How to solve 4x-60 = x,

How to solve 4x-60 = x,

4X-60=X
4X-X=60
3X=60
X=20

How to solve (360 + x) / 24 = (216 + x) / 16

(360+x)/24=(216+x)/16
Multiply both sides by 48
2(360+x)=3(216+x)
720+2x=648+3x
3x-2x=720-648
x=72
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Look at the diagram and solve the equation. Each box is 24 yuan. I bought x boxes and the total price is 360 yuan

24x=360
x=15

Given Tan (π / 4 + α) = - 1 / 2, find 4cos α (sin α + cos α) / (1-tan α)

Tan (π / 4 + α) = (1 + Tan α) / (1-tan α) = - (?) 189;, Tan α = - 3.4cos α (sin α + cos α) / (1-tan α) = 4cos α (sin α + cos α) / (1 + 3) = cos α (sin α + cos α) = (COS α, sin α + cos α) / (COS α, sin α + cos α) / (?) 178; α + Co

As shown in the figure, point P passes through the line where points B (0, - 2), C (4,0) are located, and the ordinate is - 1. Point q is on the function y = 3x image. If PQ is parallel to the Y axis, the coordinates of point q are obtained

Let the linear equation where BC is located be y = KX + B, B = - 2K = 12, y = 12x-2, where ∵ P is on BC, and ∵ 1 = 12x-2, where ∵ P (2, - 1), and ∵ PQ ∥ Y axis, where point q is on y = 3x, the abscissa of ∵ point q is x = 2, y = 32 = 1.5, q (2, 1.5)

It is known that the minimum positive period of the function f (x) = sin (2wx - π / 6) + 1 (W belongs to R, X belongs to R) is π, and the image is symmetric about x = π / 6 to find the analytic expression of F (x)
Find the analytic formula of F (x)
Minimum positive period T = 2 π / 2|w| = π w = ± 1
If the image is symmetric with respect to x = π / 6, then sin (2W * π / 6 - π / 6) = 1 or - 1
So w = - 1
About this step, I would like to ask how to see that w is equal to - 1 and not equal to 1

Sin (2W * π / 6 - π / 6) = 1, the solution is 2W * π / 6 - π / 6 = π / 2, 2W = 4, w = 2, rounding off; sin (2W * π / 6 - π / 6) = - 1, 2W * π / 6 - π / 6 = - π / 2, w = - 1. Or when w = 1, 2W * π / 6 - π / 6 = π / 6, sin (2W * π / 6 - π / 6) = 1 / 2

If a line passes through a point (- 3, - 2) and its intercept on two coordinate axes is equal, the equation of the line is______ ．

When the line passes through the origin, the slope k = - 2 − 0 − 3 − 0 & nbsp; = 23, so the equation of the line is y = 23x, that is, 2x-3y = 0. When the line does not pass through the origin, let the equation of the line be x + y + M = 0, substitute (- 3, - 2) into the equation of the line to get m = 5, so the equation of the line is x + y + 5 = 0

Simple method of 9 × 41 + 59 × 39

9×41+59×39
=9×（100-59）+59×39
=900-9×59+59×39
=900+（39-9）×59
=900+30×（60-1）
=900+1800-30
=2700-30
=2670

The line y = 2x + 8 intersects the x-axis, the y-axis is at a, B, the line L crosses the origin, the line AB intersects the point C, and the triangle AOB has an area of 1; 3

Y = 2x + 8 intersection x-axis, Y-axis at a (- 4,0), B (0,8) ∧ triangle AOB area is: 4 × 8 × 1 / 2 = 1616 × 1 / 4 = 416 × 3 / 4 = 12. Let the analytic formula of line L: y = KX case 1: 8 × (- x) × 1 / 2 = 4 ① 4 × KX × 1 / 2 = 12 ② solution: k = - 6. Therefore, the analytic formula of line L is y = - 6x case 2: 8 × (- x) ∧

Given that the function f (x) = 3x2-2x, the sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn) (n ∈ n *) are all on the graph of function f (x). (1) prove that {an} is an arithmetic sequence; (2) let BN = 3an · an + 1, tn be the sum of the first n terms of the sequence {BN}, and find the minimum positive integer m that TN < M20 stands for all n ∈ n *

It is proved that: (1) when Sn = 3n2-2n, when n ≥ 2, an = sn-sn-1 = 3n2-2n - [3 (n-1) 2-2 (n-1)] = 6n-5, when n = 1, A1 = S1 = 1, conform to the above formula, so an = 6n-5, then the sequence {an} takes 6 as the tolerance and 1 as the first term of the arithmetic sequence; (2) when (1), an = 6n-5, then BN = 3an · an + 1 = 3 (6N − 5) (6N + 1) = 12 (16N − 5 − 16N + 1), then TN = 12 [(1-17) + (17-113)+ … +(16N − 5 − 16N + 1)] = 12 (1-16n + 1) because n ∈ n *, so 16N + 1 > 0, that is, TN = 12 (1-16n + 1) < 12, and TN < M20 holds for all n ∈ n *, so M20 ≥ 12, then m ≥ 10, so the minimum positive integer m satisfying the condition is: 10