Junior high school math is going to be examed, junior high school math exam questions 。。。。。。。

1 When a 5, the solution set of the inequality (a-5) x >1 is x 1 a-5 solution: from the problem a-5 0 (from (a-5) x>1 to x 1 a-5, the sign changes, i.e., both sides of the equation are divided by a number less than 0 at the same time, and the symbol of the equation changes).

a＜52.If the solution set of the inequality m-2 1 3(m-x) is x 2, then m=2 solution: solve the inequality, yield: x -2m+6

The set of solutions for inequalities with respect to x is x 2

2m+6=2

Solution: m=2

When m=2, the solution set of the inequality is x 2

If the solution set of inequality (a-5) x<1 is x>a-1/5 is known from the inequality property 3.

a-5<0

a<51/3（x-m）>2-m

1/3)x-(1/3)m>2-m

1/3)x>2-(2/3)m

x>6-2m

So 6-2m = 2

m=2 is not necessarily true.

Analysis: According to the principle of similar triangles, aa1b a1a2b1 is obtained, and then baa1= b1a1a2; Use the Pythagorean theorem to calculate the side length of a square; Finally, the area formula of the square is used to calculate the area of the three squares, and the law is found from it, and the problem is solved

Answer: Solution: Let the areas of the square be s0, s1, s2....s2010, according to the title, get:

ad bc c1a2 c2b2, baa1= b1a1a2= b2a2x (equal isotope angles) aba1= a1b1= b2a2x=90°, baa1 b1a1a2, at right angles ado, according to the Pythagorean theorem, yields: ad= 5, cot dao= oaod= 12, tan baa1= ba1ab=cot dao, ba1= 12ab= 52, ca1= 5+ 52= 5 (1+12), the same goes for obtained: c1a2 = 5 (1+12) 1+12), from the area formula of the square, obtain: s0= (5)2, s1= 52 (1+12)2, s2= 52 (1+12)2 (1+12)2, from this, sn= 52 (1+12)2(n-1), s2010=5 (1+12)2 (2010-1), =5 (32)4018

Set the purchase price of this product to be X yuan, and Y pieces were sold in the first month.

xy*20%=6,000 xy=30,000 yuan, the second month sold: y+100 pieces, x(y+100)*10%=8,000 xy+100x=80,000 yuan.

30000 + 100 x = 80000 x = 500 yuan y = 30000 500 = 60 pieces.

Y+100 160 pcs.

Therefore, the purchase price of this product is 500 yuan, and the mall sold a total of 160 pieces in the second month.

Solution: There are three such points q on the parabola

When the point q is at the position of q1, the ordinate of q1 is 3, and the coordinates of point q1 can be obtained by substituting the parabola as (2,3);

When the point Q is at the position of the point Q2, the ordinate of the point Q2 is -3, and the coordinate of the point Q2 can be obtained by substituting the parabola as (1+ 7, -3);

When the point Q is at the position of Q3, the ordinate of the point Q3 is -3, which can be obtained by substituting the parabolic analytic formula, and the coordinates of the point Q3 are (1-7, -3);

In summary, there are three points q that satisfy the topic, namely: q1 (2, 3), q2 (1+ 7 , -3), and q3 (1-7 , -3).

y=-x2+2x+3 intersects (-1,0) and (3,0) with the x-axis, and intersects with the y-axis at (0,3), (1) a(-1,0) let p be (p,0), then, p-(-1)=0+x, so, x=p+1, q is equal to c, y=3, so -(p+1)2+2(p+1)+3=3, the solution is p=-1, or p=1, because a(-1,0), then p=1, p(1,0)q(2,3), at this time, APQC is a parallelogram.

2) a(3,0), bring in the solution as above, and find no solution.

Draw a sketch yourself, because the parabola has two focal points with the x-axis and one focal point with the y-axis, connect them according to the situation, and then draw parallel lines.

Solution: When the point q is at the position of q1, the ordinate of q1 is 3, and the coordinates of point q1 can be obtained by substituting the parabola into (2,3);

When the point Q is at the position of the point Q2, the ordinate of the point Q2 is -3, and the coordinate of the point Q2 can be obtained by substituting the parabola as (1+ 7, -3);

When the point Q is at the position of Q3, the ordinate of the point Q3 is -3, which can be obtained by substituting the parabolic analytic formula, and the coordinates of the point Q3 are (1-7, -3);

In summary, there are three points q that satisfy the topic, namely: q1 (2, 3), q2 (1+ 7 , -3), and q3 (1-7 , -3). Same as below.

Idea: The key is to do perpendicular, through the q point to do the perpendicular intersection of the x-axis at the point w, the triangle q1p1w congruent triangle cao can be proved, so p1w=ao=1, because the q point coordinates are known, and then the p1 point coordinates can be found.

The minimum is 9.

There are two in the first row of the top view, that is, there are two columns in the first column of the main view, that is, there are at least 4 cubes, and there is one in the second row of the top view.

There are three in the third row of the top view, which is the third column of the main view, which is at least 4 cubes, so the minimum is 9.

This kind of topic mostly relies on one's own spatial imagination.

Look at the first row of the main view and the first row of the top view, so that the fewness 4 and then the second row only need 1 square.

Finally, looking at the last row, you need at least 4.

then a total of 9 are needed.

The first layer looks at the top view, there are 6 squares, and the upper two layers look at the front view, except for the bottom layer, there are 3 blocks, so it is 6+3 9.

There is no way to solve this kind of problem, it depends on your space to imagine, in what circumstances you will see such a figure, in what case you will see such a figure...

The front view and top view represent 2 different faces of observation.

Since it is necessary to be satisfied, there are 6 squares on each of these two perspectives.

First of all, there must be a minimum of 12 squares.

But. Because the front view and top view are 1 whole object. If the two faces overlap each other, a maximum of three places will coincide. So 12 is to subtract 3....

That is to say, you need 6 squares for the same, but there are 3 squares that are just extra and overlapping.

So. That's 9.

Figure. It's with drawings. It's ugly. = =|||

9, draw your own picture to see, three-dimensional.

The teacher said to Xiao Ming"When I was your age, you were just three years old"From this sentence, it can be known that the teacher is 10 years older than Xiao Ming;

When you are so old as me, I am already 33 years old' From this sentence, it can be seen that the teacher's age = 33-10 = 23 years old, so Xiao Ming's age = 23-10 = 13 years old.

I'm a teacher, thank you.

Solution: The radius of the bottom rubber stool is meters.

Base area 5 5=sqm.

Cone volume cubic meters.

The sand weighs only a ton of weight.

Frequency: Approximately 57 times.

57 times can be transported to complete the beam change.

From the problem, it can be seen that point d is the intersection point of the perpendicular bisector of the three sides of the triangle, and do de ac at point e, so ae ce, find the length of ae, and then use the angle of thirty degrees to find the length of ad by the pythagorean theorem.