It is known that the parabola y = x & sup2; + (1-2a) x + A & sup2; (a ≠ 0) intersects with X axis and 2 points a (x1,0), B (X2, O) (x1 ≠ x2) ① Find the value range of a, and prove that a and B2 are on the left side of origin o ② If the plane parabola and y-axis intersect at point C, and OA + ob = oc-2, the value of a is obtained

(1) If two points intersect, then y = 0 has two,
Then △ = (1-2a) ^ 2-4a ^ 2 > 0
a＜1/4
According to Vader's theorem, if X1 + x2 = 2a-10, then x1x2 is less than 0,
So points a and B2 are all on the left side of origin o
(2) OC = a ^ 2, OA + ob = 2a-1, solution A1 = 1 + radical 2, A2 = 1-radical 2
And because of A

If the image of a linear function y = - 2x + 1 passes through the vertex of the parabola y = x2 + MX + 1 (m ≠ 0), then M=______ ．

∵ y = x2 + MX + 1, ∵ vertex coordinates are (- m2, 4 − M24), and the image of the linear function y = - 2x + 1 passes through the vertex of the parabola y = x2 + MX + 1 (m ≠ 0), ∵ 4 − M24 = - 2 × (- m2) + 1, ∵ M = 0 or M = - 4, while m ≠ 0, ∵ M = - 4

1 by 2:5 plus 2 by 3:5 plus 3 by 4:5 plus 48 by 49:5 plus 49 by 50:5 equals ()

5/1×2+5/2×3+5/3×4+…… +5/48×49+5/49×50
=5(1/1×2+1/2×3+1/3×4+…… +1/48×49+1/49×50)
=5(1/1-1/2+1/2-1/3+1/3-1/4+…… +1/48-1/49+1/49-1/50)
=5(1-1/50)
=5×49/50
=49/10

The area of a triangle is 22 square centimeters, the bottom is 8 centimeters, and the height is ()?

22*2/8=5.5

How to solve the equation 18 * (x-0.4) = 12 * (x + 0.25)

18*(x-0.4)=12*(x+0.25)
18x-18*0.4=12x+12*0.25
18x-12x=7.2+3
6x=10.2
x=1.7

The circumference of a semicircle is 10.28cm. What's its area?

Radius: 10.28 ÷ (2 + 3.14) = 2cm
Area: 3.14 × 2 × 2 △ 2 = 6.28 square centimeter

Simple calculation of 9 / 15 / 4 × 3 / 5
(5 / 6-4 / 5 + 1 / 15) × 30
5 / 9 △ 7 / 12 × 7 / 10
6 / 7 × 8 / 9 △ 6 / 7 × 8 / 9
2-9 / 2-15 / 7 / 5 / 3
If it's easy, it's easy

As shown in the figure, given that CE = 4be, CD = 3aD, the area of the shaded triangle is 6 square centimeters, and what is the area of the triangle ABC?

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Let 1, 2, 3, 4, 5, 6 be arranged as A1, A2, A3, A4, A5, A6. If AK (k < I, k = 1, 2, 3, 4, 5) satisfies | AI AK | = 1 for any AI (I = 2, 3, 4, 6), then such arrangement has ()
A. 36B. 32C. 28D. 20

If 1 is not on the front left side, then 2 must be on the left side of 1 in the order of (1) 23456, change the position of 1: (123456) (213456) (231456) (234156) (234516) (234561) (2) 3456, and the order of 1 and 2 is 21 (at the same time, 3 must be on the left side of 21): (321456) (324156) (324516) (324561) (342156) (342516) (342561) (345) 216) (345261) (345621) (3) 456 order unchanged: (432156) (432516) (432561) (435216) (435261) (435621) (453216) (453261) (453621) (456321) (4) 56 order unchanged: (543216) (543261) (543621) (546321) (564321) (5) 6 in the leftmost: (654321) a total of 32 possibilities, so B

How to prove that the product of the area of four opposite triangles formed by diagonals in any quadrilateral is equal
As shown in the figure, any quadrilateral is diagonally divided into four quadrilateral
For example, the quadrilateral ABCD and the diagonal intersect at point E,
Then the area of △ AED ×△ BEC = the area of △ AEB ×△ CED
How to prove the above conclusion?

Left = (AE * DG) * (BF * ce) / (2 * 2), right = (AE * BF) * (CE * DG)