Let a vector and B vector be two nonparallel non-zero vectors, and X (2a vector + B vector) + y (3a vector - 2b vector) = 7a, x, y belong to R, find the value of X, y

Let a vector and B vector be two nonparallel non-zero vectors, and X (2a vector + B vector) + y (3a vector - 2b vector) = 7a, x, y belong to R, find the value of X, y

Is it a 7a or a 7a vector? I'll take it as a vector 7a, pull x (2a vector + B vector) + y (3a vector - 2b vector) = 7a vector (2x + 3y-7) a vector = (2y-x) B vector. Because a vector and B vector are two nonparallel non-zero vectors, so 2x + 3y-7 = 0 and 2y-x = 0. Then I'll calculate by myself

Given a = radical 2, find the value of 2A radical (A-2) ^ 2

2A radical (A-2) ^ 2
=2√2-√(√2-2)^2
=2√2-(2-√2)
=3√2-2

A 12 cm long wire is used to circle a right triangle with an area of 6 cm 2

Let the right angle sides be x and y, then XY2 = 6x2 + y2 = (12 − x − y) 2, and the solution is x = 3, y = 4, or x = 4, y = 3. Therefore, the length of the oblique side is 12-3-4 = 5. A: the lengths of each side of the triangle are 3, 4, 5

An isosceles triangle is an axisymmetric figure whose axis of symmetry is ()
A. Straight line through vertex B. height at bottom C. straight line at waist D. vertical bisector at bottom

A. Any line not passing through the vertex is the axis of symmetry of isosceles triangle, which should be the line where the bisector of vertex angle is located, so this option is wrong; B. because the height on the bottom edge is a line segment, and the axis of symmetry is a straight line, it should be the line where the height on the bottom edge is located, so this option is wrong; C. according to the

The minimum value of polynomial x2 + y2-6x + 8y + 7 is______ ．

The original formula = (x-3) 2 + (y + 4) 2-18. When both complete square formulas take 0, the minimum value of the original formula = - 18. So the answer is: - 18

Known points a (1,1), B (_ 1,5) and vector AC = 1|2ab, vector ad = 2Ab, vector AE = - 1 / 2Ab

From a (1,1), B (- 1,5), ab = (- 1-1,5-1) = (- 2,4), AC = 1 / 2 * AB = 1 / 2 * (- 2,4) = (- 1,2), let C (x, y), then: AC = (x-1, Y-1), so X-1 = - 1, Y-1 = 2, x = 0, y = 3. So C (0,3); ad = 2 * AB = 2 * (- 2,4) = (- 4,8), let D (x, y), then: ad = (x-1, Y-1), so X-1 = - 4, Y-1 = 8

As shown in the figure, in the triangle ABC, ab = AC, the points D.E.F are bc.ab.ac And ED parallel AC, FD parallel AB, ab = 9, find the perimeter of quadrilateral AEDF

The perimeter of quadrilateral AEDF is ab = AC = 9 + 9 = 18
∵AB=AC
← B = ∠ C (equilateral and equal angle)
∵DF∥AB
■ ∠ FDC = ∠ B (same position angle)
■ ∠ FDC = ∠ C (equivalent substitution)
Ψ FD = FC (equal angle to equal edge)
Similarly, ed = EB
Perimeter of quadrilateral AEDF = AE + ed + FD + AF
=（AE+EB）+（FC+AF）
=AB+AC
=18

If the function f (x) = log3a (x-1) defined in the interval (1,2) satisfies f (x) > 0, then the value range of a is___ ．

Because x ∈ (1,2), so X-1 ∈ (0,1), from F (x) > 0, we get 0 < 3A < 1, so 0 < a < 13, so the answer is: (0,13)

How to prove the necessary and sufficient condition (vector form) of triangle heart?
It's vector form! "The intersection of the three bisectors" don't tell me.

Intersection of three bisectors

In the triangle ABC, e is the point on the extension line of BC, BP bisects the angle ABC, CP bisects the angle ace, if the angle a = 80 degrees, calculate the angle P
.......

∵∠PCE=∠∠PBC +∠P
∴2∠PCE=2∠PBC +2∠P
∴∠ACE=∠ABC+2∠P
∵∠ACE=∠ABC+∠A
∴2∠P=∠A=80°
∴∠P=40°