As shown in the figure, the image of the first-order function passing through the point Q (0,3.5) intersects the image of the positive scale function y = 2x at the point P. the equation that can represent the image of the first-order function is () A. 3x-2y+3.5=0B. 3x-2y-3.5=0C. 3x-2y+7=0D. 3x+2y-7=0

Let y = KX + B. ∵ the line passes through points P (1,2) and Q (0,3.5), K + B = 2B = 3.5, and the solution is k = − 1.5b = 3.5. Therefore, the analytic expression of the linear function is y = - 1.5x + 3.5, that is, 3x + 2y-7 = 0
Given the set a = {X / 1 ≤ log2 x ≤ 2}, B = [a, b] if a is contained in B, find the value range of a-b
From a = {X / 1 ≤ log2 x ≤ 2}, we can get a = [2,4] because a is contained in B, then a ∩ B = A should be B = [a, b], then b > a ∩ a ∈ [2,4] B ∈ [2,4], so A-B ∈ (0, - 2] has been made for you. If you don't understand, please ask!
As shown in the figure, it is known that the image intersection point a of the first-order function y = - x + 7 and the positive scale function y = 3 / 4x
It is known that the image of the first-order function y = - x + 7 and the positive scale function y = 3 / 4x intersects at point a, and intersects with the X axis at point B
The moving point P starts from point O and moves along the route of o-c-a at a speed of 1 unit length per second. At the same time, the line L starts from point B and moves to the left at the same speed. In the process of translation, the line L intersects the X axis at point R and the intersection line BA or line Ao at point Q. when point P reaches point a, both point P and line L stop moving, Set the movement time of P as T seconds
① When the value of T is, the area of the triangle with a, P and R as vertices is 8?
② When p is on OC, the triangle with vertex a, P and Q is isosceles triangle
(1) ∵ linear function y = - x + 7 and positive proportional function y=
Four
Three
The image of X intersects at point a and the X axis intersects at point B
Qi
y=-x+7
Y=
Four
Three
X
,
The solution is as follows
X=3
Y=4
,
The coordinates of point a are: (3,4);
∵y=-x+7=0,
The solution is: x = 7,
The coordinates of point B are: (7,0)
(2) When P moves on OC, 0 ≤ T < 4, Po = t, PC = 4-T, Br = t, or = 7-t,
∵ when the area of a triangle with vertices a, P and R is 8,
S trapezoid acob-s △ acp-s △ por-s △ ARB = 8,
Qi
One
Two
（AC+BO）×CO-
One
Two
AC×CP-
One
Two
PO×RO-
One
Two
AM×BR=8,
∴（AC+BO）×CO-AC×CP-PO×RO-AM×BR=16,
∴（3+7）×4-3×（4-t）-t×（7-t）-4t=16,
∴t2-8t+12=0,
The solution is: T1 = 2, T2 = 6 (rounding off),
When 4 ≤ t ＜ 7, s △ Apr=
One
Two
AP × OC = 2 (7-T) = 8, t = 3, not 4 ≤ T < 7;
To sum up, when t = 2, the area of the triangle with vertices a, P and R is 8;
② Existence. Extend CA to a line L at a point D, when l and ab intersect at Q,
∵ the linear function y = - x + 7 intersects the X axis at (7,0) and the Y axis at (0,7),
∴NO=OB,
∴∠OBN=∠ONB=45°,
∵ line L ‖ Y axis,
∴RQ=RB,CD⊥L,
When 0 ≤ T < 4, as shown in Figure 1,
RB=OP=QR=t,DQ=AD=（4-t）,AC=3,PC=4-t,
∵ if a triangle with vertices a, P and Q is an isosceles triangle, then AP = aq,
∴AC2+PC2=AP2=AQ2=2AD2,
9 + (4-T) 2 = 2 (4-T) 2, the solution is: T1 = 1, T2 = 7 (rounding off),
When AP = PQ, 32 + (4-T) 2 = (7-T) 2,
The solution is t = 4 (rounding off)
When PQ = AQ, 2 (4-T) 2 = (7-T) 2,
The solution is T1 = 1 + 3
Two
(rounding off), T2 = 1-3
Two
(omitted)
When 4 ≤ t ＜ 7, as shown in the figure (standby Figure), ad ⊥ ob is made by a, then ad = BD = 4,
Let l intersect AC with E, then QE ⊥ AC, AE = Rd = T-4, AP = 7-T,
From cos ∠ OAC=
AE
AQ
=
AC
AO
,
Get AQ=
Five
Three
（t-4）,
If AQ = AP, then
Five
Three
(T-4) = 7-T, the solution is t=
Forty-one
Eight
,
When AQ = PQ, AE = PE, that is AE=
One
Two
AP,
T-4=
One
Two
（7-t）,
The solution is: T = 5,
When AP = PQ, PF ⊥ AQ is made through P, and F,
AF=
One
Two
AQ=
One
Two
X
Five
Three
（t-4）,
In this paper, we propose a new method for the determination of the value of △ f ∠=
AF
AP
=
Three
Five
,
Get AF=
Three
Five
AP,
Namely
One
Two
X
Five
Three
（t-4）=
Three
Five
（7-t）,
The solution is: T=
Two hundred and twenty-six
Forty-three
,
In conclusion, when t = 1,5
Forty-one
Eight
,
Two hundred and twenty-six
Forty-three
Second, the triangle with a, P, Q as vertex is isosceles triangle
If set a = {x x (X-2)
Is it the wrong number? → (x-a - + 1)
From the meaning of the title
∵ a = {x (X-2) ＜ 3} a = (- 1,3)
Is that a = {B} - 1
∴ a∈（-1,4）
PS: but is the set of B right? Is it equal to
Is it possible that there is only one intersection point between positive scale function and inverse scale function in the same coordinate system
choice question
The number of intersections of the functions y = x and y = K / X in the same coordinate system is ()
A 0 B 1 C 2 D above answers are possible
Choose D
Make them x ^ 2 = K
Then when k > 0, there are two intersections
There is an intersection point when k = 0
K
If the solution set of inequality log2 (AX ^ 2 + X + 3) > 0 is r, then the value range of a is "process"?
log2(ax2+x+3)>0=log2(1)
When a = 0, the equation log2 (3) > 0 holds
When a ≠ 0, y = log2 (x) is an increasing function
So AX2 + X + 3 > 1
That is, AX2 + X + 2 > 0 holds on R
So the image can only be above the x-axis
a>0 ;△=1-8a1/8
In conclusion, the value of a is a = 0 or a > 1 / 8
^ 2 + 1 because ax > x
(log2(1)=0)
Then ax ^ 2 + X + 2 > 0
Discriminant from root 1-8a1 / 8
ax^2+x+3>0
It holds when a = 0
When a > 0, the discriminant is 1 / 12 or a = 0
There are several intersections between the positive scale function y = 3x and the inverse scale function y = x, and the coordinates of the intersection are
The coordinates of the two intersections are (1,3) and (- 1, - 3)
Find the range where y equals 5x plus 6 divided by 3x minus 4
y = （5 x + 6）/ （3 x + 4）= （5 x + 6）/ 3（x + 4 / 3）= （5 x + 20 / 3 - 2 / 3）/ 3（x + 4 / 3）= （5 x + 20 / 3）/ 3（x + 4 / 3） - （2 / 3）/ 3（x + 4 / 3）= 5（x + 4 / 3）/ 3（x + 4 / 3） - 2 / ...
How many intersections do the positive scale function y = KX and the inverse scale function y = 1 / 3x have
Please note: there is no one intersection in this problem, so the correct solution is as follows:
By combining the two analytic expressions, we can get the result
Y=KX
Y=1/3X
The results are as follows
KX=1/3X
X²=1/3K
When k > 0, the equation has two real roots, so the image has two intersections;
When k
Two, one, or none
Given a = {(x, y) / 2x + 3Y = 1}, B = {(x, y) / 3x-2y = 3}, find the intersection of a and B (important process, expressed by enumeration)
Solve the equations 2x + 3Y = 1,3x-2y = 3
The solution is a intersection B = {(x, y) / (11 / 13,2 / 39)}

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