How do you solve a math problem like this?

In this way, if there are 12 scales on the clock, then the corresponding central angle between the two adjacent scales is 30°, the minute hand goes around 360°, the hour hand travels a scale 30°, and the angular velocity of the minute hand is 12 times that of the hour hand. For example, at 3 o'clock, the minute hand is at 12, and the hour hand is at 3; And at 3:45, relative to 3 o'clock, the minute hand has gone 9 scales, 270°, then the hour hand has gone 270 12=

So given a moment, it is possible to determine the position of the minute hand.

Similarly, at 9 o'clock, the minute hand is at 12 and the hour hand is at 9, and the difference between them is 90°. At 9:15, relative to 9 o'clock, the minute hand has gone 3 more scales clockwise, 90°, then the hour hand has gone 90 12=, so the minimum angle between the hour hand and the minute hand is 90°+90°

The clock is seen as a circle.

The circle corresponds to a degree of 360°

The corresponding degrees of the minute hand scale are 360 60 = 6°

The hour scale corresponds to 360 12 = 30°

Finding the angle between the hour and minute hands is usually the main idea.

For example, the clock is 1:20 to find the minimum angle between the hour and minute hands.

At this time, the hour hand is between 1 o'clock and 2 o'clock, and at the same time, you have to think of the time hand scale as 60 parts between 1 o'clock and 2 o'clock, because every minute hand goes 1 and 2 minutes, the hour hand goes between 1 and 2 o'clock by 6 * 5 = 30 degrees

Furthermore. 9:15 could not be the minimum angle. Because it's a flat angle = = add to it; For example, n points m minutes.

Then the angle of the hour and minute hands is an absolute value of 30(n+m 60)-6m.

I don't understand, I am satisfied.

At 9:15, the minimum angle between the hour and minute hands is because at 9:15 the hour and minute hands are separated by about three 15 minutes, and the minimum angle at every 15 minutes is 30°, and the three are 90°.

But it should be 9:15, so the minimum angle is less than 90 °, 15 minutes accounts for 1 12 of an hour, and the hour hand only moves one grid per hour, that is, 30 °, then 9 o'clock is still 30 * 1 12 = from 10 o'clock, then the minimum angle between the hour hand and the minute hand is (90 °

Do it yourself! Hope!

Isn't 9:15 a flat angle? . .

Let this number be t, move 3 bits to the left, move 2 bits from right to right, and shrink it by 10 times, so there is t-t 10=, the solution is t=5, and the bottle weight a is set, and the original water weight is b. Then there is a+2b=5, a+4b=9, and the solution is a=1kg, b=2kg. So it turned out that the bottle and water weighed a total of 3kg.

(1) Set: the original digit x, then the decimal point is moved to the left by 3 digits, which is 1000x

Move 2 more bits to the right and you'll be 10x

Smaller than the original number is x-10x=

Then x has to be solved:

2) Set: the original water weight x; The bottle weighs y.

Then it's 2x+y=5;4x+y=9

Then it's OK to solve this 2-element 1st order equation.

The solution is x=2;y=1.

Therefore, the original bottle with water weighed 3 kilograms.

a²-2a+1=(a-1)²

a²-1=(a+1)(a-1)

The first term is (a-1) after (a-1) (a+1) (a-1) = (a-1) (a+1) in parentheses in the second term

a-1-(a-1)/(a+1)]=（a²-1）/（a+1）-（a-1）/（a+1）=（a²-a）/（a+1）②①=（a²-1）/（a+1）=（a+1）（a-1）/（a+1）=a-1

Substituting: a-1=2 2-1

3. Set the total work to be 150 volumes, A completes 15 per day, B 10, A completes 15 * after cooperation, B completes 9, and A completes x days alone, B alone Y days, and cooperation 8-x-y

15x+10y+（12+9）（8-x-y）=150

6x+11y=18, xy is a natural number greater than 0, so x=3, y=0, cooperate for 5 days.

of workers will do work A, and 80% of workers will do work B, so 75-(100-80)=55% will work both A and B

90% of workers will do type C work, 70% of workers will do type D work, and 90-(100-70)=60% of workers will work cd

55-(100-60)=15% of ABCD metropolis

5. Same as the first question: 379-5=374=2*11*17,245-3=242=2*11*11

It may be 11 or 22 people, and 11 people are not satisfied with the topic, because there are more than 30 apples in 34 points.

So it's 22 people.

-2=57=3*19

Therefore, the average score of 57 books must be 3 or 19 people, and 36 pens can be averaged by people.

A maximum of 3 students will be selected.

2. Either a multiple of 2 is a singular number, that is, ......If it is not a multiple of 5, we will eliminate the number of a class, that is, the sum of 1+3+7+9=20, 11+13+17+19=60, 21+23+27+29=100.........191+193+197+199=780

To remove the denominator, we have a formula (a+b)(a-b)=a2-b2

Multiply the numerator and denominator by (n+1)-n to get [ (n+1)- n] (n+1-n).

Don't set f(x)=ax2+bx+c to solve the first problem, and then substitute it to solve, if you do this, you will find that the amount of computation is relatively large, because , is a complex number. In fact, this question examines the Vedic theorem.

Since the two roots of the equation x2-x+1=0 are , so + =1....1）

Let f(x)=x2-x+1+mx+n, then.

f(α)=mα+n=β..2)

f(β)=mβ+n=α..3）

f(1)=1+m+n=1...4)

Use (2)-(3) to get m=-1

And (2) becomes n- =, i.e. n= + =1

Equation (4) is obviously satisfied, so f(x)=x2-x+1-x+1=x2-2x+2

Question 2: f(1)=a-c, f(2)=4a-c, f(3)=9a-c

Let f(3)=mf(1)+nf(2), then 9a-c=m(a-c)+n(4a-c).

(m+4n)a-(m+n)c=9a-c

So m+4n=9....1)

m+n=1...2)

The solution gives n=8 3 and m=-5 3

i.e. f(3)=1 3[-5f(1)+8f(2)].3)

There are -4 f(1) -1, -1 f(2) 5, i.e. 5 -5f(1) 20, -8 8f(2) 40

So -3 -5f(1)+8f(2) 60

i.e. -1 f(3) 20

In fact, the cultivation of this kind of equation thinking is not very difficult, because equation thinking is one of the important mathematical ideas in high school. Whenever you are thinking about a problem, you have to find a way to relate the unknown quantity to the known quantity, and the idea of equations works a lot of the time, you can try to use the idea of equations to find their relationship. The so-called equation thinking is a kind of ideology, a way of thinking when you do problems, you often use this way of thinking to do problems, and slowly this kind of equation thinking can be cultivated.

If you still have any questions, you can add me.

In the first problem, first solve the solution of the equation, i.e., ,

Then set the required quadratic function f(x)=ax2+bx+c to find f( )= ,f( )= ,f(1)=1 into the function to obtain a ternary system of equations about a, b, c, solve the value of abc can be calculated by my personal pen f(x)=x2-2x+2, you can calculate it yourself.

1. Let f(x)=ax2+bx+c

Bring in f( )= , f( )=

Then according to the relationship between the two roots of x2-x+1=0 , you can find 2, and f(1) and f(2) are used to represent f(3).

？=1-4<0 how can there be roots.

2。Let f(3) = xf(1) + yf(2).

f(3)=x(a-c)+y(4a-c)=9a-cx+4y=9

x=-5 3 y=8 3 for x+y=1

5 3 * (-4) + (8 3) * (1) 4 of 4

1.Do you also do last year's final exam papers? I don't know how to answer that question at all. 2Solution: x21+x22=23

x21+x22+2x1x2-2x1x2=23x1+x2）2-2x1x2=23

x1+x2=m x1x2=2m-1

m2-2（2m-1）=23

The solution is m1=7 m2=-3