Calculation questions in Chapters 1 and 2 of Senior 1 Compulsory 1

The 250 60 column vertical calculation and the 304 15 column vertical calculation are as follows:

Multiplication vertical calculation should pay attention to the problem:

The last digit of the two numbers is to be aligned.

Try to write the numbers with more numbers on the top and the numbers with fewer numbers below to reduce the number of multiplications.

Vertical Calculation Method:

Multiplication: multiply the i-digit of one number by the j-digit of another number.

It should be added to the i+j-1 of the product.

Division: e.g. 42 divided by 7.

Start with 4 and divide from high to low . When the division method is calculated vertically, it starts to divide from the highest digit, for example: 42 is divided from the highest ten digit 4; If you can't get rid of it, such as:

4 cannot be divided by 7, so divide by combining the highest digit and the next digit into a number until you can divide by the divisor; For example, if 42 is divided by 7 and 4 cannot be divided by 7, 4 and 2 are combined into a number 42 to divide 7, and the quotient is 6.

The mathematically prescribed order of four operations is as follows:

1. Sibling operations in a formula, if only sibling operations are included, they should be calculated in order from left to right. That is, mixed operations with only addition and subtraction, or only multiplication and division, in the order of operation from left to right.

2. Level 1 to 2 operations.

In an equation, if there are both first-level and second-level operations, then the second-level operation should be calculated first, and then the first-level operation. That is, "multiplication and division first, then addition and subtraction", referred to as "multiplication and division first, then addition and subtraction".

1 (1) 2) 3) 4) 2 (1) a= or a= (2) b= or b= (3) c= or c= (4) d= p7 exercise 1 2.∈ = = 3.

1) Set A is a true subset of set b (2) Set b is a true subset of set A (3) A = b P11 Exercise 1 a∩b= a∪b= 2. a∪b= a∩b= 3.

a∩b= a∪b= 4.Exercise A, Group 1 （1）∈ 2）∈ 3）∉ 4）∈ 5）∈ 6）∈ 2.

1）∈ 2）∉ 3）∈ 3. （1）c= （2）a= （3）b= 4.(1) a= (2) b= or (3) c= 5

1) True Subset True Subset (2) True Subset True Subset = (3) 6 a∪b= a∩b= 7. a∩b= a∩c= a∩（b∪c）= a∪（b∩c）= 8.

1) Students participating in the 100-meter run or students participating in the 200-meter run (2) Students participating in both the 100-meter run and the 200-meter run 9 10.Group B 1

4 2. d= 3.a= b= when a=3 a= a b= a b= when a=1 a= a b= a b= when a=4 a= a b= a b= when a≠1,a≠3,a≠4 a b= a b=

Group A 1(1) (2) x∈r （3） (4)

2.(1) Not equal because the defined domains are different (2) Not equal because the defined domains are different (3) Equal.

3.(1) Define the domain r domain (2) Define the domain domain (3) Define the domain r domain R (4) Define the domain r domain.

Root number 2) = 8 + 5 root number 2 f(-a) = 3a 2 + 5a + 2 f(a + 3) = 3a 2 + 13a + 14 f (a) + f (3) = 3a 2-5a + 16

5. 1.Not in 2 -3 3. 14

6.Slightly 7Outline.

8.xy=10 y=10 x l=2(x+y)=2(x+10 x) d 2=x 2+y 2=x 2+100 x 2 d = root number (x 2+100 x 2).

9.v= (d 2) 2*x=vt x=*t defines the domain t (0, d 2h (4v) ] value domain x (0,h].

Group B 1 1.[-5,0]∪[2,6) 2.[0, positive infinity) 3[0,2) (5,positive infinity).

2.Slightly 3Outline.

4.Root number (x 2+4) 3 (0 x 12) Root number 20 3=3

Classmate, brother, I went to high school four years ago, now I'll help you reminisce!

Chemical Reaction Equation:

mno2+2hcl== cl2 +h2o+mnox Here you have to add a line, the above is my analysis of the proportion of matter transformation in the chemical equation, you must have done it, and I believe you can understand it, otherwise, you will just give up my answer!

X(2)mNO2+2HCl== Cl2 +H2O+MNO According to the transformation of the substance, on the left side of the equation, the ratio of manganese dioxide to hydrochloric acid is 1 to 2, so the molar digits of the hydrochloric acid being oxidized.

3) Total amount of hydrochloric acid: 12*

The remaining amount is:

Since the volume remains unchanged after the reaction and the hydrochloric acid is completely ionized, the concentration of chloride ions is the concentration of hydrochloric acid.

c（cl-）=1/

Done, children, study hard

Don't forget to give your brother extra points quickly, if it's wrong, forget it, ask the teacher well...

Solution: Let the consumption of mno2 x grams, the oxidized HCl ymol be generated, and MnCl2 zmol

4HCl(concentrated) MNO2 = mnCl2 2H2O Cl2 --2HCl (oxidized HCl).

87 g 1mol 2mol

xg zmol ymol

x= y= z=

n(cl-)=2z=

100ml=

c(cl-)=n/v=

Answer: Slightly.

mNO2+4HCl=mnCl2+2H2O+Cl2 Under the standard condition, N(Cl2)=

According to the ratio of stoichiometric numbers is equal to the ratio of the amount of substances, so n(mno2) = so m(mno2) = n*m=

Concentrated hydrochloric acid has two directions, 1 2 is oxidized to chlorine gas, and 1 2 plays an acidifying role.

And because there is concentrated hydrochloric acid involved in the reaction, the amount of oxidized HCl is the amount of Cl- in the solution after the reaction, and the concentration is the concentration of hydrochloric acid C=(

of chlorine gas, ie.

It is known from mNO2 + 4HCl = Cl2 + mnCl2 + H2O that manganese dioxide is required, i.e., grams.

The amount of the substance of the oxidized HCl is:

The original volume is 100 ml, the remaining Cl- is, and the concentration of the amount of the substance is 4mol l

f（a）+f（3）=3xa^2-5a+2+27-15+2=3xa^2-5a+16

Question 1 of Group B.

The defined range is x: [-5,0]u[2,6) and the value range: [0,+

The third question is what range x corresponds to y: [0,2)u(5,+

This seems to be a simplification of the original form to get a two-factored form of taking the plane. Eventually, the unknowns can be eliminated.

If the two sets are equal, then there is a=a and b a=a b or a =a b and b a=a b, and a = 1, b = 0, and the substitution calculation has the original formula = 1.