The side length of equilateral triangle ABC is 2, the points P and Q are on both sides of AB and AC respectively, and PQ bisects △ ABC into two parts with equal area. The minimum length of PQ is obtained

The side length of equilateral triangle ABC is 2, the points P and Q are on both sides of AB and AC respectively, and PQ bisects △ ABC into two parts with equal area. The minimum length of PQ is obtained

The side length of equilateral triangle ABC is 2, and its area = root sign 3 points P and Q are on both sides of AB and AC respectively. PQ divides △ ABC into two parts with equal area, 1 / 2 * AP * AQ * sin60 degree = 1 / 2 root sign 3aP * AQ = 2. According to cosine theorem, there are PQ & # 178; = AP & # 178; + AQ & # 178; - 2 * AP * AQ * cos60 degree = AP & # 178; + a

Through the point m (1,1), make a straight line and the parabola x ^ 2 = 2Y intersect at two points a and B, the two tangent lines of the parabola at two points AB intersect at point P, find the trajectory equation of P, 2 find three

I can tell you about it. You can do it yourself. First, use the letters with unknowns to represent the straight line passing through M
Let the coordinates of point a be (x1, 2 / x1) and point B be (X2, 2 / x2), and then the tangent equations of points a and B can be obtained
Then the coordinates of the intersection point are obtained
Finally, these expressions are integrated and simplified
It should be able to find out

As shown in the figure, Xiaohong uses a rectangular piece of paper ABCD for origami. It is known that the width ab of the paper is 8cm and the length BC is 10cm. When Xiaohong folds, the vertex D falls at point F on the edge of BC (the crease is AE). Think about it, how long is EC at this time? Explain in the way you've learned

In RT △ ABF, ab = 8, AF = 10, ｜ BF = af2 − AB2 = 6, ｜ CF = bc-bf = 4, let CE = x, then D is a rectangle, ab = CD = 8, BC = ad = 10, BC = ad = 10, BB = af2 − AB2 = 6, CF = bc-bf = 4

It is known that propositions P: X1 and X2 are two real roots of the equation x-mx-2 = 0, and the inequality a-5a-3 ≥ | x1-x2 | holds for any real number m ∈ [- 1,1]
It is known that propositions P: X1 and X2 are two real roots of equation x-mx-2 = 0, inequality a-5a-3 ≥ | x1-x2 | holds for any real number m ∈ [- 1,1]; proposition q: inequality ax + 2x-1 > 0 has solution; if proposition p is true proposition, proposition q is false proposition, the value range of a is obtained

Proposition q is a false proposition, so ax + 2x-1 > 0 has no solution, so a

Given the coordinates of three points, find the normal vector of plane vector
How to calculate the next step
2. If any x is taken in to solve y, Z, where is the point? Coordinate point? Or the abscissa of the vector

I don't know how you do it. The correct way is to know three points ABC, vector AB = B-A, vector AC = C-A, AB x AC is the answer
How can you be x, y and Z, which is so troublesome

Fifth grade 10 multiplication vertical calculation, 10 division vertical calculation,

53×31 32×34 21×44 98×11 34×34 60×55 27×34 41×21 84×42 42×55 560÷7 490÷7 810÷9 497÷7 648÷8 321÷3 444÷4 122÷2 999÷9 147÷7

In the cube abcd-a.b.c.d. with edge length a, where m is the midpoint of AA, then the distance from point A. to plane MBD is?

Connecting AC and BD, the focus is o. connecting OM, then AOM is a right triangle, and the distance from a to MBD is the height of the long side of AOM. From the figure, we can know the length of AO and am, and further we can find OM. Then, the height of OM is equal to Ao times am?

The value process of X & sup2; + Y & sup2; + Z & sup2; - xy-yz-zx if x-3 = Y-2 = Z-1

X-3 = Y-2 = Z-1, then X-Y = 3-2 = 1y-z = 2-1 = 1z-x = 1-3 = - 2, so x & sup2; + Y & sup2; + Z & sup2; - xy-yz-xz = (2x & sup2; + 2Y & sup2; + 2Z & sup2; - 2xy-2yz-2xz) / 2 = [(x & sup2; - 2XY + Y & sup2;) + (Y & sup2; - 2yz + Z & sup2;) + (Z & sup2; - 2XZ + X & sup2;)] / 2 = [(X-Y) & su

Given that line AB, l is the vertical bisector of line AB, the perpendicular foot is C, P is any point on L, and PA = Pb is obtained

Because l is the vertical bisector of line ab
So PA = PB

The solution equation is as follows: (1) 1 / 4 x + 3 / 4 = 7 / 8 (2) 7 / 6 minus 3x = 3 / 14 (3) 6 / 5 minus 3 / 3 = 5 / 8

(1) 1 / 4 x + 3 / 4 = 7 / 8
1/4x=1/8
x=1/2
(2) 6 out of 7 minus 3x = 3 out of 14
3x=9/14
x=3/14
(3) 5 / 6 minus 1 / 3 x = 5 / 8
1/3x=5/24
x=5/8