mathematics翻译

数学 mathematics, maths（BrE）, math（AmE）被除数 dividend

除数 divisor 商 quotient 等于 equals, is equal to, is equivalent to 大于 is greater than

小于 is lesser than

大于等于 is equal or greater than

小于等于 is equal or lesser than

运算符 operator

数字 digit

数 number

自然数 natural number

公理 axiom

定理 theorem

计算 calculation

运算 operation

证明 prove

假设 hypothesis, hypotheses（pl.）

命题 proposition

算术 arithmetic

加 plus（prep.）, add（v.）, addition（n.）

被加数 augend, summand

加数 addend

和 sum

减 minus（prep.）, subtract（v.）, subtraction（n.）

被减数 minuend

减数 subtrahend

差 remainder

乘 times（prep.）, m

Mathematics HK CE Paper

Form 5

HKCEE 1993 Mathematics II

93 If f(x) = 102x, then f(4y) = 1. A. 104y . B. 102 + 4y . C. 108y . D. 40y . E. 402y .

93 2. If s = n2

[2a + (n 1)d], then d =

A. 2(s an)n(n 1) . B. 2(s an)(n 1) .

C. sn(n 1) . D. as na(n 1) . E.

4(s an)

n(n 1)

.

93 Simplify (x2 3x + 1)(x2 +3x + 1). 3. A. x4 + 1 B. x4 x2 + 1 C. x4 + x2 + 1

D. x4 3x2 23x 1

E. x4 + 3x3 23x2 +3x 1

93 a4. a

b

+

aa

b

.

A. 1a

b

B.

a 2ab b

a b

93-CE-MATHS II C. b a

2a

D. b 2ab aa b

E.

a ba b

93 If 3x2 + ax 5 (bx 1)(2 x) 3, 5.

Mathematics HK CE Paper

Form 5

HKCEE 1993 Mathematics II

93 If f(x) = 102x, then f(4y) = 1. A. 104y . B. 102 + 4y . C. 108y . D. 40y . E. 402y .

93 2. If s = n2

[2a + (n 1)d], then d =

A. 2(s an)n(n 1) . B. 2(s an)(n 1) .

C. sn(n 1) . D. as na(n 1) . E.

4(s an)

n(n 1)

.

93 Simplify (x2 3x + 1)(x2 +3x + 1). 3. A. x4 + 1 B. x4 x2 + 1 C. x4 + x2 + 1

D. x4 3x2 23x 1

E. x4 + 3x3 23x2 +3x 1

93 a4. a

b

+

aa

b

.

A. 1a

b

B.

a 2ab b

a b

93-CE-MATHS II C. b a

2a

D. b 2ab aa b

E.

a ba b

93 If 3x2 + ax 5 (bx 1)(2 x) 3, 5.

Problem-Based Learning (PBL) describes a learning environment where problems drive the learning. That is, learning begins with a problem to be solved, and the problem is posed is such a way that students need to gain new knowledge before they can solve the

Problem-based Learning in Mathematics

Kyeong Ha RohApril 2003

Problem-Based Learning (PBL) describes a learning environment where problems drive the learning. That is, learning begins with a problem to be solved, and the problem is posed is such a way that students need to g

Problem-Based Learning (PBL) describes a learning environment where problems drive the learning. That is, learning begins with a problem to be solved, and the problem is posed is such a way that students need to gain new knowledge before they can solve the

Problem-based Learning in Mathematics

Kyeong Ha RohApril 2003

Problem-Based Learning (PBL) describes a learning environment where problems drive the learning. That is, learning begins with a problem to be solved, and the problem is posed is such a way that students need to g

applied mathematics

FEDERAL PUBLIC SERVICE COMMISSION

COMPETITIVE EXAMINATION FOR RECRUITMENT TO POSTS IN BS-17 UNDER THE FEDERAL GOVERNMENT, 2013

TIME ALLOWED: THREE HOURS MAXIMUM MARKS: 100

NOTE: (i) Candidate must write Q. No. in the Answer Book in accordance with Q. No. in the Q. Paper.

(ii) Attempt FIVE questions in all by selecting TWO questions from SECTION-A and ONE

question from SECTION-B and TWO questions from SECTION-C ALL questions carry EQUAL marks.

(iii) Extra attempt of any question or any part of the att

applied mathematics

FEDERAL PUBLIC SERVICE COMMISSION

COMPETITIVE EXAMINATION FOR RECRUITMENT TO POSTS IN BS-17 UNDER THE FEDERAL GOVERNMENT, 2013

TIME ALLOWED: THREE HOURS MAXIMUM MARKS: 100

NOTE: (i) Candidate must write Q. No. in the Answer Book in accordance with Q. No. in the Q. Paper.

(ii) Attempt FIVE questions in all by selecting TWO questions from SECTION-A and ONE

question from SECTION-B and TWO questions from SECTION-C ALL questions carry EQUAL marks.

(iii) Extra attempt of any question or any part of the att

SECTION2

GENERAL INFORMATION,

CONVERSION TABLES,AND

MATHEMATICS

2.1GENERAL INFORMATION AND CONVERSION TABLES 2.3

Table2.1Fundamental Physical Constants 2.3 Table2.2Physical and Chemical Symbols and De?nitions 2.6 Table2.3Mathematical Symbols and Abbreviations 2.23 Table2.4SI Pre?xes 2.24 Table2.5Greek Alphabet 2.25 Table2.6Abbreviations and Standard Letter Symbols 2.26 Table2.7Conversion Factors 2.35 Table2.8Temperature Conversion Table 2.54 2.1.1Conversion of Thermometer Scales 2.66 2.1.2Density and Speci?c

Abstract. The notion of relative closure (X, Y)0 of a semi-Pfaffian couple (X, Y) was introduced by Gabrielov to give a description of the o-minimal structure generated by Pfaffian functions. In this paper, an effective bound is given for the number of con

DIMACSSeriesinDiscreteMathematics

andTheoreticalComputerScience

OntheNumberofConnectedComponentsoftheRelative

ClosureofaSemi-Pfa anFamily

AndreiGabrielovandThierryZell

Abstract.Thenotionofrelativeclosure(X,Y)0ofasemi-Pfa ancouple

(X,Y)wasintroducedbyGabrielovtogiveadescriptionof

用Microsoft Mathematics求极限

求极限的基本操作如下：

求极限的过程如下： 1. 点击极限图标：

2. 输入函数表达式和0

3. 点击“输出”或直接回车：

4. 得到计算结果：

也可以键盘输入极限式：limt(sin(x)/x,x,0)，然后回车。

输入：limit(((1+x^2/2-sqrt(1+x^2))/((cos(x)-e^(x^2))sin(x^2))), x, 0)

输入：limit(((1+a/x)^x), x, infinity)

输入：limit((arcTan(x)), x, infinity)

输入：limit((arcTan(x)), x, -infinity)

单侧极限

输入：limit(((e^h-1)/(e^h+1)), h, -infinity)

结果：