Xiao Hua read a book. On the first day, he read 18 pages more than 21, and on the second day, he read 16 pages less than 6, with 172 pages left. How many pages are there in this story book?

Suppose there are x pages in this story book, the number of pages read on the first day: 18x + 21, the number of pages read on the second day: 16x-6, X - (18x + 21) - (16x-6) = 172, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X -- 18x-21-16x + 6 = 172, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X -- 18x-16x + 6-21 = 172, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 1724X-15=172， 1724X-15+15=172+15， 1724X=187， 1724X÷1724=187÷1724， &A: there are 264 pages in this storybook

Known: a = 2010, B = 2008, find the square of a-2ab + B + 3a-3b-3

a²-2ab+b²+3a-3b-3
=(a-b)²+3(a-b-1)
=(2010-2008)²+3×（2010-2008-3）
=2²+3×（-1）
=4-3
=1

It takes eight hours for a bus and five hours for a car to go from place a to place B. the bus takes more time than the car %The speed of a bus is slower than that of a car %．

(1) (8-5) △ 5, = 3 △ 5, = 60%; (2) (15 − 18) △ 15, = (840 − 540) × 5, = 38, = 37.5%; so the answer is: 60, 37.5

What should we pay attention to when drawing the number axis?
I really don't know

1 the direction arrow faces positive (generally to the right) 2 the origin 3 ruler (distance per unit length)

From city a to city B, it takes five hours for the bus and four hours for the train. Five quarters of an hour after the bus starts from city a, the freight cars are opposite each other from city B. when the two cars meet, how many hours did the bus and the freight car walk respectively?

The 5 / 4-hour journey of passenger train takes up the following proportion of the whole journey:
5 / 4 △ 5 = 1 / 4
The rest of the way:
1-1 / 4 = 3 / 4
Time from departure to meeting:
3 / 4 (1 / 5 + 1 / 4)
=3 / 4 ／ 9 / 20
=5 / 3
=1 and 2 / 3 (hours)
Time from departure to meeting:
5 / 3 + 5 / 4
=35 out of 12
=2 and 11 / 12 (hours)

What is ascending power permutation?

There are also descending power permutations
"Two permutations" refers to ascending and descending power permutations. It should be noted that: (1) determine which letter is the principal element, for example, the descending power permutation of 3x2y-xy2 + x3-y3 should be X3 + 3x2y-xy2-y3 (at this time - Y3 is regarded as a constant term) according to X (x is the principal element, and the one with the largest power exponent in the X term is in the front); when ascending power permutation, the constant term should be placed in the first place, and the one with the largest power exponent in the last place; In descending order, the constant term is placed at the last place, and the one with large power index is placed at the first place
Ascending and descending are meaningless, only when the ascending and descending power of a letter (or number) is specified
The ascending power (descending power) arrangement of a letter (number) refers to the arrangement from high power to low power (from low power to high power) according to the power of the letter (number). If the letter (number) is not included, it is the 0 power of the letter (number). The power refers to the power of the letter (number)
For example, the polynomial you mentioned is arranged according to the descending power of X
The third power of X + the second power of X + X + y + the second power of Y + the fourth power of Y The last three terms do not contain x, that is, the order of X to the power of 0 can be interchanged. By the way, this formula is still arranged according to the ascending power of Y
On the contrary, the formula is as follows according to the ascending power of X:
The fourth power of Y + the second power of Y + y + X + the second power of X + the third power of X Similarly, the first three terms do not return x, that is, the order of X to the power of 0 can be interchanged. By the way, this formula is still arranged according to the descending power of Y
In general, if there are two letters in a polynomial, and you are required to arrange them according to the ascending or descending power of one letter, then the other letter is generally arranged in the opposite way. This is a habit. Of course, it is not necessary to abide by it, but it is best to abide by it, so that the idea of solving the problem will be clear, and it is not easy to miss a certain item

As shown in the figure, the inclined plane a is still placed on the horizontal ground. The sliding block B with mass m moves downward along the inclined plane surface under the combined action of external forces F1 and F2. When the F1 direction is horizontally right and the F2 direction is downward along the inclined plane surface, the inclined plane receives the friction force from the ground to the left
A. If F1 and F2 are removed at the same time, the acceleration direction of slider B must be downward along the inclined plane B. If only F1 is removed, the direction of ground friction on a may be to the right while slider B is still moving downward C. If only F2 is removed, the direction of ground friction on a may be to the right while slider B is still moving downward D If only F2 is removed, the ground friction on a will not change during the downward movement of slider B

This problem can be discussed from the following two aspects: (1) the surface of the wedge a is smooth (let the inclination angle of the slope be θ, the mass of a be Ma, and the mass of B be MB) A. if F1 and F2 are removed at the same time, the object must also make a uniformly accelerated linear motion along the inclined plane under the action of the downward component force MB GSIN θ of its gravity along the inclined plane; B. If F1 is removed, the magnitude of the external force that makes a move relative to the ground is fn2sin θ = MB GCOS θ sin θ, and the direction is to the right. As shown in Figure 1, because MB GCOS θ sin θ < (MB GCOS θ + f1sin θ) sin θ, the ground friction on a is still static friction, and its direction is still to the left, but not to the right; C. remove F2. In the process of object B still moving downward, the change of ground friction on a should be considered from the perspective of the reason why a is affected by ground friction, that is, to find out the change of external force that makes a move relative to the ground GCOS θ + f1sin θ) sin θ. As shown in Fig. 2 and Fig. 3, it has nothing to do with the existence of F2. Therefore, when F2 is removed, the friction force of a on the ground should remain unchanged while object B is still moving downward. Therefore, C is wrong; D, according to the above judgment, D is correct; therefore, under the condition that the surface of the wedge is smooth, the answer to this question should be ad What about the rough and different surface? (2) the surface of a is rough In the case of rough surface of wedge a, the downward motion of B along the inclined plane under the joint action of F1 and F2 may not be uniformly accelerated linear motion, but may also be uniform linear motion. If we fall into the discussion of the motion of B here, it will increase the difficulty of judgment. Step back, the sea is wide and the sky is wide. Can we not entangle in the force analysis of B and see what will happen to a? According to Newton's third law, these two forces react on A. wedge a actually moves to the right relative to the ground under the action of the horizontal components of these two forces. FN sin θ > fcos θ, and because f = μ FN, FN sin θ > μ FN A. withdraw F1 and F2 at the same time. From the above analysis, we can see that MB GSIN θ is greater than μ MB GCOS θ. Therefore, the resultant force of object B is downward along the wedge, and the acceleration direction must also be downward along the wedge B. If F1 is removed, n = mgcos θ, f = μ n while B is still moving downward, assuming that the direction of friction FA on a is to the left, nsin θ = fcos θ + FA, then there are: FA = nsin θ - μ NOS θ = n (sin θ - μ cos θ) > 0, so wedge a tends to move to the right relative to the ground, and the direction of friction is to the left. Therefore, B is wrong; C. whether or not the wedge is subject to ground friction due to the existence of F2 Therefore, C is wrong; D, according to the above judgment, D is correct; therefore, in the case of rough surface of wedge a, the correct option of this problem is still ad. so select ad

Read and complete the following questions: the solution of the equation x + 1 / x = 2 + 1 / 2 is X1 = 2, X2 = 1 / 2; the solution of X + 1 / x = 3 + 1 / 3 is X1 = 3, X2 = 1 / 3
The equation x + 1 / (x-1) = a + 1 / (A-1) about X is transformed into the equation x + 1 / x = C + 1 / C__ The solution of the equation is_ The mathematical idea to solve this problem is_______________ .

The solution of this problem embodies the mathematical thought of analogy
X + 1 / (x-1) = a + 1 / (A-1) push out (x-1) + 1 / (x-1) = (A-1) + 1 / (A-1)
The X in the question has the same form as the X in front of it, but its meaning is different
Consider X-1 as X and A-1 as C
The solution of this problem is X1 = a; x2 = 1 / A
Where does the landlord not understand

How to judge the difference between static friction and sliding friction?

The difference between sliding friction and static friction is here. Static friction is the force exerted on a static object, which has a relative sliding trend but does not occur relative sliding. Sliding friction has occurred relative sliding
In fact, it is very simple to judge whether there is relative motion, sliding friction; no, see whether there is relative motion trend, that is, whether your research object is forced, forced, immobile, static friction in the direction of motion

Finding the solution set of X / 2 + X / 8 + X / 12 + X / 20 + X / 30 + X / 42 greater than 6

x/2+x/8+x/12+x/20+x/30+x/42=x/2+x/8+(x/3-x/4)+(x/4-x/5)+(x/5-x/6)+(x/6-x/7)=x/2+x/8+x/3-x/7=137x/168＞6

Xiao Hua read a book. On the first day, he read 18 pages more than 21, and on the second day, he read 16 pages less than 6, with 172 pages left. How many pages are there in this story book?