How to solve the equation of 0.6 (x + 5) = 48

How to solve the equation of 0.6 (x + 5) = 48

x+5=80 x=75

60:48 = (6-x): X in this equation, what is the solution?

Firstly, 60:48 is divided into 5:4;
Then multiply the cross: 4 * (6-x) = 5x
Then: 24-4x = 5x
Then: 9x = 24
Then: x = 8 / 3

If Tan α = 1 / 3, then sin ^ 2 α - sin α cos α - cos ^ 2 α=

sin^2α-sinαcosα-cos^2α
=(sin^2α-sinαcosα-cos^2α)/(sin²a+cos²a)
=The denominator of (Tan & # 178; a-tana-1) / (Tan & # 178; a + 1) is also divided by cos & # 178; a
=(1/9-1/3-1)/(1/9+1)
=-11/10

If the distance from a point P on the bisector of angle AOB to OA is equal to 5com and Q is any point of ray ob, the relationship between the length of PQ and 5cm is obtained___________

PQ>=5cm
Theorem used:
The distance from the point on the bisector to both sides of the corner is equal

It is known that the minimum positive period of the function f (x) = sinwxcoswx + sin ^ 2wx is Wu,
(1) Find the value of F (Wu / 4), (2) find the monotone increasing interval of function f (x), (3) if x belongs to [0, Wu / 2], find the maximum value of F (x) and the corresponding x value

According to the double angle formula: Sin & # 178; a = (1-cos2a) / 22sinacosa = sin2af (x) = (sin2wx) / 2 + (1-cos2a) / 2 = (1 / 2) (sin2wx-cos2wx) + 1 / 2 = (√ 2 / 2) [(√ 2 / 2) sin2wx - (√ 2 / 2) cos2wx] + 1 / 2 = (√ 2 / 2) [cos (- - π / 4) sin2wx + sin (- - π / 4) cos2wx] + 1 / 2 =

If the intercept of the line L on the X axis is - 5 and the intercept on the Y axis is 3, the equation of L is obtained

Straight line passing (- 5,0) (0,3)
Slope k = (3-0) / (0 + 5) = 3 / 5
So y = (3 / 5) x + 3
That is: 3x-5y + 15 = 0

Use a simple method to calculate the following questions. (1) 2.9 × 0.45 + 0.29 × 4.2 + 0.029 × 13 (2) 3.84 × 9.6 + 0.96 × 61.6

（1）2.9×0.45+0.29×4.2+0.029×13，=2.9×0.45+2.9×0.42+2.9×0.13，=2.9×（0.45+0.42+0.13），=2.9×1，=2.9；（2）3.84×9.6+0.96×61.6，=3.84×9.6+9.6×6.16，=（3.84+6.16）×9.6，=10×9.6，=96．...

The line L passing through point (1,2) intersects with the positive half axis of X axis and the positive half axis of Y axis respectively with two points a and B. o is the origin of the coordinate. When the area of △ ABO is the smallest, the line L is obtained

Let Y-2 = K (x-1), x = 0, y = 2-k; y = 0, x = 1-2 / K
So the area s = 1 / 2 * | 2-k | * | 1-2 / K | = 1 / 2 * | (K + 4 / k) + 4 | > = 4 if and only if k = - 2 equals
So the L equation is y = - 2x + 4

Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n ∈ n +) are all on the graph of function y = 3x-2. (1) find the general formula of the sequence {an}. (2) let BN = 3 / ana (n + 1), tn be the sum of the first n terms of the sequence {BN}, and find the formula such that TN

(1) When Sn / N = 3n-2sn = 3N ^ 2-2nn = 1, A1 = S1 = 1n ≥ 1, an = SN-S (n-1) = 6n-5n = 1, A1 = 1, it is established that an = 6n-5 (2) BN = 3 / [(6n-5) (6N + 1)] = (1 / 2) [1 / (6n-5) - (6N + 1)] TN = (1 / 2) [1-1 / 7 + 1 / 7-1 / 13 +... - 1 / (6N + 1)] = 1 / 2 × [1-1 / (6N + 1)] = 1 / 2-1 / (12n + 2) ≤ 1 / 2 -

Set a = {x | x2-x-2 ≤ 0}, B = {x | 2x ≤ 1, then a ∩ (∁ RB) =___ ．

From the inequality in set a, we can get: (X-2) (x + 1) ≤ 0, the solution is: - 1 ≤ x ≤ 2, i.e. a = [- 1, 2]; from the inequality in set B, we can get: 2x ≤ 1 = 20, the solution is: x ≤ 0, i.e. B = (- ∞, 0], ∩ (∁ RB) = (0, + ∞), then a ∩ (∁ RB) = (0, 2]. So the answer is: (0, 2]