As shown in the figure, in the triangle ABC, ab = 7, AC = 6, BC = 8, the straight line of line BC moves along the Ba direction at the speed of 2 units per second, and always keeps the same as the original line

As shown in the figure, in the triangle ABC, ab = 7, AC = 6, BC = 8, the straight line of line BC moves along the Ba direction at the speed of 2 units per second, and always keeps the same as the original line

Y=（7-2X）*7/8

As shown in the figure, in △ ABC, ab = 7, AC = 6, BC = 8. The straight line of segment BC moves along Ba direction at the speed of 2 units per second, and always keeps parallel with the original position. When x seconds is recorded, the length ed of the part of the straight line in △ ABC is y. (1) prove: △ AED ∽ ABC; (2) try to find the functional relationship between Y and X

It is proved that: (1) be = 2x, ed = y, ∧ ABC ∧ AED, ab = 7, BC = 8, ∧ aEAb = DEBC. ∧ 7 − 2x7 = Y8, that is, y = − 167x + 8

As shown in the figure, in the triangle ABC, ab = 8, AC = 6, BC = 9, if the moving point d moves from point B to point a in the direction of BA at a speed of 2 units per second,
A straight line de ∥ BC, the length of this straight line in △ ABC is y in x seconds, write the function relation of Y with respect to x, and draw its image

As shown in the upper part of the figure above: ad = 8-2x
y/9=(8-2x)/8=1-x/4   
The functional relation of Y with respect to X is: y = 9-9x / 4 & nbsp; & nbsp; (0 & lt; = x & lt; = 4)
The function image is shown in the lower part of the figure above

How to do vector AB minus vector AC plus vector BD minus vector CD?
Simplification,
Vector AB minus vector AC plus vector BD minus vector CD

Vector AB vector AC = vector CB
Vector CB + vector BD = vector CD
Vector CD vector CD = 0 vector

D is a point on edge BC of triangle ABC, tangent BD / DC = 2 / 5, if vector AB = a, vector AC = B, then what is vector ad equal to?

AD=AB+BD=AB+2/7BC=AB+2/7(AC-AB)=5/7a+2/7b
(the above letters are vectors)

Is vector Ao + vector OA equal to 0 or o vector

O vector

Vector OA - vector od + vector ad =? (process)

OA-OD+AD
=OA+AD-OD
=OD-OD
=0 vector
Note that the zero vector, plus the arrow, the computer is not easy to play, the above other also want to add the arrow

Vector addition and subtraction: oa-od-ad =?

Vector OA vector od = vector Da
So vector OA vector od vector ad = vector Da vector ad = - 2 vector ad
You can draw triangles

Vector Ba * vector BC =?

The module length of the vector Ba is multiplied by the module length of the vector BC, and then multiplied by the cosine value of the two vectors. But note that the angle between the two vectors must be the same starting point, or let the vector Ba = (a, b) and the vector BC = (c, d), so their product is AC + BD, and their product is a real number

Vector Ba = {3,3, - 3}, vector BC = {3,1,0}, find: vector BA × vector BC

Vector BA × vector BC = 3 × 3 + 3 × 1 + (- 3) × 0 = 12