It is known that the equations 3x + 2 = 0 and X + a = 20 about X have the same solution, () (fill in the brackets with yes or no)

It is known that the equations 3x + 2 = 0 and X + a = 20 about X have the same solution, () (fill in the brackets with yes or no)

The topic is not complete
When a = 62 / 3, it is, otherwise it is not

4x5 (60-3x) = 20 help to solve this equation

4x5 (60-3x) = 20
20-60+3x=20
3x=20+60-20
3x=60
x=20

Distance formula of two points
1. On the x-axis, if the distance from point P to point a (1,2) is the root sign 5, then the coordinate of point P is______ .
2. On the y-axis, the distance from point P to point a (- 1,2) is the root sign 2, then the coordinate of point P is______ .
3. Find a point P on the x-axis so that its distance to point a (1,2) is equal to its distance to point B (- 1,1)
4. Given that the distances from point P to two coordinate axes and to a (1,1) are equal, calculate the coordinates of point P

1. On the x-axis, if the distance from point P to point a (1,2) is the root sign 5, then the coordinate of point P is_ (0,0) or (2,0)_____ 2. On the y-axis, if the distance from point P to point a (- 1,2) is root sign 2, then the coordinate of point P is__ (0,1) or (0,3)____ . 3. Find a point P on the x-axis so that the distance from it to point a (1,2) is the same as that from it to point B (-)

The formula of base area of rectangle

First of all, a rectangle has no bottom area. It should be a cuboid. The bottom area is nothing more than length multiplied by width

Finding monotone interval of F (x) = xlnx-1

If f (x) is derived, f ′ (x) = 1 + LNX
Let f '(x) = 0, then x = 1 / E
And it is easy to know: when x0, the function increases monotonically

Simple operation of solving two mathematical problems (with process)
1.1-1/2-1/4-1/8-… -1/128
2.1/1*2+1/2*3+1/3*4+1/4*5+1/5*6

1,1-1/2-1/4-1/8-… -1/128 =1/2-1/4-1/8-… -1/128 =1/4-1/8-… -1/128 =1/8-1/16-.-1/128=.=1/64-1/128=1/1282,1/1*2+1/2*3+1/3*4+1/4*5+1/5*6 =1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6 =1-1/6 =5/6

As shown in the figure, in △ ABC, ∠ ACB = 90 °, AC = BC, P is a point in △ ABC, and ∠ APB = ∠ APC = 135 ° (1) verify: △ CPA ∽ APB; (2) try to find the value of Tan ∠ PCB

(1) ∵ in △ ABC, ∠ ACB = 90 °, AC = BC, ∵ BAC = 45 °, that is ∠ PAC + ∠ PAB = 45 °, and in △ APB, ∠ APB = 135 °, ∵ PBA + ∠ PAB = 45 °, ∵ PAC = ∠ PBA, ∵ APB = ∠ APC, ∵ CPA ∽ APB. (2) ∵ ABC is isosceles right triangle, ∵ caab = 12, ∵ CPA ∵ apab = papB = caab = 12, let CP = k, then PA = 2K, Pb = 2K And in △ BCP, ∠ BPC = 360 ° - ∠ APC - ∠ APB = 90 ° and 〈 Tan ∠ PCB = PBPC = 2

Given that the square of a + AB = 3, the square of AB + B = 7, find the value of (1) the square of a = the square of 2Ab + B, and (2) the square of A

Excuse me, wait a minute
a^2+ab=3,ab+b^2=7
a^2+ab+ab+b^2=7+3=10=a^2+2ab+b^2=(a+b)^2
a^2=2ab+b^2
Then,

Y = 32x ^ 2, please enter the answer dy / DX

64x
This is equivalent to deriving a ^ X

Let f (x) = 2Sin (ω x + π 6) (ω > 0) have f (x1) ≤ f (x) ≤ f (x2) for any x ∈ R, and the distance between point a (x1, f (x1)) and point B (X2, f (x2)) is 20, then the minimum value of ω is ()
A. π2B. πC. 2πD. π4

∵ f (x) = 2Sin (ω x + π 6) (ω > 0) for any x ∈ R, there are f (x1) ≤ f (x) ≤ f (x2), | f (x1) = - 2, f (x2) = 2, and | ab | = (X2 − x1) 2 + [f (x2) − f (x1)] 2 = (x2 − x1) 2 + 16 = 20, | x2-x1 | = 2 ≥ T2, | t = 2 π ω≤ 4, | ω ≥ π 2