As shown in the figure, the vertices a and C of the rectangular oabc are on the positive half axis of the X and Y axes respectively, and the point D is the midpoint of the diagonal ob, and the inverse scale function y = K The image of X (x > 0) in the first quadrant passes through point D and intersects with AB and BC at e and f respectively. If the area of quadrilateral BEDF is 1, then the value of K is______ .

As shown in the figure, the vertices a and C of the rectangular oabc are on the positive half axis of the X and Y axes respectively, and the point D is the midpoint of the diagonal ob, and the inverse scale function y = K The image of X (x > 0) in the first quadrant passes through point D and intersects with AB and BC at e and f respectively. If the area of quadrilateral BEDF is 1, then the value of K is______ .

Connect of, EO,
∵ point D is the midpoint of diagonal ob, and the area of quadrilateral BEDF is 1,
∴S△BDF=S△ODF，S△BDE=S△ODE，
The area of the quadrilateral foed is 1,
If e, m and D are on the inverse scale function image, then s △ OCF = K
2，S△OAE=k
2，
If D is taken as DG ⊥ Y-axis at point G and DN ⊥ X-axis at point n, then s ⊥ ondg = K,
If ∵ D is the intersection point of the diagonal of rectangular ABCO, then s rectangle ABCO = 4S ∵ ondg = 4K,
Since the function image is in the first quadrant, k > 0, then K
2+k
2+2=4k，
The solution is: k = 2
3．
So the answer is: 2
3．

As shown in the figure, the coordinates of vertex C of lozenge oabc are (3,4), and vertex A is on the positive half axis of X axis. The inverse scaling function y = K If the image of X (x > 0) passes through vertex B, the value of K is () A. 12 B. 20 C. 24 D. 32

Pass point C as CD ⊥ OA,
∵ the coordinates of C are (3,4),
∴CD=4，OD=3，
∵CB∥AO，
The ordinate of B is 4,
∴OC=
CD2+OD2=5，
∴AO=OC=5，
∵ the quadrilateral Coab is a diamond,
The abscissa of B is 8,
∴k=8×4=32，
Therefore, D

As shown in the figure, the vertex o of the parallelogram oabc is at the coordinate origin, and the vertices a and C are in the inverse scale function y = K On the image of X (x > 0), the abscissa of point a is 4, the abscissa of point B is 6, and the area of parallelogram oabc is 9, then the value of K is______ .

Pass point C as CD ⊥ X-axis at point D, point a as AE ⊥ X-axis at point E, point B as BF ⊥ x-axis, AF ∵ x-axis, intersecting point F, connecting AC, ∵ quadrilateral oabc is a parallelogram,

As shown in the figure, the image with known positive scale function and inverse scale function passes through a (3,3) 1) The analytic expressions of positive and inverse proportional functions 2) After the straight line OA is shifted downward, it is intersected with the inverse proportional function B (6, m), and the value of M and the analytic formula of this first order function are obtained 3) The analytic formula of the quadratic function whose image of the first order function in question (2) intersects with the x-axis and y-axis respectively at C, D, passing points a, B and D 4) Under the condition of ground (3), is there a point E on the image of quadratic function, so that the area S1 of quadrilateral OECD and quadrilateral s meet the following requirements: S1 = 2 / 3S? If so, find the coordinates of point E; if not, please explain the reasons

(1)Y=X,Y=9/X
(2)m=2/3
Where is the picture

Known inverse proportional function y = k The image of X passes through point a (- 2,3) (1) The analytic expression of the inverse proportional function is obtained; (2) The image of the positive scale function y = k ′ x passing through point a and the inverse scale function y = k Do you have any other intersections in the image of X? If yes, find out the coordinates of intersection points; if not, explain the reasons

(1) ∵ point a (- 2,3) at y = K
In the image of X,
∴3=k
−2，∴k=-6；
The analytic formula of inverse proportional function is y = - 6
x；
(2) Yes
∵ the images of positive and negative scaling functions are symmetric about the origin, and point a is on their images,
The symmetry point B (2, - 3) of a (- 2,3) about the origin is also on their image,
The coordinates of the other intersection point are (2, 3)

It is known that a first-order function intersects the x-axis at the point (- 3,0) and the image of a positive scale function intersects (- 2,1). The analytic expressions of the two functions are obtained

Let the positive proportional function be y = KX
∵ image passing (- 2,1)
∴1=-2k
∴k=-0.5
ν positive proportional function: y = - 0.5x
Let the first degree function be y = KX + B
∵ image passing (- 3,0) (- 2,1)
∴0=-3k+b
1=-2k+b
∴k=1
B=3
ν first order function: y = x + 3

The image of positive scale function passes through the point (2. - 4), passes through a point a on the image as the vertical line of X axis, and the perpendicular foot is B (4.0). Find the coordinates of point a and the area of △ AOB

∵ positive proportional function
ν let y = ax
Replace x = 2, y = - 4 into the
The result is: - 4 = 2A
The solution is a = - 2
Then the function expression is y = - 2x
∵B（4.0）
ν replace x = 4 into the
Results: y = 4 × (- 2)
The solution is y = - 8
∴A（4,-8）
ν s △ AOB = 4 × 1 - 8 × 1 / 2 = 16

As shown in the figure, the first-order function image intersects the point a in the second quadrant, points B (- 6,0) on the x-axis, the area of △ AOB is 15, and ab = Ao (1) Find the analytic formula of positive proportion function; (2) Finding the analytic formula of function of degree one

Let a (XA, ya)
S=1/2*6*YA=15
YA=5
Because AB = Ao
So XA = - 6 / 2 = - 3
A(-3,5)
Let the analytic expression of positive proportional function be y = K1X
-3k1=5
k1=-5/3
y=-5/3x
Let the analytic expression of the function of first degree be y = k2x + B
A (- 3,5) B (- 6,0) on a straight line
-3k2+b=5
-6k2+b=0
k2=5/3 b=10
y=5/3x+10

The image with positive scale function y = KX passes through point a (k, 2K) (1) Find the value of K; (2) If point B is on the x-axis and ab = Ao, find the analytic formula of line ab

(1)

The image with positive scale function y = KX passes through point a (k, 2K) (1) Find the value of K; (2) If point B is on the x-axis and ab = Ao, find the analytic formula of line ab

(1)