What to do with this problem, thank you math students 10

First pass the score, the negative sign in parentheses is negative negative to positive, positive and negative to negative, and then the integers are added, the fractions are added, and the full 24 is converted to the integer remainder is expressed by n 24.

8.5 7< 5 6, so turn on the absolute symbol =5 6-5 7. Open the brackets again, and the whole question is equivalent to 5 6-5 7-5 6-2 7=-1

10.Replace decimals with fractions, brackets in the whole question, negative and negative to positive, which is equivalent to 1 8 + 3 and 1 4-3 and 1 8 + 11 and 2 3-1 4. The same denominator is counted together. It's simple, you'll do the math yourself.

Well, 10 questions are too simple for me to do the math, in fact, the fractions for the denominator are exactly canceled out, and the final answer = 11 and 2 3

In the end, the friend said that the method is the most conventional, but I don't recommend this method, in fact, the operation of [with fractions] is very simple, you must not honestly turn it into [fake fractions], the simple and unmistakable method is to divide [with fractions] into [integers] + [true fractions], such as the problem of [3 and 1 4] is actually = 3 + 1 4 ; In the same way, [3 and 1 8] = 3+1 8

Having said so much, not for the score, I just hope that the landlord will not learn mathematics so deadly. (because I love math).

Two hundred and fifteen out of twelve.

Analysis: First of all, let Xiaoqian's money be x yuan, Xiaoling's money is y yuan, x, y are non-negative integers, and the equation system can be obtained according to the meaning of the question

x+2=n(y-2)y+n=2(x-n), eliminate y, and 2n=1+ can be obtained

152y-7, which is solved by analyzing n as a positive integer, and the answer can be obtained. Answer: Solution: Let all the money of Xiaoqian be x yuan, and the number of money owned by Xiaoling be y yuan, and x, y are non-negative integers.

From the question setting, we get: x+2=n(y-2)y+n=2(x-n), and subtract x to obtain: (2y-7)n=y+4, that is: 2n=(2y-7)+152y-7=1+

152y-7.

152y-7 is a positive integer, 2y-7 has values of 1, 3, 5, 15, and y can only be 4, 5, 6, 11

The values of n are 8, 3, 2, 1; The values of x are 14, 7, 6, 7

That is, there are 4 possible values of n.

Therefore, D. was chosen

f(x)=√3sin2x-cos2x=2[sin2x·(√3/2)-cos2x·(1/2)]

2sin[2x-(π/6)]

Translate 3 units to the right to get g(x)=2sin[2(x- 3)-(6)].

2sin[2x-(2π/3)-(/6)]=2sin[2x-(5π/6)]

(4) a(0,4) b(-2,-1) c(3,0)ab linear equation: (y-4) (x-0)=(-1-4) (-2-0)2y-8=3x

3x-2y+8=0

The distance from c to the straight line ab is the height on the edge of ab:

d=|3×3-2×0+8|/√[3²+(2)²]=17/√13

It was =3 Chongzen code 2*5 7+5 7*5 2-1 2*5 7

3 2+5 2-1 2)*5 Sanna 7

7 8*5 hits high 7