How to solve the equation 8 (2x-5) = 4 (3x + 2)

How to solve the equation 8 (2x-5) = 4 (3x + 2)

16x-40=12x+8
16x-40-12x-8=0
4x-48=0
4x=48
x=12

2.2x-5 / 8 = 4 / 3x to solve the equation

2.2X-5/8=4/3X
2.2X-4/3X=5/8
11/5X-4/3X=5/8
13/15X=5/8
X=5/8*15/13
x=75/104

41-3x = 8 solution equation

What is the power of the logarithm of the base two of nine
Why is 3 ^ [log3 (4)] equal to four

9^[log3(2)]
=(3^2)^[log3(2)]
=3^[2log3(2)]
=3^[log3(2^2)]
=3^[log3(4)]
=4

There is 45 kg of oil in barrel a and 24 kg in barrel B. how many kg of oil can be poured from barrel a to barrel B so that there is as much oil in barrel a and barrel B?

Total weight 45 + 24 = 69kg
Average weight of two barrels 69 / 2 = 34.5kg
The weight poured from the nail barrel is 45-34.5 = 10.5kg
Pour 10.5 kg of oil from barrel a to barrel B, so that there is as much oil in barrel a and barrel B

A-A 1 / 2 = 3, find the square of A-A 1 / 2

Square of a - one of the squares of a = (a + 1 / a) (A-1 / a) = 3 (a + 1 / a)
（a+1/a）²=（a-1/a）²+4=9+4=13
A + 1 / a equals the positive and negative root sign 13
The value is plus or minus 3 root sign 13

If the function f (x) = X3 + AX-2 is an increasing function in the interval (1, + ∞), then the value range of real number a is ()
A. [-3，+∞）B. （-3，+∞）C. [0，+∞）D. （0，+∞）

F ′ (x) = 3x2 + A, according to the relationship between the function derivative and the monotonicity of the function, f ′ (x) ≥ 0 is constant on [1, + ∞), that is, a ≥ - 3x2 is constant, only if a is greater than the maximum value of - 3x2, and the maximum value of - 3x2 on [1, + ∞) is - 3, so a ≥ - 3. That is, the value range of number a is [- 3, + ∞)

How many square kilometers is 400 square decimeters

400 square decimeters = 4 square meters
4 square meters = (4 △ 1000000) square kilometers = 0.000004 square kilometers
So 400 square decimeter = 0.000004 square kilometer

99 times 99 plus 199 is 100 times 100. What do you find?

n*n+(n+100)=(n+1)*(n+1)

Given the curve C: y = x ^ 3, find the tangent equation of the point P with abscissa 1 on the curve C; whether there are other common points between the tangent and the curve C

F '(x) = 3x & sup2;, f' (1) = 3, P (1,1) on the curve C,
So the tangent equation of P is Y-1 = 3 (x-1), that is y = 3x-2
Simultaneous y = x & sup3;, y = 3x-2,
We obtain X & sup3; - 3x + 2 = x (X & sup2; - 1) - 2 (x-1) = (x-1) & sup2; (x + 2) = 0,
Solution
X = 1 or x = - 2,
So there is another intersection (- 2, - 8)