1 / 5 (X-9) = 1 / 2x-9 fine calculation process

1 / 5 (X-9) = 1 / 2x-9 fine calculation process

1 / 5 (X-9) = 1 / 2x-9 multiplied by 10
2（x-9）=5x-90
2x-18=5x-90
5x-2x=90-18
3x=72
x=24

Square of (x-2x) - 6 (x-2x) + 9

(x²-2x)²-6(x²-2x)+9
=[(x²-2x)-3]²
=(x²-2x-3)²
=[(x+1)(x-3)]²
=(x+1)²(x-3)²
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(2x-9) square - (X-6) square = 0

6 = 1 / 2x square-x
How to solve it

5 X-6 = 0
Multiply 2 on both sides of the equation to get x-2x-12 = 0
In the equation, a = 1, B = - 2, C = - 12
From the root formula, (- B plus minus root b-4ac) / 2a is obtained
There are two values for X

Let f (x) = x2-2ax + 2 (a ∈ R), when x ∈ [- 1, + ∞), f (x) ≥ A is constant, and the value range of a is obtained

F (x) = x2-2ax + 2 = (x-a) 2 + 2-a2f (x) the symmetry axis of the image is x = A. in order to make f (x) ≥ a constant on [- 1, + ∞), it only needs that the minimum value of F (x) on [- 1, + ∞) is greater than or equal to A. when (1) a ≤ - 1, f (- 1) is the minimum, and the solution is - 3 ≤ a ≤ - 1 & nbsp; (2) When a ≥ - 1, f (a) is the smallest, and the solution a ≥ − 1F (a) = 2 − A2 ≥ A is - 1 ≤ a ≤ 1. To sum up, - 3 ≤ a ≤ 1

Let D, e, f be the points on BC, CA, AB, and DC = 2bd, CE = 2EA, AF = 2fb, then the positions of AD + be + CF and BC?
Let D, e and f be the points on BC, Ca and AB, respectively, and the vector DC = 2, the vector BD, the vector CE = 2, the vector EA and the vector AF = 2, the positions of the vector AD + the vector be + the vector CF and the vector BC?
(vector problem)

According to the vector formula of fixed score point, ad → = AC → + 2Ab → / 1 + 2 = 1 / 3aC → + 2 / 3AB →, be → = 1 / 3bC → + 2 / 3ba →, CF → = 1 / 3CA → + 2 / 3CB →,
Add the above three formulas to get ad → + be → + CF → = - 1 / 3bC →,
So reverse parallel

A and B two ropes, a rope length of 63m, B rope length of 29m, two ropes minus the same length, the result of the remaining length of B rope is 1 / 3 of the remaining length of a rope, a and B rope length of each how many meters? Minus how many meters? (Guide: think about the two ropes minus the same length, explain what

Do you want to explain
Solution: let x meter be subtracted
29-x=1/3（63-x）
29-x=21-1/3x
2/3x=8
x=12
A: 63-12 = 51m
B: 29-12 = 17 meters
I hope you understand. Take it

The area of a circle is derived from the area of a parallelogram

We know that the area of a circle is equal to the square of the radius times the circumference
How is this formula derived?
Because the periphery of a circle is an arc, not a straight line, it can not be solved by the area method of a rectangle. But it also gives us space to think
So, we draw a circle on the cardboard, divide the circle into several equal parts, cut it out, and then use these similar isosceles triangle pieces of paper to put together to form an approximate parallelogram. If the score is more, each part will be thinner. The more close the figure is to a rectangle. The length of the rectangle is equal to half of the circumference of the circle, that is, C / 2, and the width is equal to the radius r of the circle, Because the area of rectangle = length × width, the area of circle s = C × R △ 2
And because C = 2 π R, s = π R & #

A cylindrical steel pipe with an empty middle is 12 cm in diameter from the outside and 8 cm in diameter from the inside. If the mass of steel per cubic centimeter is 7.8 g, what is the mass of this steel pipe?

Outer radius = 12 △ 2 = 6cm
Inner radius = 8 △ 2 = 4cm
Steel pipe weight
=Steel pipe volume × 7.8
=3.14 × (6-178; - 4-178;) × length × 7.8
=489.84 × length
Your topic is short of a length condition, please add
I'm glad to answer for you. I wish you progress in your study!

1. The plane passing through any two generatrix of the cylinder is α, and the centers of the upper and lower bottom surfaces of the cylinder are o and O1. The position relationship between oo1 and α is judged and proved
2. Try to prove that there is only one plane parallel to the known plane through a point outside the plane

Oo1 is parallel to or on plane α,
It is proved that: oo1 is parallel to any generatrix, and any line outside the plane parallel to a line in the plane is parallel to the known plane, so oo1 is parallel to plane α. When oo1 passes through plane α,. Oo1 is on plane α
2. Two planes parallel to the same plane are parallel. Suppose that there are two planes parallel to the known plane through a point outside the plane, then the two planes are also parallel to each other, but the two planes intersect, which is contradictory. Therefore, there is only one plane parallel to the known plane through a point outside the plane