The side area of each cylinder below is calculated. The perimeter of 1 bottom surface is 3.14dm, the height is 5DM, the diameter of 2 bottom surface is 4cm, the radius of 3 bottom surface is 5DM, and the height is 4dm 14 DM, 5 DM high, 4 cm diameter, 5 DM radius and 4 DM high

The side area of each cylinder below is calculated. The perimeter of 1 bottom surface is 3.14dm, the height is 5DM, the diameter of 2 bottom surface is 4cm, the radius of 3 bottom surface is 5DM, and the height is 4dm 14 DM, 5 DM high, 4 cm diameter, 5 DM radius and 4 DM high

Note that the side of the cylinder is a rectangle
1. The perimeter of the bottom surface is the length of one side rectangle, and the other side of the rectangle when the cylinder is high, so the side area s = 3.14 * 5 = 15.7
2.S=3.14*4*10=125.6
3.S=2*3.14*5*4=125.6

A 2.5-long ladder AB leans against a vertical wall AC, and the distance from the foot of the ladder to the bottom of the wall C is 0.7m
(1) How high is the top a of the ladder from the ground?
(2) If the top of the ladder descends 0.4m along the wall, does foot B also move out by 0.4m

Let two right sides of a right triangle be a and B, and the hypotenuse be C, then A2 + B2 = C2
(1) That is: 0.7 * 0.7 + B2 = 2.5 * 2.5, B = 2.4m
(2) 4 m, i.e. B '= b-0.4 = 2.4-0.4 = 2m
The new Pythagorean theorem of right triangle can obtain a '2 + B' 2 = C '2 and a' = 1.5m because the length of C '= C hoof and ladder is constant
The displacement of the foot is equal to a '- a = 1.5-0.7 = 0.8m.. so the foot moves more than 0.4m

The ladder AB is against the wall. The distance from the lower end a of the ladder to the wall root o is 2m, and the distance from the top B of the ladder to the ground is 7m. Now move the bottom end a outward to "a", so that
The distance from the lower end a 'of the ladder to the root o of the wall is 3M. At the same time, the top B of the ladder drops to B'. Is BB 'equal to AA' by calculation

From the topic, we can get: ab = a'B '#ab # ab # 178; = a'B' \35; 178; ab = a 'B' \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35;178; = 53-9 = 44b'o = 2 √ 11 BB '=

It is known that a = - (x1 + x2);
How to decompose the - 2x ^ 2-ax + 1 = 0 factor into (x ^ 2-x1x-1) * (x ^ 2-x2x-1);
The original question is as follows:
a = -(x1 + x2);
b = (x1*x2);
(x^2-ax+b)=0;
x^2*(x^2 - ax + b-2) - ax + 1;
After simplification, we get - 2x ^ 2 - ax + 1;
Why does factorization result in (x ^ 2-x1x-1) * (x ^ 2-x2x-1);

I see!
There is a trap in this problem, that is, x ^ 2-ax + B = 0. When you simplify the formula, you substitute the value of the formula, so there is a mistake
If you don't care x ^ 2 - ax + B = 0, expand the brackets and decompose the factor to get the answer

Factorization: - A + 2a2-a3=______ ．

-A + 2a2-a3 = - A (a2-2a + 1) -- (extract common factor) = - A (A-1) 2. - (complete square formula) so the answer is: - A (A-1) 2

Factorization (a + b) - 4 (a + b) + 4 A's cube + A's cube + 1 / 4A 6xy-9x's cube, Y-Y's cube 32A (x's cube + X's Cube) - 2A

(a + b) - 4 (a + b) + 4
=(a+b－2)²
Cube of a + square of a + 1 / 4A
=1／4a(4a²＋4a＋1)
=1／4a(2a＋1)²
Cube of 6xy-9x-y-y
=﹣y(y²－6xy＋9x²)
=﹣y(3x－y)²
32A (x + x) - 2A
=2a×[16(x²＋x)²－1]
=2a{[4(x²＋x)]²－1}
=2a(4x²＋4x＋1)(4x²＋4x－1)
=2a(2x＋1)²(4x²＋4x－1)

Xsquare-ax-2a cube < 0

(a+(a^2+8a^3)^0.5)/2>x>(a-(a^2+8a^3)^0.5)/2
Solving the equation x-square-ax-2a cube = 0
parabola
X squared-ax-2a cube is open up, so only the lower segment is less than 0

Factorization of the square of AX + the square of 2A + the third power of a

The square of AX + the square of 2A + the third power of a
=a(x²+2ax+a²)
=a(x+a)²

Factorization: a) factorization y ^ 2-34y-35 b) therefore, factorization x ^ 2 (2x + 3) ^ 2-34x (2x + 3) - 35

x^2(2x+3)^2-34x(2x+3)-35=[x(2x+3)]^2-34x(2x+3)-35
=[x(2x+3)-35][x(2x+3)+1]
=(2x^2+3x-35)(2x^2+3x+1)
=(2x-7)(x+5)(2x+1)(x+1)
y^2-34y-35=（y-35)(y+1)

Factorization: - 3ax ^ 2-4ax + A

-ax(3x+4)+a
=a(1-x)(3x+4)