Chapter 2 the straight line
line是什么意思中文翻译
line是什么意思中文翻译line既能做名词也能做动词，那么你知道line做名词和动词分别都是什么意思吗?下面店铺为大家带来line的英语意思解释和英语例句，欢迎大家一起学习!line作名词的意思：线条;排;行列;界线line作动词的意思：排队;用线标出;沿…排列成行;给…安衬里line的英语音标：英 [lain] 美 [laɪn]line的时态：现在分词： lining过去式： lined过去分词： linedline的英语例句：1. Megamalls and fast food restaurants line the highway system.公路系统沿线有大型商场和快餐店。
2. Their first car rolls off the production line on December 16.他们的第一辆车于12月16日下线。
3. Elliott crossed the finish line just half a second behind his adversary.埃利奥特跨过终点线时仅比对手落后半秒钟。
4. Her ear, shoulder and hip are in a straight line.她的耳朵、肩膀和髋部呈一条直线。
5. The Chief Constable's clipped tones crackled over the telephone line.电话线那头噼里啪啦地传来警察局长短促的声音。
6. The train backed out of Adelaide Yard on to the Dublin-Belfast line.火车倒出了阿德莱德调车场，开上了都柏林-贝尔法斯特线。
7. There was silence at the other end of the line.电话那头没有声音。
8. Every line etched on her face told a story of personal anguish.她脸上的每条皱纹都讲述了一次痛苦的经历。
MACRS折旧法
The yearly recovery allowance, or depreciation expense, is determined by multiplying each asset’s depreciable basis by the applicable recovery percentage shown in Table 11A-2. Calculations are discussed in the following sections.
TAX DEPRECIATION LIFE
For tax purposes, the entire cost of an asset is expensed over its depreciable life. Historically, an asset’s depreciable life was determined by its estimated useful economic life; it was intended that an asset would be fully depreciated at approximately the same time that it reached the end of its useful economic life. However, MACRS totally abandoned that practice and set simple guidelines that created several classes of assets, each with a more-or-less arbitrarily prescribed life called a recovery period or class life. The MACRS class life bears only a rough relationship to the expected useful economic life.
了不起的盖茨比-CHAPTER TWO
affectation vanity Behind the times
Mr.Wilson:
“Works pretty slow, don’t he?” “No, he doesn’t,” said Tom coldly. “And if you feel that way about it, maybe I’d better sell it somewhere else after all.” “I don’t mean that,” explained Wilson quickly. “I just meant--"他干得很慢，是不是？" "不，他干得不慢，"汤姆冷冷地说，"如果你有这样的 看法，也许我还是把它拿到别处去卖为好。" "我不是这个意思，"威尔逊连忙解释，"我只是说……" “Wilson? He thinks she goes to see her sister in New York. He’s so dumb he doesn’t know he’s alive.” "威尔逊？他以为她是到纽约去看她妹妹。他蠢得要命， 连自己活着都不知道。"
PART 2
Characters introdyrtle） Mr.Wilson McKees Catherine
The dog vendor
spousal relationship
McKees
Tom:
and his determination to have my company bordered on violence. 他硬要我陪他的做法近乎暴力行为 "Go and buy ten more dogs with it.” 给你钱。拿去再买十只狗。 “Daisy! Daisy! Daisy!” shouted Mrs. Wilson. “I’ll say it whenever I want to! Daisy! Dai——” Making a short deft movement, Tom Buchanan broke her nose with his open hand. "黛西！黛西！黛西！"威尔逊太太大喊大叫，"我什么时候想 叫就叫！黛西！黛……" 汤姆· 布坎农动作敏捷，伸出手一巴掌打破了威尔逊太太的鼻 子。
写直线运动和曲线运动的小作文
写直线运动和曲线运动的小作文Straight-line motion is a type of motion where an object moves in a straight path, without changing its direction. This type of motion can be observed in various everyday scenarios. For example, when a car travels on a straight road, it moves in a straight line. Similarly, when a ball is thrown vertically upwards and then falls back down, its motion can be considered straight-line motion.直线运动是物体沿直线路径移动的一种运动方式，不改变其方向。
这种运动方式可以在日常生活中的各种场景中观察到。
例如，当一辆汽车在一条笔直的道路上行驶时，它沿直线运动。
同样，当一个球被垂直向上抛出然后再回落时，它的运动可以被视为直线运动。
In straight-line motion, the object's velocity remains constant if there are no external forces acting on it. This means that the object covers equal distances in equal intervals of time. The equation that describes straight-line motion is given by:在直线运动中，如果没有外力作用于物体，其速度保持恒定。
这意味着物体在相等的时间间隔内覆盖相等的距离。
Learning_English_lsn't_a_Straight_Line_学习英语不是一帆风顺的
他山之石If you ’re in a class,following a textbook or using any kind of curriculum 1,youmay find the topics and other materials 2in some ually,things move from eas -ier to more difficult as time goes by,but it doesn ’t mean that you always have to fol -low the book.There will be times that you feel bored 3and want to find new information that ’s not in the book.That would be a good way to make learning much more interesting.There will also be times that you ’ll have to go back and review 4some material that you didn ’t know clearly the first time around.That ’s good and normal 5.Sometimes when things don ’t go well,you have to stop and check your plan to find outthe problems.It will be great if you can practice English for 10hours a day,but that usually won ’t happen 6.Because no one can learn one subject for such a long time and doesn ’t feel tired and bored.It is very difficult for you to remember everything in your learning.You need more time to find out what you have forgotten and try to remember it again.You should know that there will be some time when you feel like a great English speaker,and there will be time when you feel like you couldn ’t finish a sentence.Those ups and downs are normal,too.Just keep working and be patient 7,and you ’llget better and better.Learning English Isn ’t a Straight Line学习英语不是一帆风顺的苗淑华选译4▌▍▎▏Notes 他山之石如果你在课堂上依照课本或遵循任何一种课程学习，你可能会发现（单元）主题和其他材料内容是按某种顺序排列的。
直线折旧法（Straight-line
直线折旧法（Straight-line Depreciation Method）
什么是直线折旧法？
直线折旧法是指将资产原值扣除净残值后，在预计使⽤年限内平均摊销。
年折旧费为购置成本扣除残值后，再除以预计使⽤年限。
直线折旧法的计算公式
每年折旧费=（成本－残值）/估计有效使⽤年限
直线折旧法举例
下⾯举例说明⼀台服务器的折旧计算过程：
服务器购置成本：4500美元
预计使⽤年限：4 年
净残值：500美元
年折旧费 = (4500美元 - 500美元)/4 年 = 1000美元/每年。
因此，应对这台服务器每年计提1000美元折旧费，共计提4年。
使⽤年限届满时，通常会将这台服务器淘汰。
⼀位采购员为公司购买了⼀辆汽车，价格为 $18,000，汽车估计服务8年，残值为$2,000。
减掉残值后，两年后这辆汽车的价值是多少？使⽤直线折旧，
A $ 8,000
B $10,000
C $12,000
D $14,000
答案：C
解释：每年折旧额=（采购价-残值）/服务年数
=（$18,000-$2,000）/ 8
= $2000
两年之后的价值= 采购价-残值 - 每年折旧额X2年
=$18,000-$2,000- 2X$2,000
= $12，000。
外研社美国文学史及选读(第三版)(第二册)教学课件0 Part V-Introduction
Waste Land, the most significant American poem of the 20th century, helped to establish a modern tradition of literature rich with learning and allusive thought.
ICnhatprtoerd3uction
American society. Early in the century Ezra Pound and T. S. Eliot published works that would change the nature of American poetry, but their impact (and that of other modernist writers) on the general reading public was slight. The genteel tradition and popular romanticism still dominated the nation’s literary tastes.
高一数学必修一第五章知识总结
高一数学必修一第五章知识总结1.本章主要内容是关于一次函数及其应用。
The main content of this chapter is about linear functions and their applications.2.一次函数的定义是f(x)=ax+b。
The definition of a linear function is f(x)=ax+b.3.一次函数的图像是一条直线。
The graph of a linear function is a straight line.4.一次函数的斜率表示为a，截距表示为b。
The slope of a linear function is represented by a, and the intercept is represented by b.5.一次函数的斜率可以表示为Δy/Δx。
The slope of a linear function can be represented asΔy/Δx.6.斜率为正的一次函数是递增的，斜率为负的一次函数是递减的。
A linear function with a positive slope is increasing, while a linear function with a negative slope is decreasing.7.一次函数的零点就是方程f(x)=0的解。
The zero point of a linear function is the solution tothe equation f(x)=0.8.一次函数的图像经过坐标原点时，截距为零。
The graph of a linear function passes through the origin when the intercept is zero.9.一次函数的图像是一条直线，其斜率决定了直线的倾斜程度。
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Chapter 2 : The Straight Line2.1 The Gradient of Straight LineDefinition : The gradient of a straight line is defined as Gradient of Straight Line, m =rise run = increase in y coordinateincrease in x coordinate----- (1) If the straight line makes an angle θ with the positive direction of the x -axis and this angle is a measure of the slope of the line with respect to the horizontal x -axis .In general , the gradient of a line passing through the points A(x 1,y 1) and B(x 2 , y 2) is m = tan θ =2121y y x x -- , where x 1≠ x 2 and 0 0 ≤ θ < 18002.2 Sign of Gradientundefined .Type 1 :_________Gradient AB =• angle of inclination , θ =• The line is ________ to the x -axis .Type 2 :_________Gradient AB =• angle of inclination , θ =• The line is __________ to the y-axis .xyB(x 2 , y 2 ) A (x 1 , y 1 )θθ riserunType 3__________Gradient AB =• angle of inclination , θ =• A line sloping upwards to the right has a positive gradient .Type 4__________Gradient AB =• angle of inclination , θ =• A line sloping upwards to the left has a negative gradient .Example 1:Find the gradient of the line joining the points and also its angle of inclination.Example 2 :Find the gradient of the line joining the points on the curve y = 2x 2 – 3x + 1 whose abscissae are 1 and 3 . [ans : 5](a) ( 3 , 5 ) , ( - 4 , 12 )(b) (10,8) , ( 4 , - 4 )(c) ( 0 , 0 ) , (-1 , 3 )(d) ( -3 , 2 ) , ( -2 , 3 )*Example 3: ( Collinear Points : three points lie on the same straight line )(a) Given that three points A(0, t-2 ) , B(t+1 , t- 4) , C(t-5,2) . Find the values of t if the three points are collinear .[ans : t = 2 , 3 ](b) Show that the points A (0,-3 ) , B (4,-2 ) , C (16,1 ) are lie on the same line .2.3 The Equation of the Straight Line❒ General Form of the Equations of Straight LinesIn general , an equation of the form ax + by + c = 0 ( a , b not both zero ) represents a straight line in the rectangular coordinate plane , see figure 1. ❒ There are two special types of straight lines : (a) (b)❒ In the following section , we will discuss other forms of straight lines when some conditions are given .( I ) Gradient-intercept Form 斜截式• y-intercept : The point where the straight line cuts the y-axis , the coordinate is ( 0 , c ) . 1 : Given a straight line , gradient m and y-intercept c , find its equation . Let P(x,y) be any point lie on the straight line passing through the point A ( 0,c ) . gradient of AP , m =y cx -- y – c = mxi.e. y = mx + c is the equation of straight line . xy xyxy hkx =k(vertical line) e.g x=1, x=-1/2y =h(horizontal line) e.g y=0, y=3/4xy A(0,c)P(x,y)Example 4 :Write down the gradient , m and y-intercept , c of the following equations . (a) y = 3x + 4 gradient , m = y-intercept , c = (b) 2x – 3y + 8 = 0(c) 4x – 5y = 0(d) 5223x y +=-(e)2213y x -= (f) 1702x y -=Example 5 :Find the equation of each of the following lines . (a)[ans : x -y + 3 = 0] (b)[ans : 4x + y = 2 ](II) Point-slope form 点斜式Given a fixed point Q (x 1 , y 1 )Let P ( x , y ) be any point on the line , then gradient of PQ , m = 11y y x x -- i.e. y – y 1 = m ( x – x 1 )x y P(x,y) Q(x 1,y 1)(III) Two – Point Form 两点式As the straight line l passing through two points A ( x 1 , y 1 ) and B (x 2 , y 2 ) .let P(x,y) be any point on line l , the gradient AB = gradient BP .gradient BP = 11y y x x -=-m or gradient AB = 2121y y x x --∴ 11y y x x -=- 2121y y x x --Example 6:Find the equation of each of the following straight lines given its gradient , m and the coordinates of a point on the line . (a) m = 2 , (0,1) Equation of line isy – 1 = 2 ( x – 0 ) i.e. y = 2x + 1(b) m = 2/5 , ( 1 , -2 ) (c) m = -1 , ( -3 , 4 ) (d) m = 3 , ( 3 , 0 ) (e) m = -1/3 , ( 1 , -1/2 ) (f) m = 0 , ( 3 , 5 )Example 7 :Find the equation of the straight line which the line makes an angle of 450 with the positive x-axis and passing through the point P ( - 4 , 5 ) .Example 8 : Find the equation of the straight line passing through the points A and B . (a) A( -2, 5 ) , B ( 3 , 8 )[ ans :3x-5y+31=0](b) A(a,0 ) , B(2a , 3a ) (c) A(0,-1 ) , B ( 5,0)x yB (x 1,y 1) l P (x , y) A (x 2, y 2)The vertices of a triangle ABC are A(3,5) , B(-2,-5) and C(-5,-1) . Find the equation of the side AB and also the median of BC through vertex A .Example 10Find the equation of the straight line passing through point ( -1 , 2) and through the point of intersection of the lines 3x-4y=18 and 5x + 6y = - 8 .Form IV : Intercept Form 截距式Let a : the x – interceptb : the y – intercept Example 11 :from two point form of the straight line : Find the equation of line which through A(0,5) and B(-1,0).000y b b x a a a---==--ay = -bx + abbx + ay = ab divide ab both sides : 1x ya b+=Example 11:Interpret the general equation of line 3x + 2y – 7 = 0 into the following form : (a) intercept form (b) slope-intercept form B (0 , b )A (a, 0)yxFind the equation of the following lines . (a)(b)(c)(d)Example 13:Find equation of the line which passing through the point (2 , 2 ) and cutting off intercepts on the axes whose sum is 9 . [ans: 3x + 6y -18=0 or 6x + 3y-18= 0]1350yxP (3,-2)yx4P (3,5)Common Practise 1:1. Write down the equations of the lines :(i) gradient 5/6 , and through the point ( -3 , -1 ) (ii) inclined at 600 to the x-axis , and through ( 4 , -8 ) (iii) through the points ( 3 , 8 ) , ( 6 , 12 )(iv) through the points (,),(2,)2c c ct ct t t2. Find the equation of the chord joining the points on the curve y = 2x 3 + 1 whose absicissae are 1 and -2 .3. Find in terms of m , the equation of the straight line passing through the point (0,m ) with gradient m . Find the value of m if this line passes through the point (2 , 6 ) .4. If the mid-points of the sides BC , CA and AB of the triangle ABC be (1, 3 ) , ( 5,7 ) and ( -5 , 7 ) , find the equation of the side AB .5. A , B are the points ( 3 , 7 ) , ( -11 , -1 ) . Find the equation of the line through the point ( 4 , -6 ) and bisecting AB .6. P ( a , b ) lies on the line 6x – y = 1 and Q ( b , a ) lies on the line 2x – 5y = 5 . Find the equation of PQ .7. Find the equation of the line PQ which joining the points of intersection of the line y = x – 6 and the curve y 2 = 8x .Answer : 1. (i) 5x – 6y + 9 = 0 ; (ii) 3 x – y = 4 (2 +3) ; (iii) 4x – 3y + 12 = 0 ; (iv) x + 2t 2 y = 3ct 2. y = 6x – 3 3. y = m x + m , m = 2 4. x - y + 12= 0 5. 9x + 8y + 12 = 0 6. x + y – 6 = 0 7. y = x – 62.4 Equations of Parallel LinesConsider 2 lines L AB : y = m 1 x + c 1 L PQ : y = m 2 x + c 2 as shown in the diagram , since AB is parallel to PQ ,∴ AB PQ θ=θ∵ tan tan AB PQ θ=θ∴ m 1 = m 2Example 14: Find the equation of the line which parallel to the given line and passing through the given point .(a) y = 2x + 5 , ( 3, - 4 ) (b) 4x – 5y + 8 = 0 , (-2 , 7 )(c) 3x + 2y – 3 = 0 , ( 1 , 0 ) (d) 6x + y = 7 , ( 2 , 3 )L 1 and L 2 parallel ⟺ m 1 = m 2 P θ θxyQB AExample 15l 1 and l 2 with gradients m 1 and m 2⊥ BC , in ⊿ADBl 1 has positive gradient , m 1 =ADDB= tan θ --- (1) l 2 has negative gradient ( i.e. m 2 < 0 ) ⊿ADC , we have m 2 = −ADCD -------(2) but ⊿ADB ~ ⊿CDA ⇒ AD DBCD AD= from (1) and (2) ⇒ 21m -=(1AD = )Example 17 Example 182.5.1 Perpendicular Bisector Key steps to find the equation of perpendicular bisector :Step 1 : Find gradient of AB , m = 2121y y x x -- Step2 : Find gradient perpendicular to AB , 1m m'=- Step 3 : Find midpoint of AB , 1212(,)22x x y y M ++= Step 4 : Equation of perpendicular bisector of AB , y – y 1 = m ' ( x – x 1 )Example 20Find the equation of perpendicular bisector of AB given that A ( 1 , 2 ) and B ( 5 , -1 ) .[ans : 6y = 8x – 21 ] Example 21A triangle has vertices A(1,2) , B(3,4 ) and C(0,3) , calculate the coordinates of circumcentre of the ⊿ABC .Perpendicular bisector of ABPoint of concurrency of triangle(a) Centroid•Medians : is a segment joining any vertex to themidpoint of theopposite side . •Centroid : The centroid is the point of concurrency (intersection) of thethree medians of a triangle . (b) Orthocenter•H is called the orthocentre oftriangle ABC .•Orthocenter is the point ofconcurrency of the threealtitudes of a triangle .(c) circumcenter•Circumcenter is the point ofconcurrency of the threeperpendicular bisectors of atriangle .Example 22:===== 如果你想学会游泳，你必须下水；如果想成为解题能手，你必须解题。