Given the position of a and B on the number axis as shown in the figure, if point P is on the number axis and PA + Pb = 6, find the corresponding position of P

Given the position of a and B on the number axis as shown in the figure, if point P is on the number axis and PA + Pb = 6, find the corresponding position of P

(1) Because AB = 4
As shown in the figure, the numbers represented by two points a and B on the number axis are - 2 and 6 respectively. The point C on the number axis satisfies AC = BC, and the point D is on the extension line of the line AC. if ad = 32AC, then BD=______ The number represented by point D is______ .
The numbers represented by a and B are - 2 and 6, ab = 6 - (- 2) = 8, AC = BC = 12ab = 12 × 8 = 4, ad = 32AC = 32 × 4 = 6, OD = ad-ac = 6-2 = 4, BD = 6-4 = 2, and the number represented by D is 4
We know that point a on the number axis represents the number - 3, point B represents the number 5, and point P represents the number x (1). When PA = Pb, point P represents the number X=_____________ .
(2) Express PA and Pb with algebraic expression containing X: (it is required that there is no sign of absolute value in the result)
PA is expressed as:
Pb is expressed as:
(3) When PA + Pb = 10, find the number represented by X
(4) If point a and point B move to the left at the speed of 5 and 20 unit lengths per second respectively, and point P also moves to the left from the origin at the speed of 1 unit length per second_____________ When PA = Pb, the number represented by X is______________ .
The examination questions of the seventh month in the high school attached to Jida should be accurate
(1) When PA = Pb, | x + 3 | = | X-5 |, the number x represented by point P=____ 1_________ .
(2) Express PA and Pb with algebraic expression containing X: (it is required that there is no sign of absolute value in the result)
PA is expressed as: PA = x - (- 3) = x + 3
Pb is expressed as: Pb = X-5
(3) When PA + Pb = 10, find the number represented by X
|x+3|+|x-5|=10
X = - 3-1 = - 4 or x = 5 + 1 = 6
(4) If point a and point B move to the left at the speed of 5 and 20 unit lengths per second respectively, and point P also moves to the left from the origin at the speed of 1 unit length per second_____________ When PA = Pb, the number represented by X is______________ .
Let PA = Pb when the time is t
PA=|-t-(-3-5t)|,PB=|-t-(5-20t)|
There is | 3 + 4T | = | 19t-5|
3+4t=19t-5 3+4t=5-19t
t=8/15 t=2/23
So when the time is 8 / 15 seconds or 2 / 23 seconds, PA = Pb, then x = - t = - 8 / 15 or - 2 / 23
Mathematical answer group for you to answer, I hope to help you.
We know that point a on the number axis represents the number - 3, point B represents the number 5, and point P represents the number X
(1) When PA = Pb, the number represented by point P is x = 1
(2) Express PA and Pb with algebraic expression containing X: (it is required that there is no sign of absolute value in the result)
PA is expressed as: 3 + X (when x ＞ - 3), - 3-x (when x ＜ - 3)
Pb is expressed as: X-5 (when x > 5), 5-x (when x < 5)
(3) When PA + Pb = 10
Mathematical answer group for you to answer, I hope to help you.
We know that point a on the number axis represents the number - 3, point B represents the number 5, and point P represents the number X
(1) When PA = Pb, the number represented by point P is x = 1
(2) Express PA and Pb with algebraic expression containing X: (it is required that there is no sign of absolute value in the result)
PA is expressed as: 3 + X (when x ＞ - 3), - 3-x (when x ＜ - 3)
Pb is expressed as: X-5 (when x > 5), 5-x (when x < 5)
(3) When PA + Pb = 10, find the number represented by X.
X-5 + X + 3 = 10, x = 6 (when x > 5),
5-x + (- 3-x) = 10, x = - 4 (when x ＜ - 3)
(4) If point a and point B move to the left at the speed of 5 unit length / s and 20 unit length / s respectively, and point P also moves to the left from the origin at the speed of 1 unit length / s, (5 + 3) / (20-5) = 8 / 15, PA = Pb, then the number represented by X is X - 8 / 15
I wish you progress in your study and make progress! (*^__ ^*) put it away
It is known that f (x) = 1 / 3x + 1 / 2aX + 2bx + C (a, B, C ∈ R), and the function f (x) has a maximum in the interval (0,1),
If the minimum value is obtained in the interval (1,2), then the value range of Z = (a + 3) + B is
Tick (COS ^ 2x) = - cosx, the value range of X
In (0,2 π), the value range of √ (COS ^ 2x) = - cosx, X
If you simplify the left side, it will become | cosx | = - cosx, then there should be cosx
The known function f (x) = x ^ 3-2ax ^ 2-3x, X ∈ R
When x ∈ (0, + ∞), f (x) ≥ ax is constant
Function f (x) = x ^ 3-2ax ^ 2-3x
And f (x) ≥ ax
That is x ^ 3-2ax ^ 2-3x ≥ ax
x^3-2ax^2-3x-ax≥0
x（x^2-2ax-3-a）≥0
The equation holds: x ^ 2-2ax-3-a ≥ 0
====》（-2a）^2-4x1x(-3-a)≤0
The solution is x =
Known FX= cosx.cos (x - π / 3) find the minimum positive period of FX and make FX
F (x) = cosx (cosxcos π / 3 + sinxsin π / 3) = 1 / 2 * cos & # 178; X + √ 3 / 2 * sinxcosx = 1 / 2 * (1 + cos2x) / 2 + √ 3 / 2 * sin2x / 2 = √ 3 / 4 * sin2x + 1 / 4 * cos2x + 1 / 4 = 1 / 2 * (sin2xcos π / 6 + cos2xsin π / 6) + 1 / 4 = 1 / 2 * sin (2x + π / 6) + 1 / 4, so t = 2 π / 2 = π, that is sin (2x +
Given the function f (x) = - 3x ^ 2 + 2ax-1, X ∈ [0,1], Let f (a) be its minimum, find the expression of F (a), and find the maximum of F (a)
f(x)=-3x^2+2ax-1
=-3(x-a/3)^2+a^2/3-1
Axis of symmetry x = A / 3
Discussion on the value classification of a
① When a ≤ 0 / 3
Monotone decreasing of F (x) on X ∈ [0,1]
The minimum value of F (x) f (1) = - 3 + 2a-1 = 2a-4
In this case, f (a) = 2a-4a / 3 > 0 3 / 2 > a > 0
The minimum value of F (x) = 2a-4
Notice that f (a) = f (1) a / 3 > = 1 / 2 3 > a > = 3 / 2
The minimum value of F (x) is f (0) = - 1
In conclusion, the maximum value of F (a) is - 1
f(a)=-3a^2+2a^2-1=-a^2-1
A new function about a is obtained: F (a) = - A ^ 2-1
The maximum value of this function is - 1
You can draw the picture below
Given a ≥ 1, the function f (x) = 4x + 9 / x + 1 + 4 (x ∈ [0,1]), find the range of F (x)
Function f (x) = 4x + 9 / (x + 1) + 4 (x ∈ [0,1]), find the range of F (x)
∵f(x)=4x+9/(x+1)+4 ∴f'(x)=4-9/(x+1)²
Let f '(x) = 0: X1 = - 5 / 2, X2 = 1 / 2
When ∵ x ∈ [0,1] ∵ x ∈ [0,1 / 2], f (x) is simple decreasing; X ∈ [1 / 2,1], f (x) is simple increasing
Ψ f (x) min = f (1 / 2) = 12 f (x) max = f (1) = 25 / 2
∴f(x)∈[12,25/2]
What does it have to do with a? Follow me up..... I have the wrong number
Function f (x) = 4x + 9 / x + 1 + 4 (x ∈ [0,1]), find the range of F (x)
If the function y = 1 / 3x3-1 / 2ax2 + (A-1) x + 1 is a decreasing function in the interval (1,4) and an increasing function in the interval (6, + ∞)
Try to find the value range of real number a
Let y = f (x), then f (x) = x ^ 3 / 3 - ax ^ 2 / 2 + (A-1) x + 1 is a continuous function!
f’=x^2-ax+a-1
It is a decreasing function in the interval (1,4)
in other words
f’（1）=0
So a > = 7
So the value range of a is [5,7]
It is a closed interval! The understanding of Building 1 is not appropriate!
∵ function f (x)=
One
Three
X3-
One
Two
ax2+(a-1)x+1
∴f′（x）=x2-ax+（a-1）=（x-1）[x-（a-1）]
Moreover, the function f (x) is a decreasing function in the interval (1,4) and an increasing function in the interval (6, + ∞),
∴4≤a-1≤6
∴5≤a≤7
Derivation y '= x ^ 2-ax + A-1
The problem is that when x is in the interval (1,4), y ` 0 holds
The separation parameters a > 1 + X, X in the interval (1,4) and AA > 5
…… It's really hard to make do with it on the computer
The derivative of function f (x) f '(x) = x2-ax + A-1
Let f ′ (x) = 0, the solution is x = 1 or x = A-1
When A-1 ≤ 1, i.e. a ≤ 2, the function f (x) is an increasing function on (1, + ∞)
When A-1 > 1, i.e. a > 2, the function f (x) is an increasing function on (- ∞, 1),
It is a decreasing function in (1, A-1) and an increasing function in (A-1, + ∞)
According to the meaning of the title
When x ∈ (1,4), f ′ (x) ＜ 0,