Four math problems in the second year of junior high school can be added soon .

1. There are a total of 30 questions in the mathematics competition, 2 points will be added for a correct answer, 1 point will be deducted for a wrong answer, no points will be scored if you do not answer, a student took 49 points after the exam, and the number of unanswered questions is the base, how many questions are wrong?

Answer X question incorrectly, and fail to answer Y question.

Then 30*2 - 3x - 2y = 49

i.e. 3x + 2y = 11

Because y is an odd number, it is easy to get y = 1 and x = 3

So answer 3 questions incorrectly.

2. There are 23 stories in a storybook, and the number of pages of each story is 1 page, 2 pages, 3 pages, ......23 pages. How many pages are even-numbered on the first page of each story in this book?

The first story p. 1.

Second story pages 1 + 1.

The third story on page 1 + 1 + 2.

The nth story page 1 + 1 + 2 + n - 1), i.e. page 1 + n * (n-1) 2.

To be an even number, n*(n-1) 2 is an odd number.

That is, n*(n-1) is 2*odd.

Because n and n-1 must be odd and even.

Therefore, it is required that the even number must be in the form of "2*odd".

It is easy to launch, and within 23 of them, there are 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, a total of 12.

i.e. the 2nd story, the 3rd story, the 6th story, the 7th story...

3. The five lights A, B, C, D, and E were all on at the beginning, and one person turned on the animal light switch in turn, and asked him how many lights were still on after pressing 2003 times?

2003 5 = 400 surplus 3

The lights that have been pressed 400 times are on, and 3 lights are turned off, so only 2 lights are on.

4. A string of numbers is arranged in a row, the first two numbers are 1, each from the third number is the sum of the first two, how many even numbers are there until 500 numbers?

The number is: odd, odd, even, odd, odd, even, odd, odd, odd, even...

500 3 = 166 remainder 2

So the number of even numbers up to 500 is 166.

1) 49 points must be scored with 25 or more correct answers. Assuming that the answer is 25 correct and the answer is 1 incorrect, then the answer 4 is not answered, which does not meet the meaning of the question. 26 correct answers, 3 incorrect answers, and 1 unansweredIn line with the topic.

2) The first few stories have a total number of pages and the number of first pages.

As you can see from the figure, there are 12 even numbers on the first page. It's hard to explain the rules. Except for the first story, from the second story onwards, the number of even pages and odd pages are 2 2 intervals.

3) After pressing 16 times, the lights are all turned off, that is, after pressing 2003 times (16 times out of 125 times and 3 times left), 3 lights are on.

4) 1 = 1 odd.

1 = 1 odd.

1+1 =2 even.

2+1 =3 odd.

3+2 =5 odd.

5+3 =8 even.

8+5 =13 odd.

13+8 =21 odd.

21+13 =34 even.

34+21 =55 odd.

55+34 =89 odd.

89+55 =144 even.

Taking 3 odd and even numbers as a group, there are 166 groups of 500 numbers and 2 odd numbers. That is, there are 166 even numbers.

These 5 points are really rare, and the college entrance examination is not so difficult!

1.26 correct, 3 wrong, 1 none, 2, 11.

There are also two lights that are on.

Liberation Brigade: 1. Set up the need to add water x kg to He quietly, then.

20*16%=（20+x)*10%

Solution. x=12

2. Set A and B to make x and y parts of the Zen cover every hour.

60/x=90/y

x+y=45

Solution. x=18, y=27 thanks.

I don't understand the process.

Method 1: Connect to the AP

ab=ac, e, f are the midpoints of ac and ab, respectively, ae=af

pe⊥ac,pf⊥ab

aep=∠afp=90°

ap=ap，ae=af,∠aep=∠afp∴△aep≌△afp

PE=PF method 2 connects EF, AB=AC, E, and F are the midpoints of AC and AB, respectively, AF=AE

AEF is an isosceles triangle with AFE= AEF PE AC and PF AB

afb=90°=∠afe+∠pfe,∠aec=90°=∠aef+∠pef

pef=∠pfe

PEF is an isosceles triangle.

pe=pf

Attestation: Connect to the AP

ab=ac, pe, and pf are the perpendicular bisectors of ab and ac, respectively, ae=af

aep=∠afp=90°，ap=ap

aep≌△afp

pe=pf

This problem is too simple, connect AP, use HL to prove that the triangle APE and the triangle APF are congruent, if you have to use "the distance from the point on the perpendicular bisector of the line segment to the two breakpoints of the line segment is equal", then connect PB and PC, then the application theorem has PB PA PC, prove that the triangle APB and the triangle APC congruence (SSS), and then obtain PE PF by the congruent triangle corresponding to the high equality.

The first one won't ...

The second OQ is variable, and its range is either non-negative (when your drawing is inaccurate) or positive (when your drawing is accurate).

The first question is simple, ignore.

2) Solution: The coordinates of the point q remain unchanged

Proof: Let the analytic formula of the straight line mb be y=nx-4(n≠0) point m(m+4,-m-8).

On the straight line mb, -m-8=n(m+4)-4 is sorted to get (m+4)n=-m-4

m＞0，∴m+4≠0．

The solution is n=-1

The analytic formula for the straight line mb is y=-x-4

1)x=2,y=1,z=1,3x-y+z=6

2)a^2+ab+b^2=(a+b)^2-ab=36-(9-6)=33

3)4a-b+11=0;1/3b-4a-3=0;From this, we can see that a=1 4, b=12Let's count the ridge bridge yourself.

4) Know 1>a>0, 1-2a+a 2 a-1-root number(a 2-2a+1) a 2-a=(1-a) 2 (a-1+1 a)=, and then count it yourself.

Summary: The four questions are all a type of year answer, that is, the polynomial in the quadratic root and the absolute value symbol, and the value after opening the symbol is asked by the wild wisdom question.

I don't know. Is the root number x-2y the root number (x-2y) or the root number x-2y?

1.Solution: According to the problem: y=2x+1=kx+b y=-x-8=kx+b and y=2x+1=kx+b to get x 2

y=-x-8=kx+b gives y= 7

i.e.: (2,5)( 1,—7).

y=4x--3

2.Solution: The image needs to be in the first quadrant.

x＞02x+3＞0

Solution: 0 x 2 3

1.Since it is an intersection point, then the abscissa and vertical coordinates are all equal! For the previous known condition, taking x=2 into the two lines yields 2k+b=5;The next two lines intersect, bring y=-7 into the two lines, and get x=-1, so k-b=7,.

Solve these two binaries about k and b again, and it's good to get in touch!

2. Draw an x, y coordinate system, when x=0, then y=3, mark the coordinates; When y=0, then x=, the coordinates are marked. Then the straight line connecting the two points is this point, and the value between x in the first quadrant is the range: x belongs to (0,.

1.When k is taken as a value, there is a solution to the fractional equation 6 x-1=x+k x(x-1)-3 x?

2.If there is a positive integer solution for equation 1 x-1=2 x-a, find the value of a.

3.The distance between A and B is 48 km, and the time taken by a steamer to travel from A to B is equal to the time taken by the steamer to complete half of the distance between A and B in the reverse flow, and the speed of the current is known to be 4 km h, so find the speed of the ship in still water.

4.^2÷(x=y)*^3

1.The fractional equation 1 (x-2) + k (x+2)=4 (x 2-4) with the root x=-2 has the root x=-2, find k

x+2+k(x-2)=4

Substituting k = 12The solution of the equation for x x+1 x=c+1 c is x1=c, x2=1 c, if the solution of x-3 x=c-3 c is x1=c, x2=-3 c, then the solution of (1)x+2 x=a+2 a is 2)x+3 (x-1)=a+3 (a-1).

1) x = a or x = 2 a, (2) x = a or x = 3 (a-1).

Original question = 1-1 2+1 2-1 3....1/99-1/100

2): According to (1), it is obtained:

1-1/2+1/2-1/3+..1/n-1/(n+1)

1-1/(n+1)

2)1/1*2+1/2*3+1/3*4+..1/n(n+1)==（1-1/2)+(1/2-1/3）+（1/3-1/4)+.1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)

1.(3/2x) +2 = 0

2.(x-x/1) +x+2/1) = 1

3.(x+1/1) -x +3x+x/2) = -1

This classmate, please describe the problem clearly.

What the hell do you want?