There are three math problems in the first year of junior high school, and the process is written wi

1.Solution: Let's say they played for x minutes.

From the title: 1 - 3 5 + (1 50 + 1 30) = 1

Solve the equation to obtain: x=

Because < 10, they were able to finish in the required time.

2.Solution: Let the first number be x.

From the inscription: (x+4)+(x-4)+(x*4)+(x4)]=100

Solve the equation to obtain: x=16

From this, the first number is 16+4=20, the second number is 16-4=12, the third number is 16*4=64, and the fourth number is 16 4=4

3.(1) Scheme 1: 3000x2 + 30000 = 36000 (yuan) (2) yuan).

Plan 2: 14x3000=42000 (yuan) Plan 1: 1500x2+30000=33000 (yuan).

36,000 yuan is less than 42,000 yuan Plan 2: 1500x14=21000 (yuan).

So choose option one. 21,000 yuan is less than 33,000 yuan.

So choose option two.

I don't know if it's right or wrong

1-3/5 = 2/5

30*2/5 = 12

A: No, you cannot. 2 questions and 3 questions will not.

Question 1: (1-1 50 * 30) = 2 5

2 5 divided by (1 50 + 1 30) = 30 4 minutes.

30 4+30 = minutes.

Isn't there one?

Assuming that the open number is a, if sqrt(a) is used to represent the root number a, then the root of [sqrt(x)-sqrt(a x)] 2=0 is sqrt(a).

The deformation yields sqrt(a)=(x+a x) 2, so you just need to set an initial value that is approximately equal to (x+a x) 2, substitute it into the above formula, you can get a more approximate value, and then substitute it to get a more accurate value ......In this way, a value of (x+a x) 2 with sufficient accuracy is obtained.

For example, calculate sqrt(5).

Let the initial value be 21)sqrt(5)=(2+5 2)2=

2）sqrt(5)=(

3）sqrt(5)=(

The results obtained in these three steps are less than different from sqrt(5).

1. (1) Proof: Because angle 1 = angle bac, so ab||EF (Wrong Angle) because ab||cd,ab||ef, so cd||ef2) because ef||cd, so angle 1 = angle acd (isotope angle), angle 1 = angle 3 + angle caf (complementary angle theorem), so angle acd = 45 degrees.

Because angle fab = angle cab - angle caf, angle cab = angle 1, so angle fab = 45-15 = 30, angle b = 180 - angle 2 - angle fab = 105

The rest will be answered later, and it will be troublesome to enter.

I wrote down the handwritten and made it into ** and sent it to the space, and you can check it yourself.

The idea of solving the problem is:

The condition is known to be quadratic, and the answer is required to be quadratic, so the primary is to be converted into a quadratic. Then square (m+1 m).

The solution is as follows: m+1 m=3 (known condition).

m+1 m) 2=9 (both sides of the equation squared at the same time) m 2+2+1 m 2=9 (left side of the equation).

m 2 + 1 m 2 = 7 (move 2 to the right of the equation, i.e. 9-2 = 7).

The ratio of the number of plants of the two flower seedlings of B is 1:2, and the ratio of the number of plants of the two flower seedlings of B and C is 3:4, A:B:C=3:6:8

3a×3+6a×2+8a×1=29000，a= 300，b 600，c 800

2.There are two solutions.

AOC is set to x, 1) c is between ab, and 2) c is the same outside ab.

1.Original = (-49 50) * (48 49) * (47 48) *1/2)=-1/50

1 2 to 1 50 49 items in total, so it is negative.

2.Original = 5 4 * 4 5 - 4 5) * (9 4) +1 4) * 4 5

11...1)*(99...9)=(11..1)0(88..8) 9 top indication.

n 1*n 9 = (n-1) 1 1 0 (n-1) 8 1 9

For example, 111111*999999=111110888889

==(-49/50)*(48/49)*(47/48)*.1/2)=-1/50

2.Original = [(5 4)*(5 4)]-4 5)*(9 4)]+1 4)*(4 5)]

1+9 5-1 5=2 and 3 5

Reasoning questions 11*99 1089

11...1 (n 1) * 99....9 (n 9) (n 1 1) 0 (n 1 8) 9

1: The title is not clear.

2: Got 81-25

Regularity: 1 (n-1 1) 0 (n-1 8) 9

The first question agrees with Sunshine Xiaoqiang's answer.

The third question is as follows.

x-a 0 is x a

3-2x -1 i.e. x 2

If there is no solution to the inequality of a2, otherwise the solution of the group of inequalities is a x 2, according to this equation the solution of 5 integers can be 1, 0, -1, -2, -3, so 4 a -3If you don't count the 0 solution, the integer solution is 1, -1, -2, -3, -4, so the range of a is 5 a -4

I don't know if my answer is right or not, but your teacher's -4 a -5 must be wrong, and there can't be a number that can be both greater than 4 and less than 5

Question 2 (4x+a) 3 1

Then 4x+a 3 then 4x 3 a, i.e., x (3 a) 4- (2x+1) 3 0, then 2x+1 0, i.e., x The street of the preceding tense is the solution of the latter tense, i.e., (3 a) 4, then 3 a, 2, i.e., a 5, is a 5

∵∠1=∠2，∠3=∠4 bm=bm

Triangles are colateral, corresponding to equal angles].

cbm dbm[ gets:cm=dm

cm=dmm on ab, [m and ab in a straight line], 3= 4180°- 3=180°- 4

cma=∠dma

am=am, the two sides and their angles correspond to equal triangle congruence].

cma≌△dma【

ac=ad

One-year profit: 4 20 (1-10%) million yuan, one-year interest: 40 15% = 60,000 yuan, two-year profit: 26 2 = 520,000 yuan.

Principal and interest for two years: 40 + 6 2 = 520,000 yuan It takes two years for the first question to be less conditional.

Is it expressed algebraically: total students: 2x+3y-2 people planting trees x+4y x+8y+2 more than 2 classes.

Classmates, you must have less conditions in the first question, but the second class must plant more trees than the first class, and 4x+2 more trees are set to n years.

40 (1 + 15n 100) = n * (4 * (1-10%) n = 2 two years.