Everybody help me with the math. Math everybody helps

5 and 6 10 and 11 49 and 50 101 and 100 4 and 16 5 and 45 16 and 64 100 and 25 5 and 7

11 and 13 2 and 19 17 and 23

2.Write the greatest common factor of the numerator and the parent in the following fractions.

3.Write out two numbers as required so that their greatest common factor is 1.

1) Both numbers are composite: (22) and (25) 2) Both numbers are prime: (2) and (3) 3) A prime number and a composite number:

3) and (4) 4) an odd number and an even number: (1) and (2) 4Approximation.

1.Find the greatest common factor for each of the following groups.

5 and 6 10 and 11 49 and 50 101 and 100 4 and 16 5 and 45 16 and 64 100 and 25 5 and 7

11 and 13 2 and 19 17 and 23

2.Write the greatest common factor of the numerator and the parent in the following fractions.

3.Write out two numbers as required so that their greatest common factor is 1.

1) Both numbers are composite: (4) and (9) 2) Both numbers are prime: (2) and (7) 3) A prime number and a composite number:

2) and (9) 4) an odd number and an even number: (9) and (2) 4Approximation.

Question 1: 1, 1, 1, 1, 4, 5, 16, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Question 3: 4 and 9 3 and 5 9 and 16 1 and 2

Question 4: 1 5 5 9 7 5 3 4

Wrong! Set the travel time of the first car to t

then the travel time of car B is .

t+1-7+4 = t-2 < t

1 : [Car B departs one minute earlier than Car A].

7 : [but stayed in place B for seven minutes].

4 : [Finally, car B arrives at C four minutes later than car A].

Thus, car B travels less time than car A, and the speed should be greater than that of car A, but [the speed of car A is 80% of the speed of car A] contradictory!

Multiply x at the same time

Move items merge items of the same kind.

Same as divide by x=

Test: When x= the denominator is not 0, x= is the solution of the original equation.

-13 separate term or in denominator?

Calculate the right square:

mx 2-60x+25=(nx) 2-10nx+25, so the coefficients of the same power are equal.

m=n^260=10n

So n=6m=36

12.Suppose that when B runs the whole 60-meter race, C runs a total of x meters, and since the speed of the two remains the same, the distance covered in the whole course is also proportional:

That is, 50 : 40 = 60: x

Solution: x = 48 meters, 60 from the end point 48 = 12 meters Answer: When B crosses the finish line, C is 12 meters away from the end point.

13.It only takes x months to complete the annual production plan, so there are:

360 : 2 = 1800 : x

Solution: x = 10 months.

A: Actually, it only takes 10 months to complete the annual production plan.

14.Set the last month gas x cubic meters, so there are:

60 * x 60) * = 60 solution: x = 70 cubic meters.

A: Wang Fang's family used 70 cubic meters of gas last month.

15.If you buy big in shopping mall A and get small free, the discount is 10 = (80% off) 9% off in shopping mall B.

C shopping mall is full of 30 yuan and 20% off.

Therefore, it is recommended that A go to shopping mall B (cost yuan).

B go to C shopping mall (cost 40 yuan).

c Go to a shopping mall (cost 40 yuan).

d Go to a shopping mall (cost yuan).

15 questions on piecewise functions and mathematical modeling. It is more cumbersome, but according to the preferential adjustment of the three shopping malls and customer needs, the expression can be solved smoothly.

14 questions, a typical piecewise function model. List the analytical expressions of the piecewise functions, and substitute the meaning of the question to get the solution.

13 questions, proportional solution, the key is the impact of the current production speed on the original plan. 360 in 2 months, which is equal to 180 in a month, the original plan is 1800 for the whole year (the original monthly speed is 150), and the total amount remains the same under the new speed of production that is, the new time is obtained. Let the new time be x months, then there is 150 180 = x 12

12 questions, proportional questions, although not difficult, but it takes a little time. The key point is that if A's 60m running time (set to t) is used as a benchmark, then B runs 50m at time t, and C runs 40m. Let the speed of the three be v1, v2, v3, then:

t=60/v1=50/v2=40/v3

From this, v3:v2=4:5 v3=40 t v2=50 t

So with the same speed.

That is, when B reaches the end point, it is 8m ahead of C.

From this, it can be seen that when B reached the finish line, C ran 52m

The topics are in the first year of junior high school, and these questions are all types of mathematical modeling. Mathematical modeling in junior high school is very important, so I hope you pay attention to it during class.

Let's shoot it again, I can't see clearly.

Don't you have an answer behind us, we have it anyway.

Hello landlord] I don't know if you are in high school, have you learned permutations and combinations? I'd better use the popular solution. 1.

Needless to say, the probability of getting a qualified product for the first time is 80%, and the probability of the second P2 is 2 10 * 8 9 = 8 45, which is obviously P1 P22This is a matter of ratios, i.e., even if the unknown number is taken as any real number, it has no effect on the result.

Then: x x 1-( 2m 2mx)=x+(1-2m)x+(1+2m) 0, then x x 1 2m 2mx, ratio method. Thank you