3x-60=x

3x-60=x

3x-60=x
3x-x=60
2x=60
x=30

How to solve the equation 3x = x + 60

3x-x=60
2x=60
x=30

There is a point charge of negative sixth power with Q of - 3 * 10. When it moves from point a to point B in the electric field, it overcomes the electric field force to do work of - fourth power J of 6 * 10;
1. With B as the zero potential energy point, what is the electric potential energy EPA of the charge at point a?
2. If the electric potential energy at point C is 0, what is the electric potential energy EP2 of the charge at point a?

According to the definition of potential difference, UAB = φ a - φ B = Δ EP / Q = 200V. If point B is defined as 0 potential, then the potential of point a is 200V, EPA = - 6 * 10 ^ - 4J. A to C, and the total work of electric force is 3 * 10 ^ - 4J, then UAC = - 100V

12. Given proposition p: "for any x ∈ R, there exists m ∈ R, 4 * x + 2 * XM + 1 = 0". If proposition p is not a false proposition, then the value range of real number is -- problem complement

Non P is a false proposition, so p is a true proposition, so the equation has a solution in the range of real number, that is, (2m) ^ 2-4 * 4 * 1 > = 0, M > = 2 or m

-1 + 2-3 + 4-5 + 6-7 + 8-9, fill in the nine palace, satisfy! The product of the three numbers are negative; the sum of the absolute values of the three numbers are equal
Three numbers must satisfy every row, every column and every diagonal row: the product of three numbers is negative; the sum of absolute values of three numbers is equal.

+2 -9 +4 -7 -5 -3 +6 -1 +8

There are two functions f (x) = asin (KX + Wu / 3), G (x) = btan (KX - Wu / 3). The sum of their periods is 3 Wu / 2
There should be a specific process, not jumping,

The minimum positive period of F (x) = 2 π / | K |, and the minimum positive period of G (x) = π / | K |,
3 π / | K | = 3 π / 2, | K | = 2, k = soil 2,
Insufficient conditions

Solving linear equation with absolute value
2|2x-1|-1=2

2|2x-1|-1=2
|2x-1|=3/2
2x-1=-3/2,2x-1=3/2
x=-1/4,x-5/4

Function f (x) = 3 ^ x (0

The domain of inverse function is the domain of F (x)
0

What is the original function with derivative (SiNx) ^ 5?
You're not right on the first floor. It's a composite function

The number of times is so high, obviously it needs to be reduced
Original formula = ∫ [1 - (cosx) ^ 2] ^ 2 * sinxdx
=-∫[1-(cosx)^2]^2d(cosx)
Let t = cosx, then the original = - ∫ (1-T ^ 2) ^ 2DT
=-∫(t^4-2t^2+1)dt
=-(1/5)t^5+(2/3)t^3-t+C
=-(1/5)(cosx)^5+(2/3)(cosx)^3-cosx+C

It is known that the function f (x) = 1-A / (x power of 3) + 1 is an odd function. (1) find the value of a; (2) prove that f (x) is an increasing function on R; (3) find the range of function f (x) when x belongs to [- 2,2]; (4) find the solution set of inequality f [Log1 / 2 (3-x)] + F [1 / 3. Log2 (3-x) - 2 / 3] > = 0. (4)

(1) Obviously, if f (x) is defined as R, then f (- x) = 1 - [a * 3 ^ X / (3 ^ x + 1)] because f (x) is an odd function, then f (- x) = - f (x) has 1 - [a * 3 ^ X / (3 ^ x + 1)] = A / (3 ^ x + 1) - 1, that is, a (3 ^ x + 1) / (3 ^ x + 1) = 2, and a = 2 because 3 ^ x > 0