between the second power of 8x + 2x-5 and another polynomial is the second power of 5x - x + 3, then the other polynomial is () If the difference between the second power of 8x + 2x-5 and another polynomial is the second power of 5x - x + 3, then the other polynomial is ()

If the difference between the second power of 8x + 2x-5 and another polynomial is the second power of 5x - x + 3, then the other polynomial is () If the difference between the second power of 8x + 2x-5 and another polynomial is the second power of 5x - x + 3, then the other polynomial is ()

8x^2+2x-5-(5x^2-x+3)
=3x^2+3x-8
Another polynomial is (3x ^ 2 + 3x-8)

When the value of integer n is, the N + 2 power of polynomial X - the 2-N power of 2x + 2 is cubic

N + 2 = 3 or 2-N = 3
The solution is n = 1 or n = - 1

To solve difficult mathematical problems, you can only use 1 ~ 10 numbers and add, subtract, multiply and divide

3, 3, 7, 7, add, subtract, multiply and divide, and you get 24

The pulley block lifts the weight for 10m in 10s, the tension at the free end of the rope is 200N, and the mechanical efficiency of the pulley block is 85%
(1) The power of pulling force;
(2) The gravity of the object;
(3) When lifting 580n weight with this pulley block, what is the required tension?
Number of shares 3

1. The length of the free end of the rope is s = 10 × 3 = 30m
The work of pulling force w = f · s = 200 × 30 = 6000J
Power P = w / T = 6000 △ 10 = 600W
2. The total weight of lifting is 200 × 3 = 600N
Gravity g = 600 × 85% = 510n
3. Pulley weight g '= 600-510 = 90N
Total weight g = 580 + 90 = 670n
Required tension f = 670 △ 3 = 223.33n

An is the arithmetic sequence S13 = 56, find the square of (A7) - a3-a13=
A is the subscript

S13=(a13+a3-2d)13/2=(2a7)*13/2)=56=(2a1+12d)*13/2
The results are as follows
a1+6d=56/13=a7.1)
a3+a13=a1+2d+a1+12d=2a7+2d
a7^2-a3-a13
=a7^2-2a7-2d
=(30*56/13)-2d
I can't find out

When an object moves in a straight line with uniform acceleration, the velocity of the object increases from V1 = 1m / s to V2 = 1.8m/s from T1 = 0 to T2 = 2S, then the velocity of the object increases at T3 = 12s

Acceleration = change of velocity / time of speed change a = (v2-v1) / (t2-t1)
A = (1.8m / s-1m / s) / (2s-0s) = 0.4m / (s * s)
The expression of speed is: v = V0 + A * t (V0 is the speed at the timing, that is, the initial speed)
v=1m/s+0.4m/(s*s)*12s=5.8m/s

The product of three continuous natural numbers is 336. The three natural numbers are______ ．

336 = 2 × 2 × 2 × 3 × 7 = 8 × 6 × 7, so the three natural numbers are 6, 7 and 8 respectively

When k is the value, the solution of the equation 5 (X-Y) = 3x-k + 2 is not greater than 17?

5 (x-k) - 3x + K-2 = 0, that is: 5x-5k-3x + k = 2 (5x-3x) - (5k-k) = 22x-4k = 22 (x-2k) = 2x-2k = 1x = 1 + 2K ① when X & gt; 0, 1 + 2K & gt; 02k & gt; - 1K & gt; - 1 / 2 ② when X & lt; 0, 1 + 2K & lt; 02k & lt; - 1K & lt; - 1 / 2 ③ when X & lt; = - 1, 1 + 2K & lt; = - 12K & lt; = - 2K

On the number axis, point a represents - 4. If the origin 0 is moved to the negative direction by 1.5 counting units, then the number represented by point a on the new number axis is - 4_______ .

-2.5

Six cards labeled 1, 2, 3, 4, 5 and 6 are put into three different envelopes. If each envelope contains two cards, and the cards labeled 1 and 2 are put into the same envelope, there are different ways to put them
(A) 12 species (b) 18 species (c)
36 species (d) 54 species
This is a math problem in this year's college entrance examination. It puzzles me. I choose 18 kinds of math problems, which are calculated one by one with stupid method. Many experts in my class, including math class representatives, choose 36. I'm not convinced~

A12 species