The price for a newspaper booth to buy a daily newspaper from a newspaper office is 0.30 yuan per copy, and the selling price is 0.50 yuan per copy. Newspapers that can't be sold can be returned to the newspaper office at the price of 0.10 yuan per copy. Experience shows that in a month (30 days), only 150 newspapers can be sold in 20 days, and 200 newspapers can be sold in the other 10 days. Suppose that the number of newspapers bought from the newspaper office every day must be the same, then this problem is solved How many newspapers do kiosks buy every day to maximize their monthly profits? What is the maximum profit?

The price for a newspaper booth to buy a daily newspaper from a newspaper office is 0.30 yuan per copy, and the selling price is 0.50 yuan per copy. Newspapers that can't be sold can be returned to the newspaper office at the price of 0.10 yuan per copy. Experience shows that in a month (30 days), only 150 newspapers can be sold in 20 days, and 200 newspapers can be sold in the other 10 days. Suppose that the number of newspapers bought from the newspaper office every day must be the same, then this problem is solved How many newspapers do kiosks buy every day to maximize their monthly profits? What is the maximum profit?

Let the kiosk buy x newspapers from the newspaper office every day, and the monthly profit is y (yuan). (1 point) according to the title, y = 20 × (0.50-0.30) × 150 + 20 × (0.10-0.30) (X-150)) + 10 × (0.50-0.30) X. (150 ≤ x ≤ 200) (4 points), that is, y = - 2x + 1200. (150 ≤ x ≤ 200)

Given the sequence {an}, the first n terms and Sn satisfy s (n + 1) = 2 & micro; Sn + 1, (& micro; is a constant greater than 0),
(1) Find the value of & micro
(2) The general term formula of the sequence {an}
(3) Let the sum of the first n terms of the sequence {n × an} be TN, and try to compare the size of (TN) / 2 and Sn

(1) When n = 1, S2 = 2 & micro; * S1 + 1 = 2 & micro; * a1 + 1, S2 = 2 & micro; + 1; when n = 2, S3 = 2 & micro; * S2 + 1, then S2 + a3 = 2 & micro; * S2 + 12 & micro; + 1 + 4 = 2 & micro; * (2 & micro; + 1) + 1 is solved to & micro; = 1 (& micro; = - 1 rounding off) (2) s (n + 1) = 2Sn + 1 ∧ s (n + 1)

Given a, B ∈ positive real number and a + B = 2, then the value range of a ^ 2 + B ^ 2 + AB is?

a^2+b^2≥2ab
(a+b)^2 = a^2+b^2+2ab ≥4ab
∴ 0 < ab ≤ 1
a^2+b^2+ab = (a+b)^2 - ab = 4 -ab
Value range [3,4]

If 64 × 8 & # 179; = 2, find the value of X

64=2^6
8=2^3
8^3=2^(3*3)
64*8^3=2^(6+3*3)=2^15=2^x
x=15

1. (1 + x) & #179; - 27 / 8 = 0 the solution process is 2. If x = √ 3 + 1, then X=_________
3. If a and √ 2 are opposite to each other, then a = (); 2 - √ 3 = () help!

1. (1 + x) & #179; - 27 / 8 = 0 the main process of solving
(1+x)³=27/8
1+x=3/2
x=1/2
2. If x = √ 3 + 1, then x=____ √3+1__ Or - √ 3-1___
3. If a and √ 2 are opposite to each other, then a = (- √ 2); l √ 2 - √ 3 = (√ 3 - √ 2)

(X & # 178; + Y & # 178; - 1) &# 179; = x & # 178; Y & # 179;, find how much y equals when x equals

First, open both sides to the third power, X & # 178; + Y & # 178; - 1 = YX ^ 2 / 3Y & # 178; - YX ^ 2 / 3 = 1-x & # 178; (y-0.5x ^ 2 / 3) ^ 2-0.25x ^ 4 / 3 = 1-x & # 178; y-0.5x ^ 2 / 3 = under the positive and negative root sign (1-x & # 178; + 0.25x ^ 4 / 3) y = under the positive and negative root sign (1-x & # 178; + 0.25x ^ 4 / 3) + 0.5x ^ 2 / 3

Find the value of the following formula X: 1 / 4 (2x + 3) &# 179; = 2 * 5 & # 178; (5x-0.1) &# 179; + 0.008 = 0 PS. &# 178; is quadratic. &# 179; is cubic

1.1/4 (2x + 3) cube = 2 * 5 cube
Cube of (2x + 3) = 8 * 5
2x+3=2*5
2x+3=10
2x=7
x=3.5
2. (5x-0.1) cubic + 0.008 = 0
Cube of (5x-0.1) = 0.008
5x-0.1=0.2
5x=0.3
x=0.06

Solve the equation M & # 178; X & # 178; - 28 = 3mx (m ≠ 0)
With the formula method solution!

Solve the equation M & # 178; X & # 178; - 28 = 3mx (m ≠ 0) M & # 178; X & # 178; - 3mx = 28 (MX) & # 178; - 3mx + (3 / 2) & # 178; = 28 + (3 / 2) & # 178; (MX-3 / 2) & # 178; = 121 / 4mx-3 / 2 = ± 11 / 2mx = 3 / 2 ± 11 / 2mx = - 4 or MX = 7, so x = - 4 / m or x = 7 / m, if you don't understand, have a good study

It is known that a root of the equation x2 + 3mx + M2 = 0 about X is x = 1, then M = 1___ ．

Substituting x = 1 into the equation, we can get M2 + 3M + 1 = 0, and the solution is m = - 3 ± 52

If the quadratic equation of one variable (M & # 178; - 1) x & # 178; + 3mx + 1 = 0, there must be a solution. Please give me the process, thank you
If (M & # 178; - 1) x & # 178; + 3mx + 1 = 0, there must be a solution
Please give me the process, thank you

∵ is a quadratic equation of one variable
∴m²-1≠0
m≠±1
△=(3m)²-4(m²-1)
=9m²-4m²+4
=5m²+4>0
The discriminant is always greater than 0, so there must be two solutions