Given vector group A1, A2 It is proved that vector group B1 = A1, B2 = a1 + A2 ，br=a1+a2+… +AR is also linearly independent

Given vector group A1, A2 It is proved that vector group B1 = A1, B2 = a1 + A2 ，br=a1+a2+… +AR is also linearly independent

Suppose there is a set of real numbers K1 So that k1b1 + +Krbr = 0, namely & nbsp; & nbsp; k1a1 + K2 (a1 + A2) + +kr（a1+… +ar）=（k1+… +kr）a1+（k2+… +kr）a2+… +Krar = 0. Because vector group A1, A2 So K1 + +kr＝0… kr−1+kr＝0...

If the two base lengths of the trapezoid are 1 and 4, and the two diagonal fields are 3 and 4, the area of the trapezoid is 1
The height of the trapezoid is 12cm, and the length of the two diagonals is 15cm and 20cm respectively

1、6
2、150
Auxiliary line is to translate a diagonal to construct a triangle with the same area as trapezoid

Given the function f (x) = | lgx |, if 0 < a < B, and f (a) = f (b), then the value range of a + 2b is

f(a)=f(b),0

A grain bin full of wheat is a cone on the top and a cylinder on the bottom. The circumference of the bottom of the cylinder is 6.28 meters, the height is 2 meters, and the height of the cone is 0.6 meters. If each cubic meter of wheat weighs 750 kg, how many kg of wheat is there in the bin?

The bottom area of the grain storage is: 3.14 × (6.28 △ 3.14 △ 2) 2, = 3.14 (M2); the volume of the grain storage is: 3.14 × 2 + 3.14 × 0.6 × 13, = 6.28 + 0.628, = 6.908 (M3); the weight of the wheat storage is: 750 × 6.908, = 5181 (kg). A: the weight of the wheat storage is 5181 kg

The line L passing through the point P (2,1) intersects the ellipse x ^ 2 / 2 + y ^ 2 = 1. The trajectory equation of the midpoint of the chord cut by the ellipse is obtained

Straight line L: y = K (X-2) + 1; ellipse: x ^ 2 / 2 + y ^ 2 = 1, intersection a (x1, Y1), (X2, Y2), midpoint (x0, Y0);
L is substituted into the elliptic equation: (2k ^ 2 + 1) x ^ 2 + 4K (1-2k) x + 8K (k-1) = 0;
Weida's theorem: 2x0 = X1 + x2 = 4K (2k-1) / (2k ^ 2 + 1), 2y0 = Yi + y2 = 2 (1-2k) / (2k ^ 2 + 1); the division of the two formulas: k = - 1 / 2 * x0 / Y0, which is simplified by formula 2, obtains the midpoint equation: (Y-1 / 2) ^ 2 / (3 / 4) + (x-1) ^ 2 / (3 / 2) = 1, which is a translated ellipse
Point P is on the directrix of a given ellipse. There should be other simple methods

The area of a triangle is 40 square centimeters, the height is 5 centimeters, and what is the bottom

Area formula bottom × height × 1 / 2 = area
Bring in data base × 5 × 1 / 2 = 40
So height = 16cm

All odd numbers are prime numbers, all even numbers are composite numbers, {is this sentence right or wrong?} give reasons

Wrong
1 is not prime, 2 is not composite

Use two identical triangles (3 cm, 4 cm, 6 cm) to make a parallelogram. How many kinds of parallelogram can be made at most? What's the perimeter and the minimum?

At most 3 kinds of figures can be spelled out, the maximum circumference is 20 cm, and the minimum is 14 cm

It is known that AB is the diameter of ⊙ o, point P is a moving point on the extension line of AB, the tangent of ⊙ o is made through P, the tangent point is C, the bisector of ⊙ APC intersects AC at point D, then ⊙ CDP is equal to ()
A. 30°B. 60°C. 45°D. 50°

As shown in the figure, connect OC, ∵ OC = OA, PD bisects ∵ APC, ∵ cpd = ∵ DPA, ∵ a = ∵ ACO, ∵ PC is the tangent line of ⊙ o, ∵ OC ⊥ PC, ∵ CPO + ∵ cop = 90 °, ∵ cpd + ∵ DPA + (∵ a + ∵ ACO) = 90 ° and ∵ DPA + ∵ a = 45 ° respectively, that is ∵ CDP = 45 °. Therefore, select C

As shown in the picture, there is a section on both sides of a river that is parallel. There is a tree every 5 meters on the South Bank of the river, and a telegraph pole every 50 meters on the north bank. Xiaoli stands at point P 15 meters away from the south bank and looks at the North bank. She finds that two adjacent telegraph poles A and B on the north bank are just covered by two trees C and D on the south bank, and there are three trees between the two trees

Let p be PF ⊥ AB, intersect CD at e, intersect AB at F, as shown in the figure: let the river width be x M. ∵ ab ∥ CD, ∵ PDC = ∥ PBF, ∥ PCD = ∥ PAB, ∥ PDC ∥ PBA, ∥ ABCD = PFPE, ∥ ABCD = 15 + x15, CD = 20 m, ab = 50 m, ∥ 2050 = 1515 + X according to the theme, the solution is: x = 22.5 (m)