The latest edition of Beijing Normal University third grade mathematics volume I "how many small trees" teaching design

The latest edition of Beijing Normal University third grade mathematics volume I "how many small trees" teaching design


Teaching objectives:
1. Explore and master the oral calculation method of integral ten, integral hundred and integral thousand multiplied by one digit, and be able to calculate correctly
2. Combined with the specific situation, in the process of discussing and solving practical problems, cultivate students' awareness and ability to put forward and solve problems
Key points of teaching: To explore and master the oral calculation method of integral ten, integral hundred and integral thousand multiplied by one digit, and to be able to calculate correctly
Teaching difficulties: combined with specific situations, cultivate students' awareness and ability of raising and solving problems in the process of discussing and solving practical problems
Teaching equipment: wall chart, digital card
Teaching design:
1
Context introduction:
Students, today is the first day of school. As soon as you enter the campus, you will see beautiful flowers blooming in the flower beds. They are decorating our beautiful campus. Do you like such a beautiful environment? (like) to have a beautiful environment, you need flowers and trees
2
Explore new knowledge: show the wall chart, let's have a look, what's here?
1. Observe carefully and speak out the meaning
2. Who can ask questions based on this picture?
3. Try to solve the problem



Teaching plan of mathematics "how many are there altogether" in the first volume of the first grade book of Beijing Normal University Edition


1. Combined with specific situations, explore independently, experience the process of preparing multiplication formula, understand the meaning of each multiplication formula. Be able to use the multiplication formula to calculate, and be able to solve simple practical problems. Communicate the relationship between new and old knowledge, and initially develop the ability of analogy. Teaching focus: explore and prepare the multiplication formula, understand each multiplication formula



Outline of the review of the eighth grade physics Volume I of PEP


The first chapter "sound phenomenon" review outline 1. The occurrence and transmission of sound 1. All sound producing objects are vibrating. When the vibration stops, the sound also stops. The sound source of vibrating objects. 2



As shown in the figure, it is known that ﹥ 3 = ﹥ 1 + ﹥ 2, and it is verified that ﹥ a + ﹥ B + ﹥ C + ﹥ d = 180 °


It is proved that GH ∥ EB, ∫ 3 = ∧ 1 + ∧ 2 = ∧ eGk + ∧ fgk, ∧ 1 = ∧ eGk, ∧ 2 = ∧ fgk, ∧ GH ∥ CF, ∧ be ∧ CF, ∧ a + ∧ B = ∧ BMD, ∧ C + ∧ d = ∧ ANC, ∧ a + ∧ B + ∧ C + ∧ d = ∧ BMD, ∧ ANC = 180 ° through G



A round flower bed covers an area of 50.24 square meters. Now we need to build a circular path with a width of 1.5 meters outside the flower bed. What is the area of the path


50.24/3.14=16(m)
( 4 )*( 4 )=16
Small circle: r = 4m
Big circle: 4 + 1.5 = 5.5 (m)
S-ring = 3.14 * (square of R-square of R)
=3.14*(30.25-16)
=745 (cm 2



The interval of function f (x) = lgx-x + 1 must have a zero is a (0.1, 0.2) B (0.2, 0.3) C (0.3, 0.4) d (0.4, 0.5)


Because f (0.1) = lg0.1-0.1 + 1 = - 1-0.1 + 1 = - 0.1 < 0.01
f(0.2)=lg0.2-0.2+1=lg2-0.2>0
So f (0.1) f (0.2) < 0
So the zero interval is a (0.1,0.2)
Answer: choose a



A 6 cm long, 5 cm wide and 3 cm high rectangular frame can be made with a long () wire
A. 90 cm B. 28 cm C. 126 cm D. 56 cm


(6 + 5 + 3) × 4 = 14 × 4 = 56 (CM)



Decomposition factor: 2 (3a & sup2; - b) - A (3b-4)
Given M & sup2; - 3M + 1 = 0, find 1 / 2 of M & sup2; + M & sup2


3a^2-b)-a(3b-4)
=6a^2-2b-3ab+4a
=2a(3a+2)-b(3a+2)
=(2a-b)(3a+2)
M ^ 2-3m + 1 = 0, find m ^ 2 + 1 / m ^ 2
m^2-3m+1=0
m-3+1/m=0
m^2+1/m^2=(m+1/m)^2-2=3^2-2=7



The three sides of the triangle are 10, 17 and 21. What's the area, please?


84
Cosine theorem gives cosa = 15 / 17
sinA=8/17
Area formula = (1 / 2) * c * b * Sina = 1 / 2 * 17 * 21 * 8 / 17 = 84



Given that a > 0 and a is not equal to 1, f (x) = x ^ 2-A ^ x, when x belongs to (- 1,1), all f (x)


We can change f (x) & lt; 1 / 2 to: x ^ 2-1 / 2 & lt; a ^ x is constant on (- 1,1), x ^ 2-1 / 2 ∈ (- 1 / 2,1 / 2) due to a & gt; 0 and a ≠ 1, and discuss the classification of a: 0